Why didn't Voyager 1 and Voyager 2 crash on into Jupiter or Uranus when they approached near to these massive...
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I was reading about gravity assist and I read that both the Voyagers used gravitational force of these planets to speed up.
But I wanted to know why weren't they completely attracted by their gravitational field?
One thing I thought about it was that the escape velocity of these planets were known to NASA and these space crafts used the propeller fuel to escape but how did NASA calculated escape velocity?
$$v_{escape} = sqrt{frac{2GM}{R}}$$
But how could they predict mass?
Radius I guess can be predicted by parallax method.
voyager gravity-assist flyby escape-velocity
edited Jan 29 at 14:08
Glorfindel
1771110
asked Jan 28 at 16:05
GarimaGarima
10915
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show 12 more comments
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I was reading about gravity assist and I read that both the Voyagers used gravitational force of these planets to speed up.
But I wanted to know why weren't they completely attracted by their gravitational field?
One thing I thought about it was that the escape velocity of these planets were known to NASA and these space crafts used the propeller fuel to escape but how did NASA calculated escape velocity?
$$v_{escape} = sqrt{frac{2GM}{R}}$$
But how could they predict mass?
Radius I guess can be predicted by parallax method.
voyager gravity-assist flyby escape-velocity
edited Jan 29 at 14:08
Glorfindel
1771110
asked Jan 28 at 16:05
GarimaGarima
10915
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63
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I kind of feel like we should have some kind of a canonical "how does gravity work?" question that we could link to whenever these kinds of questions based on a fundamental misunderstanding of orbital mechanics come up. Unfortunately, searching for that phrase only turns up questions about zero-gravity cat litter boxes and other random stuff like that. :P
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– Ilmari Karonen
Jan 28 at 17:41
8
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That said, searching for "gravity assist" instead does turn up this question, which I highly recommend the OP should read.
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– Ilmari Karonen
Jan 28 at 17:43
4
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I have a question that branches off this one... How do you kind of crash into something????
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– Happy Koala
Jan 28 at 18:58
4
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Never mind these unhelpful/unkind comments above, it's a good question! You've taken the time to find out that escape velocity is important and added the equation (I converted it to MathJax for you). Your question has certainly attracted several great answers below. Welcome to Space!
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– uhoh
Jan 29 at 0:48
4
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@Baldrickk: In fact, we do have space.stackexchange.com/questions/22834/… and, more generally, space.stackexchange.com/questions/32776/…. But while playing KSP is a good way to learn basic orbital mechanics, it does require an investment of (at least) several hours before paying off. As much as it might improve question quality here, I don't think it would be fair or reasonable to tell new users to "come back after you've landed a kerbal on the Mun". :)
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– Ilmari Karonen
Jan 30 at 15:14
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show 12 more comments
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I was reading about gravity assist and I read that both the Voyagers used gravitational force of these planets to speed up.
But I wanted to know why weren't they completely attracted by their gravitational field?
One thing I thought about it was that the escape velocity of these planets were known to NASA and these space crafts used the propeller fuel to escape but how did NASA calculated escape velocity?
$$v_{escape} = sqrt{frac{2GM}{R}}$$
But how could they predict mass?
Radius I guess can be predicted by parallax method.
voyager gravity-assist flyby escape-velocity
edited Jan 29 at 14:08
Glorfindel
1771110
asked Jan 28 at 16:05
GarimaGarima
10915
$endgroup$
I was reading about gravity assist and I read that both the Voyagers used gravitational force of these planets to speed up.
But I wanted to know why weren't they completely attracted by their gravitational field?
One thing I thought about it was that the escape velocity of these planets were known to NASA and these space crafts used the propeller fuel to escape but how did NASA calculated escape velocity?
$$v_{escape} = sqrt{frac{2GM}{R}}$$
But how could they predict mass?
Radius I guess can be predicted by parallax method.
voyager gravity-assist flyby escape-velocity
voyager gravity-assist flyby escape-velocity
edited Jan 29 at 14:08
Glorfindel
1771110
asked Jan 28 at 16:05
GarimaGarima
10915
edited Jan 29 at 14:08
Glorfindel
1771110
asked Jan 28 at 16:05
GarimaGarima
10915
edited Jan 29 at 14:08
Glorfindel
1771110
edited Jan 29 at 14:08
Glorfindel
1771110
edited Jan 29 at 14:08
Glorfindel
1771110
1771110
asked Jan 28 at 16:05
GarimaGarima
10915
asked Jan 28 at 16:05
GarimaGarima
10915
asked Jan 28 at 16:05
GarimaGarima
10915
10915
63
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I kind of feel like we should have some kind of a canonical "how does gravity work?" question that we could link to whenever these kinds of questions based on a fundamental misunderstanding of orbital mechanics come up. Unfortunately, searching for that phrase only turns up questions about zero-gravity cat litter boxes and other random stuff like that. :P
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– Ilmari Karonen
Jan 28 at 17:41
8
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That said, searching for "gravity assist" instead does turn up this question, which I highly recommend the OP should read.
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– Ilmari Karonen
Jan 28 at 17:43
4
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I have a question that branches off this one... How do you kind of crash into something????
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– Happy Koala
Jan 28 at 18:58
4
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Never mind these unhelpful/unkind comments above, it's a good question! You've taken the time to find out that escape velocity is important and added the equation (I converted it to MathJax for you). Your question has certainly attracted several great answers below. Welcome to Space!
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– uhoh
Jan 29 at 0:48
4
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@Baldrickk: In fact, we do have space.stackexchange.com/questions/22834/… and, more generally, space.stackexchange.com/questions/32776/…. But while playing KSP is a good way to learn basic orbital mechanics, it does require an investment of (at least) several hours before paying off. As much as it might improve question quality here, I don't think it would be fair or reasonable to tell new users to "come back after you've landed a kerbal on the Mun". :)
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– Ilmari Karonen
Jan 30 at 15:14
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show 12 more comments
63
$begingroup$
I kind of feel like we should have some kind of a canonical "how does gravity work?" question that we could link to whenever these kinds of questions based on a fundamental misunderstanding of orbital mechanics come up. Unfortunately, searching for that phrase only turns up questions about zero-gravity cat litter boxes and other random stuff like that. :P
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– Ilmari Karonen
Jan 28 at 17:41
8
$begingroup$
That said, searching for "gravity assist" instead does turn up this question, which I highly recommend the OP should read.
$endgroup$
– Ilmari Karonen
Jan 28 at 17:43
4
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I have a question that branches off this one... How do you kind of crash into something????
$endgroup$
– Happy Koala
Jan 28 at 18:58
4
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Never mind these unhelpful/unkind comments above, it's a good question! You've taken the time to find out that escape velocity is important and added the equation (I converted it to MathJax for you). Your question has certainly attracted several great answers below. Welcome to Space!
$endgroup$
– uhoh
Jan 29 at 0:48
4
$begingroup$
@Baldrickk: In fact, we do have space.stackexchange.com/questions/22834/… and, more generally, space.stackexchange.com/questions/32776/…. But while playing KSP is a good way to learn basic orbital mechanics, it does require an investment of (at least) several hours before paying off. As much as it might improve question quality here, I don't think it would be fair or reasonable to tell new users to "come back after you've landed a kerbal on the Mun". :)
$endgroup$
– Ilmari Karonen
Jan 30 at 15:14
63
63
$begingroup$
I kind of feel like we should have some kind of a canonical "how does gravity work?" question that we could link to whenever these kinds of questions based on a fundamental misunderstanding of orbital mechanics come up. Unfortunately, searching for that phrase only turns up questions about zero-gravity cat litter boxes and other random stuff like that. :P
$endgroup$
– Ilmari Karonen
Jan 28 at 17:41
$begingroup$
I kind of feel like we should have some kind of a canonical "how does gravity work?" question that we could link to whenever these kinds of questions based on a fundamental misunderstanding of orbital mechanics come up. Unfortunately, searching for that phrase only turns up questions about zero-gravity cat litter boxes and other random stuff like that. :P
$endgroup$
– Ilmari Karonen
Jan 28 at 17:41
8
8
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That said, searching for "gravity assist" instead does turn up this question, which I highly recommend the OP should read.
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– Ilmari Karonen
Jan 28 at 17:43
$begingroup$
That said, searching for "gravity assist" instead does turn up this question, which I highly recommend the OP should read.
$endgroup$
– Ilmari Karonen
Jan 28 at 17:43
4
4
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I have a question that branches off this one... How do you kind of crash into something????
$endgroup$
– Happy Koala
Jan 28 at 18:58
$begingroup$
I have a question that branches off this one... How do you kind of crash into something????
$endgroup$
– Happy Koala
Jan 28 at 18:58
4
4
$begingroup$
Never mind these unhelpful/unkind comments above, it's a good question! You've taken the time to find out that escape velocity is important and added the equation (I converted it to MathJax for you). Your question has certainly attracted several great answers below. Welcome to Space!
$endgroup$
– uhoh
Jan 29 at 0:48
$begingroup$
Never mind these unhelpful/unkind comments above, it's a good question! You've taken the time to find out that escape velocity is important and added the equation (I converted it to MathJax for you). Your question has certainly attracted several great answers below. Welcome to Space!
$endgroup$
– uhoh
Jan 29 at 0:48
4
4
$begingroup$
@Baldrickk: In fact, we do have space.stackexchange.com/questions/22834/… and, more generally, space.stackexchange.com/questions/32776/…. But while playing KSP is a good way to learn basic orbital mechanics, it does require an investment of (at least) several hours before paying off. As much as it might improve question quality here, I don't think it would be fair or reasonable to tell new users to "come back after you've landed a kerbal on the Mun". :)
$endgroup$
– Ilmari Karonen
Jan 30 at 15:14
$begingroup$
@Baldrickk: In fact, we do have space.stackexchange.com/questions/22834/… and, more generally, space.stackexchange.com/questions/32776/…. But while playing KSP is a good way to learn basic orbital mechanics, it does require an investment of (at least) several hours before paying off. As much as it might improve question quality here, I don't think it would be fair or reasonable to tell new users to "come back after you've landed a kerbal on the Mun". :)
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– Ilmari Karonen
Jan 30 at 15:14
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show 12 more comments
8 Answers
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why weren't they completely attracted by their gravitational field?
How much a trajectory is changed, depends on 3 factors:
- the mass of the planet,
- the speed of the spacecraft,
- the distance between spacecraft and planet
Voyager's speed and distance were chosen to make sure Voyager wouldn't enter orbit around the planet. Voyager's speed before approaching Jupiter was higher than Jupiter's escape speed.
In fact, as long as your trajectory doesn't touch the planet's atmosphere, you're good: because you're arriving from another planet your speed is always higher than escape speed.
But how could they predict mass?
When a planet has moons, you can calculate the planet's mass to a good degree of accuracy. This uses Kepler's third law:
$$ M = frac{4pi^2r^3}{GT^2}$$
G is the universal gravitational constant, 6.6726 x 10-11N-m2 /kg2
r = orbital radius of the moon
T = orbital period of the moon
So you only need r and T, both of which you can observe from Earth pretty well.
these space crafts used the propeller fuel to escape
No. When Voyager 1 was launched from Earth, the last stage of the rocket put it on an escape trajectory, then the stage was jettisonned. That was the last time a rocket was used to significantly change Voyager's speed.
It traveled to Jupiter without further propulsion, so its speed slowly decreased as it got further away from the Sun.
Here's a plot of Voyager 2's speed:
Then as it got close to Jupiter, Voyager's speed increased drastically as it was pulled in by Jupiter's gravity. At Voyager's closest approach, Voyager's trajectory was bent. This put the spacecraft on course to Saturn.
As Voyager sped away from Jupiter, Jupiter's gravity slowed it down. All without using rocket propulsion.
edited Jan 29 at 16:57
NathanOliver
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answered Jan 28 at 16:41
HobbesHobbes
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+1. Heck, exact mass value isn't really that important for orbital mechanics . Gravitational Parameter (GM) is what the orbital equations generally care about and can be measured easily from observations to better accuracy than G or M individually.
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– notovny
Jan 28 at 16:56
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While accurate as far as it goes, I kind of feel like this answer fails to fully address some of the OP's deeper misconceptions, such as that "these space crafts used the propeller fuel to escape". It seems like the OP imagines gravity as some kind of a dissipative "suction" force that would capture incoming spacecraft unless they actively used their engines to escape it, rather than as the conservative force that it is.
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– Ilmari Karonen
Jan 28 at 17:48
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The gravity assist did not change velocity relative to the planet, only to the sun, so it had nothing to do with escape velocity.
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– Nathan Tuggy
Jan 28 at 18:28
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well, there wasn't really anywhere else to go after Neptune
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– 12Me21
Jan 30 at 3:20
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@Criggie that question has been answered here: space.stackexchange.com/questions/10195/…
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– Hobbes
Jan 30 at 8:57
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To add to the answers @Hobbes & @Steve Linton posted, the mission designers indeed knew Jupiter's gravity field quite well from the orbits of Jupiter's moons. But before the Voyagers arrived they got additional measurements from the close flybys of two other spacecraft, Pioneers 10 and 11.
@Steve Linton correctly describes the effect of the "sideways" part of a spacecraft's velocity with respect to the planet it's approaching. If the sideways component is large enough, the probe will miss the planet. To increase that sideways component, on approach you aim the spacecraft [via trajectory adjustment maneuvers, where you use the spacecraft's rocket engine(s)] to precisely tune the components of velocity, one straight toward the planet and one sideways. That combination sets the "aim point", the point to the side of the planet where the spacecraft would go if the planet had no gravity. Trajectory designers call that miss distance "b", a part of the b-plane aim point.
This diagram helps to see the effect of increasing b if the approach velocity is held constant. For typical interplanetary trajectories the approach velocity doesn't depend on the chosen b. [Sorry for the pixellation—this is a very old diagram I pulled out of my presentation archives]
The brownish-tan (in the original file it was orange!) circle represents the planet, and the different colors of curves represent the trajectories followed by spacecraft with different b, all coming in at the same approach velocity and parallel to the black dashed line through the center of the planet. The thinner, dot-dashed lines represent what the spacecrafts' trajectories would be if the planet had no gravity. If b isn't large enough, as with the red and brownish-tan (formerly orange!) trajectories, they impact the planet. The thick, dashed red and brownish-tan lines show how those two trajectories would continue if they didn't run into anything, as would be the case if the planet were the same mass but a lot smaller in size (i.e., a lot denser).
The green and blue lines are for trajectories whose b are large enough that they miss the planet. They point out a couple of characteristics of these hyperbolic orbits: 1) the larger the b, the farther from the planet is the closest-approach distance; and 2) the larger the b, the smaller is the angle the planet "bends" the trajectory. Again, this is for a fixed approach velocity. The red trajectory is bent by almost 180°, while the blue one is bent only ~135-140°.
This bend is key to making gravity assists so useful.
answered Jan 28 at 21:18
Tom SpilkerTom Spilker
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Let's try and understand how gravity works in space. This is kind of key idea to understanding lots of issues in space travel and astronomy.
So imagine a space probe, or rock, which is heading in from deep space, aimed almost, but not quite towards a planet. We can break that motion into two parts -- the part towards the planet and the "sideways" part. The planet's gravity accelerates the part towards the planet, but it does nothing to the sideways part, so the space probe is still not heading quite towards the planet. This keeps on happening, so the probe is moving increasingly fast, still almost but not quite heading at the planet.
After a while the probe whizzes past the side of the planet, going very fast. It can't hit it because it has never lost the "sideways" part of its motion (called angular momentum, more or less). It is still being pulled towards the planet but still moving around it, due to all the speed it has picked up. Then that speed starts to carry it away from the planet again (possibly with its course having been bent around to a greater or lesser extent). Now the gravity is slowing it down. However it is slowing it down just exactly as much as it speeded it up coming in so if it came in from far away at a certain speed, it will eventually escape equally far away at the same speed.
If someone can add some diagrams to this, that would be great.
answered Jan 28 at 20:08
Steve LintonSteve Linton
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Lucidly explained, certainly. It does leave me wondering whether you've been quite clear as to how the spacecraft is speeded up, though, as you seem to have it departing the planet at the same speed it had when it arrived?
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– Ed999
Jan 29 at 14:18
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@Ed999: That is correct (in the planet's frame of reference). And that is a crucial insight in how and when gravity assists can work. You use them to gain speed in the Sun's frame of reference.
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– MSalters
Jan 29 at 14:56
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I think the OP's basic misunderstanding is that the planet is moving. Gravity assist doesn't work otherwise. The point to get across is that the probe spends a longer time falling into the planet (because its closing from behind a moving planet) and a shorter time escaping from the planet (because it slips out the side).
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– Oscar Bravo
Jan 29 at 15:32
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@Ed999 I was just trying to explain not being "completely attracted". Explaining gravity assist is a whole other problem. Very briefly MSalters and Oscar Bravo have it right. It's a bit like bouncing a tennis ball of the front of an oncoming train. Relative to the train, the ball bounced back at more of less the same speed it arrived at, but relative to the ground it's now moving much faster.
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– Steve Linton
Jan 29 at 19:40
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@SteveLinton I confess that I don't understand the o/p's use of the term completely attracted (unless he thinks a planet and a black hole are identical in mass?), but I like the analogy of the tennis ball and the train, which really does get the point across.
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– Ed999
Jan 31 at 2:54
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At the heart of this question is how the gravity assist 'works', at least an intuitive understanding of it.
When a space probe uses gravity assist from a planet, it gets some energy from it, which (yes) incidentally slows the planet down slightly.
How does it do it?
Imagine a tennis ball in flight towards an ideal moving tennis racket. The ball rebounds from the racket with (ideally, more or less) no energy dissipated in ball and strings, and so leaves the racket at the same velocity with respect to the racket that it arrived with. As the racket was moving, this added the racket velocity to the ball velocity.
Now you can't just bounce a space probe off a planet. But you can swing it round the back in a hyperbolic orbit. It comes in with a certain velocity with respect to the planet, and leaves with the same relative velocity, just like the ball and racket. If you fling the probe near the trailing edge of the planet in its orbit, 'behind' the planet, then the probe will be accelerated in the direction of its orbit. However, if you fling the probe ahead of the planet, the probe will be slowed down, and the planet speeded up just a tad. Which is what the Parker Solar Probe is going to do with Venus 7 times to drop into a lower orbit around the sun.
So I like to think of a gravity assist as 'bouncing' the probe off the planet. But you can only add speed if you end up travelling in the same orbital direction as the planet. So you have to plan your mission to suit.
answered Jan 29 at 14:45
Neil_UKNeil_UK
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Let's focus on the "why weren't they completely attracted by their gravitational field" question here.
You've probably seen Newton's cannonball diagram before:
In this diagram, the distance from the object to the planet is fixed, and it's the speed that changes, and gives different results:
- too slow (A or B, below orbital speed): it will indeed be pulled towards the planet
- between orbital speed and the escape velocity (C and D), it will enter orbit around the planet
- beyond escape velocity (E), it will escape the planet's gravity.
However, orbital speed and escape velocity depend on the distance from the planet. So given the same speed, you can get the same results by varying the distance from the planet:
- too close: A or B, fall on the planet
- C or D, enter orbit
- far enough: E, the trajectory gets modified ("bent"), but the probe can continue its journey
So it's all a matter of getting the probe to approach Jupiter at the right combination of speed and distance (which is complicated by the fact that said distance is influenced by gravity on the approach, and that the planet is moving) to get the result you want: crash into the planet, enter orbit, or just perform a "fly-by" and continue somewhere else.
In the latter case, you want to take advantage of the operation to change to a specific direction and/or gain speed, which complicates things a bit further (in terms of picking the exact initial speed/trajectory to achieve the desired result), but it doesn't change the fact that you just need to be the right combination of fast enough and far enough to avoid crashing on the planet.
Of course, to compute all this, you indeed need to know the mass of the planet, but as others have written, there have been ways to compute this for quite a while.
answered Jan 29 at 17:11
jcaronjcaron
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Excepting of course that you cannot shoot anything to orbit, as the illustration seems to imply. Everything you shoot will either land again, or escape, never orbit.
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– Stian Yttervik
Jan 30 at 13:06
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@StianYttervik I'm not sure I understand what you mean? You do understand the context of Newton's cannonball? It's shot horizontally, from a high altitude where there is no atmosphere, so as long as you shoot the object at the right speed for that altitude, it will definitely orbit the planet. What did I miss?
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– jcaron
Jan 30 at 13:57
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Well, of course you'll also need to move the cannon out of the way after shooting it, otherwise the ball will hit it after completing its orbit. (If the planet is rotating, that'll usually do the job, especially if you shoot the ball with enough velocity to put it into an elliptical orbit with an apoapsis above the cannon's altitude, as in trajectory D in the image. Of course, even so, the ball may still hit the cannon eventually, after further rotation brings the cannon back to the ball's periapsis just as the ball is passing it.)
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– Ilmari Karonen
Jan 30 at 15:42
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Newton discovered gravity (and invented mathematical laws to describe it) a long time before Pioneer 10 flew!
It has been possible to calculate the mass of the planets with a fair degree of accuracy for a very long time, based on their effects on each other, as well as on their moons, and on passing NASA space probes. There were very well worked out laws of planetary motion in the pre-spaceage, which gave good values for the masses of the planets.
What seems to be missing from the other answers here is simply this bit of the explanation: a spacecraft steals energy from the planet by this type of slingshot orbit, since some of the planet's rotational motion is transfered to the vehicle (as motion): in effect, the planet is slowed down slightly, and the spacecraft is speeded up slightly.
You couldn't even begin to measure the effect on the planet's rotation: it's too minute to be measurable by our current techniques. But, in the math, the effect does occur, even though in practice we can't observe it.
The question seems to imply the spacecraft is slowed down in space, on its approach to a planet, but there is nothing in the vacuum of space to cause such an effect. And it will accelerate as it dives toward the planet, due to the planet's gravity pulling on it. If it misses the planet's atmosphere (which extends only a very short distance into space), the slingshot effect will actually increase the craft's speed, stealing rotational energy as momentum.
I'd like to recommend some further reading: the novel 2010: Odyssey 2 by Arthur C Clark contains a description of this type of slingshot orbit (slightly modified, but highly dramatic) which gives a scientifically accurate account of one, set at Jupiter.
answered Jan 29 at 14:53
Ed999Ed999
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Of course we knew the mass fairly well before Pioneer 10, and had reasonable estimates for J2 & J4, but after Pioneers 10 & 11 we knew the mass exquisitely, had better estimates for J2 and J4, and also measured J3 and J6. See Anderson & Null's papers agupubs.onlinelibrary.wiley.com/doi/pdf/10.1029/JA079i025p03661 (unfortunately behind a paywall) and adsabs.harvard.edu/full/1976AJ.....81.1153N (free!) . Those measurements were important for reducing the Voyagers' post-flyby TCM ∆Vs for the trajectory to Saturn. This decreased the propellant load, allowing a lower launch mass.
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– Tom Spilker
Jan 29 at 19:42
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If you want a complete explanation of how the masses of the other planets have been calculated, I think that should be a separate question. But for this question, it doesn't really matter. Increasing the mass of the planet increases the escape velocity, but it also increases the gravitational force, which means that the probe would be traveling faster when it approaches the planet.
Escape velocity is the velocity one needs to reach a point infinitely far away from the planet. That is, the amount of energy to move an object from the surface of the planet to infinitely far away is equal to the kinetic energy of that object moving at escape velocity. Since energy is conserved, an object that goes from being on the surface with the kinetic energy of the escape velocity to being infinitely far away must have gravitational potential energy equal to the kinetic energy of escape velocity (keep in mind that the kinetic energy of escape velocity depends on the mass of the object, so there's an "of that object" in all of this).
Gravitational force falls off quickly with distance, so once you get to interplanetary distances, you can consider that distance to pretty much be "infinity" for planetary gravitational potential energy purposes. Thus, pretty much any probe launched from Earth will, when it reaches another planet, be traveling faster than that planet's escape velocity (it started out with the gravitational potential energy of "infinitely" far away, and that potential energy was converted to the kinetic energy of escape velocity). None of this depends on the mass of the other planet. The mass will affect the shape of the probe's path, but not that it has escape velocity.
As an analogy, suppose you have a dip in the ground, and the ground is the same height on both sides of the dip. And let's say this ground has no friction. If you roll a marble across the ground from outside the dip, it will come out the other side. It doesn't matter how deep the dip is; making the dip deeper means the marble needs more energy to climb out of it, but it also means that it gained more energy when it was falling into the dip. Similarly, if you throw something at a planet, then unless it's a direct hit, it will continue traveling at least as far away as it started (barring interactions with other bodies, of course).
answered Jan 29 at 17:17
AcccumulationAcccumulation
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Your problem appears to be too much science fiction and not enough science.
In science fiction, gravity sucks things towards the planet, sort of like a magnet. When far away you don't feel it; then you get close enough, and you are pulled down to the surface. And it holds things near the planet like glue or an elastic band.
In this science fiction model, stuff in space floats because it is far away. You go to space, you float.
In the real world, the science model, in space you fall just like you do in atmosphere. Orbit isn't far away, it is fast away -- things orbit the planet because they are going extremely fast.
Things in (low) orbit are amazingly close to the planet, from a distance they are almost skimming the surface. They are just moving so fast that the gravity of the planet pulling it down isn't fast enough; they "miss" the ground on each orbit.
Gravity doesn't stick or yank; gravity accelerates. When you go down a gravity well, you gain speed. This speed is mostly towards the planet, but it takes relative precision to actually hit the planet; even a modest velocity "horizontal" to the gravity well will make you miss.
And when you are going away from the planet, gravity doesn't "stick", it decelerates. The amount it decelerates on the way out is the same as it accelerated on the way in, at least in its frame of reference.
So you can set a trajectory that skims the surface of a planet (without an atmosphere, or outside it), and you'll only lose the speed you gained while falling in.
The slingshot maneuver uses the fact that the planet's "stationary" frame of reference isn't the same as the sun. So you approach the planet from "behind" in its orbital track, then leave roughly perpendicular to its orbital track.
If the planet was moving 13 km/s and you approached it from the side at 20 km/s.
In the planet's perspective you are moving sqrt(20^2 + 13^2) = 23.9 km/s.
You then leave the planet say, parallel to the orbital track in the same direction. You leave at 23.9 km/s.
But in the sun's frame of reference, you approached the planet at 20km/s and left at (13+23.9) = 36.9 km/s, a massive speed boost.
answered Jan 29 at 20:55
YakkYakk
25016
$endgroup$
add a comment |
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8 Answers
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$begingroup$
why weren't they completely attracted by their gravitational field?
How much a trajectory is changed, depends on 3 factors:
- the mass of the planet,
- the speed of the spacecraft,
- the distance between spacecraft and planet
Voyager's speed and distance were chosen to make sure Voyager wouldn't enter orbit around the planet. Voyager's speed before approaching Jupiter was higher than Jupiter's escape speed.
In fact, as long as your trajectory doesn't touch the planet's atmosphere, you're good: because you're arriving from another planet your speed is always higher than escape speed.
But how could they predict mass?
When a planet has moons, you can calculate the planet's mass to a good degree of accuracy. This uses Kepler's third law:
$$ M = frac{4pi^2r^3}{GT^2}$$
G is the universal gravitational constant, 6.6726 x 10-11N-m2 /kg2
r = orbital radius of the moon
T = orbital period of the moon
So you only need r and T, both of which you can observe from Earth pretty well.
these space crafts used the propeller fuel to escape
No. When Voyager 1 was launched from Earth, the last stage of the rocket put it on an escape trajectory, then the stage was jettisonned. That was the last time a rocket was used to significantly change Voyager's speed.
It traveled to Jupiter without further propulsion, so its speed slowly decreased as it got further away from the Sun.
Here's a plot of Voyager 2's speed:
Then as it got close to Jupiter, Voyager's speed increased drastically as it was pulled in by Jupiter's gravity. At Voyager's closest approach, Voyager's trajectory was bent. This put the spacecraft on course to Saturn.
As Voyager sped away from Jupiter, Jupiter's gravity slowed it down. All without using rocket propulsion.
edited Jan 29 at 16:57
NathanOliver
1032
answered Jan 28 at 16:41
HobbesHobbes
90.2k2255406
$endgroup$
7
$begingroup$
+1. Heck, exact mass value isn't really that important for orbital mechanics . Gravitational Parameter (GM) is what the orbital equations generally care about and can be measured easily from observations to better accuracy than G or M individually.
$endgroup$
– notovny
Jan 28 at 16:56
27
$begingroup$
While accurate as far as it goes, I kind of feel like this answer fails to fully address some of the OP's deeper misconceptions, such as that "these space crafts used the propeller fuel to escape". It seems like the OP imagines gravity as some kind of a dissipative "suction" force that would capture incoming spacecraft unless they actively used their engines to escape it, rather than as the conservative force that it is.
$endgroup$
– Ilmari Karonen
Jan 28 at 17:48
4
$begingroup$
The gravity assist did not change velocity relative to the planet, only to the sun, so it had nothing to do with escape velocity.
$endgroup$
– Nathan Tuggy
Jan 28 at 18:28
4
$begingroup$
well, there wasn't really anywhere else to go after Neptune
$endgroup$
– 12Me21
Jan 30 at 3:20
2
$begingroup$
@Criggie that question has been answered here: space.stackexchange.com/questions/10195/…
$endgroup$
– Hobbes
Jan 30 at 8:57
|
show 12 more comments
$begingroup$
why weren't they completely attracted by their gravitational field?
How much a trajectory is changed, depends on 3 factors:
- the mass of the planet,
- the speed of the spacecraft,
- the distance between spacecraft and planet
Voyager's speed and distance were chosen to make sure Voyager wouldn't enter orbit around the planet. Voyager's speed before approaching Jupiter was higher than Jupiter's escape speed.
In fact, as long as your trajectory doesn't touch the planet's atmosphere, you're good: because you're arriving from another planet your speed is always higher than escape speed.
But how could they predict mass?
When a planet has moons, you can calculate the planet's mass to a good degree of accuracy. This uses Kepler's third law:
$$ M = frac{4pi^2r^3}{GT^2}$$
G is the universal gravitational constant, 6.6726 x 10-11N-m2 /kg2
r = orbital radius of the moon
T = orbital period of the moon
So you only need r and T, both of which you can observe from Earth pretty well.
these space crafts used the propeller fuel to escape
No. When Voyager 1 was launched from Earth, the last stage of the rocket put it on an escape trajectory, then the stage was jettisonned. That was the last time a rocket was used to significantly change Voyager's speed.
It traveled to Jupiter without further propulsion, so its speed slowly decreased as it got further away from the Sun.
Here's a plot of Voyager 2's speed:
Then as it got close to Jupiter, Voyager's speed increased drastically as it was pulled in by Jupiter's gravity. At Voyager's closest approach, Voyager's trajectory was bent. This put the spacecraft on course to Saturn.
As Voyager sped away from Jupiter, Jupiter's gravity slowed it down. All without using rocket propulsion.
edited Jan 29 at 16:57
NathanOliver
1032
answered Jan 28 at 16:41
HobbesHobbes
90.2k2255406
$endgroup$
7
$begingroup$
+1. Heck, exact mass value isn't really that important for orbital mechanics . Gravitational Parameter (GM) is what the orbital equations generally care about and can be measured easily from observations to better accuracy than G or M individually.
$endgroup$
– notovny
Jan 28 at 16:56
27
$begingroup$
While accurate as far as it goes, I kind of feel like this answer fails to fully address some of the OP's deeper misconceptions, such as that "these space crafts used the propeller fuel to escape". It seems like the OP imagines gravity as some kind of a dissipative "suction" force that would capture incoming spacecraft unless they actively used their engines to escape it, rather than as the conservative force that it is.
$endgroup$
– Ilmari Karonen
Jan 28 at 17:48
4
$begingroup$
The gravity assist did not change velocity relative to the planet, only to the sun, so it had nothing to do with escape velocity.
$endgroup$
– Nathan Tuggy
Jan 28 at 18:28
4
$begingroup$
well, there wasn't really anywhere else to go after Neptune
$endgroup$
– 12Me21
Jan 30 at 3:20
2
$begingroup$
@Criggie that question has been answered here: space.stackexchange.com/questions/10195/…
$endgroup$
– Hobbes
Jan 30 at 8:57
|
show 12 more comments
$begingroup$
why weren't they completely attracted by their gravitational field?
How much a trajectory is changed, depends on 3 factors:
- the mass of the planet,
- the speed of the spacecraft,
- the distance between spacecraft and planet
Voyager's speed and distance were chosen to make sure Voyager wouldn't enter orbit around the planet. Voyager's speed before approaching Jupiter was higher than Jupiter's escape speed.
In fact, as long as your trajectory doesn't touch the planet's atmosphere, you're good: because you're arriving from another planet your speed is always higher than escape speed.
But how could they predict mass?
When a planet has moons, you can calculate the planet's mass to a good degree of accuracy. This uses Kepler's third law:
$$ M = frac{4pi^2r^3}{GT^2}$$
G is the universal gravitational constant, 6.6726 x 10-11N-m2 /kg2
r = orbital radius of the moon
T = orbital period of the moon
So you only need r and T, both of which you can observe from Earth pretty well.
these space crafts used the propeller fuel to escape
No. When Voyager 1 was launched from Earth, the last stage of the rocket put it on an escape trajectory, then the stage was jettisonned. That was the last time a rocket was used to significantly change Voyager's speed.
It traveled to Jupiter without further propulsion, so its speed slowly decreased as it got further away from the Sun.
Here's a plot of Voyager 2's speed:
Then as it got close to Jupiter, Voyager's speed increased drastically as it was pulled in by Jupiter's gravity. At Voyager's closest approach, Voyager's trajectory was bent. This put the spacecraft on course to Saturn.
As Voyager sped away from Jupiter, Jupiter's gravity slowed it down. All without using rocket propulsion.
edited Jan 29 at 16:57
NathanOliver
1032
answered Jan 28 at 16:41
HobbesHobbes
90.2k2255406
$endgroup$
why weren't they completely attracted by their gravitational field?
How much a trajectory is changed, depends on 3 factors:
- the mass of the planet,
- the speed of the spacecraft,
- the distance between spacecraft and planet
Voyager's speed and distance were chosen to make sure Voyager wouldn't enter orbit around the planet. Voyager's speed before approaching Jupiter was higher than Jupiter's escape speed.
In fact, as long as your trajectory doesn't touch the planet's atmosphere, you're good: because you're arriving from another planet your speed is always higher than escape speed.
But how could they predict mass?
When a planet has moons, you can calculate the planet's mass to a good degree of accuracy. This uses Kepler's third law:
$$ M = frac{4pi^2r^3}{GT^2}$$
G is the universal gravitational constant, 6.6726 x 10-11N-m2 /kg2
r = orbital radius of the moon
T = orbital period of the moon
So you only need r and T, both of which you can observe from Earth pretty well.
these space crafts used the propeller fuel to escape
No. When Voyager 1 was launched from Earth, the last stage of the rocket put it on an escape trajectory, then the stage was jettisonned. That was the last time a rocket was used to significantly change Voyager's speed.
It traveled to Jupiter without further propulsion, so its speed slowly decreased as it got further away from the Sun.
Here's a plot of Voyager 2's speed:
Then as it got close to Jupiter, Voyager's speed increased drastically as it was pulled in by Jupiter's gravity. At Voyager's closest approach, Voyager's trajectory was bent. This put the spacecraft on course to Saturn.
As Voyager sped away from Jupiter, Jupiter's gravity slowed it down. All without using rocket propulsion.
edited Jan 29 at 16:57
NathanOliver
1032
answered Jan 28 at 16:41
HobbesHobbes
90.2k2255406
edited Jan 29 at 16:57
NathanOliver
1032
edited Jan 29 at 16:57
NathanOliver
1032
edited Jan 29 at 16:57
NathanOliver
1032
1032
answered Jan 28 at 16:41
HobbesHobbes
90.2k2255406
answered Jan 28 at 16:41
HobbesHobbes
90.2k2255406
answered Jan 28 at 16:41
HobbesHobbes
90.2k2255406
90.2k2255406
7
$begingroup$
+1. Heck, exact mass value isn't really that important for orbital mechanics . Gravitational Parameter (GM) is what the orbital equations generally care about and can be measured easily from observations to better accuracy than G or M individually.
$endgroup$
– notovny
Jan 28 at 16:56
27
$begingroup$
While accurate as far as it goes, I kind of feel like this answer fails to fully address some of the OP's deeper misconceptions, such as that "these space crafts used the propeller fuel to escape". It seems like the OP imagines gravity as some kind of a dissipative "suction" force that would capture incoming spacecraft unless they actively used their engines to escape it, rather than as the conservative force that it is.
$endgroup$
– Ilmari Karonen
Jan 28 at 17:48
4
$begingroup$
The gravity assist did not change velocity relative to the planet, only to the sun, so it had nothing to do with escape velocity.
$endgroup$
– Nathan Tuggy
Jan 28 at 18:28
4
$begingroup$
well, there wasn't really anywhere else to go after Neptune
$endgroup$
– 12Me21
Jan 30 at 3:20
2
$begingroup$
@Criggie that question has been answered here: space.stackexchange.com/questions/10195/…
$endgroup$
– Hobbes
Jan 30 at 8:57
|
show 12 more comments
7
$begingroup$
+1. Heck, exact mass value isn't really that important for orbital mechanics . Gravitational Parameter (GM) is what the orbital equations generally care about and can be measured easily from observations to better accuracy than G or M individually.
$endgroup$
– notovny
Jan 28 at 16:56
27
$begingroup$
While accurate as far as it goes, I kind of feel like this answer fails to fully address some of the OP's deeper misconceptions, such as that "these space crafts used the propeller fuel to escape". It seems like the OP imagines gravity as some kind of a dissipative "suction" force that would capture incoming spacecraft unless they actively used their engines to escape it, rather than as the conservative force that it is.
$endgroup$
– Ilmari Karonen
Jan 28 at 17:48
4
$begingroup$
The gravity assist did not change velocity relative to the planet, only to the sun, so it had nothing to do with escape velocity.
$endgroup$
– Nathan Tuggy
Jan 28 at 18:28
4
$begingroup$
well, there wasn't really anywhere else to go after Neptune
$endgroup$
– 12Me21
Jan 30 at 3:20
2
$begingroup$
@Criggie that question has been answered here: space.stackexchange.com/questions/10195/…
$endgroup$
– Hobbes
Jan 30 at 8:57
7
7
$begingroup$
+1. Heck, exact mass value isn't really that important for orbital mechanics . Gravitational Parameter (GM) is what the orbital equations generally care about and can be measured easily from observations to better accuracy than G or M individually.
$endgroup$
– notovny
Jan 28 at 16:56
$begingroup$
+1. Heck, exact mass value isn't really that important for orbital mechanics . Gravitational Parameter (GM) is what the orbital equations generally care about and can be measured easily from observations to better accuracy than G or M individually.
$endgroup$
– notovny
Jan 28 at 16:56
27
27
$begingroup$
While accurate as far as it goes, I kind of feel like this answer fails to fully address some of the OP's deeper misconceptions, such as that "these space crafts used the propeller fuel to escape". It seems like the OP imagines gravity as some kind of a dissipative "suction" force that would capture incoming spacecraft unless they actively used their engines to escape it, rather than as the conservative force that it is.
$endgroup$
– Ilmari Karonen
Jan 28 at 17:48
$begingroup$
While accurate as far as it goes, I kind of feel like this answer fails to fully address some of the OP's deeper misconceptions, such as that "these space crafts used the propeller fuel to escape". It seems like the OP imagines gravity as some kind of a dissipative "suction" force that would capture incoming spacecraft unless they actively used their engines to escape it, rather than as the conservative force that it is.
$endgroup$
– Ilmari Karonen
Jan 28 at 17:48
4
4
$begingroup$
The gravity assist did not change velocity relative to the planet, only to the sun, so it had nothing to do with escape velocity.
$endgroup$
– Nathan Tuggy
Jan 28 at 18:28
$begingroup$
The gravity assist did not change velocity relative to the planet, only to the sun, so it had nothing to do with escape velocity.
$endgroup$
– Nathan Tuggy
Jan 28 at 18:28
4
4
$begingroup$
well, there wasn't really anywhere else to go after Neptune
$endgroup$
– 12Me21
Jan 30 at 3:20
$begingroup$
well, there wasn't really anywhere else to go after Neptune
$endgroup$
– 12Me21
Jan 30 at 3:20
2
2
$begingroup$
@Criggie that question has been answered here: space.stackexchange.com/questions/10195/…
$endgroup$
– Hobbes
Jan 30 at 8:57
$begingroup$
@Criggie that question has been answered here: space.stackexchange.com/questions/10195/…
$endgroup$
– Hobbes
Jan 30 at 8:57
|
show 12 more comments
$begingroup$
To add to the answers @Hobbes & @Steve Linton posted, the mission designers indeed knew Jupiter's gravity field quite well from the orbits of Jupiter's moons. But before the Voyagers arrived they got additional measurements from the close flybys of two other spacecraft, Pioneers 10 and 11.
@Steve Linton correctly describes the effect of the "sideways" part of a spacecraft's velocity with respect to the planet it's approaching. If the sideways component is large enough, the probe will miss the planet. To increase that sideways component, on approach you aim the spacecraft [via trajectory adjustment maneuvers, where you use the spacecraft's rocket engine(s)] to precisely tune the components of velocity, one straight toward the planet and one sideways. That combination sets the "aim point", the point to the side of the planet where the spacecraft would go if the planet had no gravity. Trajectory designers call that miss distance "b", a part of the b-plane aim point.
This diagram helps to see the effect of increasing b if the approach velocity is held constant. For typical interplanetary trajectories the approach velocity doesn't depend on the chosen b. [Sorry for the pixellation—this is a very old diagram I pulled out of my presentation archives]
The brownish-tan (in the original file it was orange!) circle represents the planet, and the different colors of curves represent the trajectories followed by spacecraft with different b, all coming in at the same approach velocity and parallel to the black dashed line through the center of the planet. The thinner, dot-dashed lines represent what the spacecrafts' trajectories would be if the planet had no gravity. If b isn't large enough, as with the red and brownish-tan (formerly orange!) trajectories, they impact the planet. The thick, dashed red and brownish-tan lines show how those two trajectories would continue if they didn't run into anything, as would be the case if the planet were the same mass but a lot smaller in size (i.e., a lot denser).
The green and blue lines are for trajectories whose b are large enough that they miss the planet. They point out a couple of characteristics of these hyperbolic orbits: 1) the larger the b, the farther from the planet is the closest-approach distance; and 2) the larger the b, the smaller is the angle the planet "bends" the trajectory. Again, this is for a fixed approach velocity. The red trajectory is bent by almost 180°, while the blue one is bent only ~135-140°.
This bend is key to making gravity assists so useful.
answered Jan 28 at 21:18
Tom SpilkerTom Spilker
9,7062152
$endgroup$
add a comment |
$begingroup$
To add to the answers @Hobbes & @Steve Linton posted, the mission designers indeed knew Jupiter's gravity field quite well from the orbits of Jupiter's moons. But before the Voyagers arrived they got additional measurements from the close flybys of two other spacecraft, Pioneers 10 and 11.
@Steve Linton correctly describes the effect of the "sideways" part of a spacecraft's velocity with respect to the planet it's approaching. If the sideways component is large enough, the probe will miss the planet. To increase that sideways component, on approach you aim the spacecraft [via trajectory adjustment maneuvers, where you use the spacecraft's rocket engine(s)] to precisely tune the components of velocity, one straight toward the planet and one sideways. That combination sets the "aim point", the point to the side of the planet where the spacecraft would go if the planet had no gravity. Trajectory designers call that miss distance "b", a part of the b-plane aim point.
This diagram helps to see the effect of increasing b if the approach velocity is held constant. For typical interplanetary trajectories the approach velocity doesn't depend on the chosen b. [Sorry for the pixellation—this is a very old diagram I pulled out of my presentation archives]
The brownish-tan (in the original file it was orange!) circle represents the planet, and the different colors of curves represent the trajectories followed by spacecraft with different b, all coming in at the same approach velocity and parallel to the black dashed line through the center of the planet. The thinner, dot-dashed lines represent what the spacecrafts' trajectories would be if the planet had no gravity. If b isn't large enough, as with the red and brownish-tan (formerly orange!) trajectories, they impact the planet. The thick, dashed red and brownish-tan lines show how those two trajectories would continue if they didn't run into anything, as would be the case if the planet were the same mass but a lot smaller in size (i.e., a lot denser).
The green and blue lines are for trajectories whose b are large enough that they miss the planet. They point out a couple of characteristics of these hyperbolic orbits: 1) the larger the b, the farther from the planet is the closest-approach distance; and 2) the larger the b, the smaller is the angle the planet "bends" the trajectory. Again, this is for a fixed approach velocity. The red trajectory is bent by almost 180°, while the blue one is bent only ~135-140°.
This bend is key to making gravity assists so useful.
answered Jan 28 at 21:18
Tom SpilkerTom Spilker
9,7062152
$endgroup$
add a comment |
$begingroup$
To add to the answers @Hobbes & @Steve Linton posted, the mission designers indeed knew Jupiter's gravity field quite well from the orbits of Jupiter's moons. But before the Voyagers arrived they got additional measurements from the close flybys of two other spacecraft, Pioneers 10 and 11.
@Steve Linton correctly describes the effect of the "sideways" part of a spacecraft's velocity with respect to the planet it's approaching. If the sideways component is large enough, the probe will miss the planet. To increase that sideways component, on approach you aim the spacecraft [via trajectory adjustment maneuvers, where you use the spacecraft's rocket engine(s)] to precisely tune the components of velocity, one straight toward the planet and one sideways. That combination sets the "aim point", the point to the side of the planet where the spacecraft would go if the planet had no gravity. Trajectory designers call that miss distance "b", a part of the b-plane aim point.
This diagram helps to see the effect of increasing b if the approach velocity is held constant. For typical interplanetary trajectories the approach velocity doesn't depend on the chosen b. [Sorry for the pixellation—this is a very old diagram I pulled out of my presentation archives]
The brownish-tan (in the original file it was orange!) circle represents the planet, and the different colors of curves represent the trajectories followed by spacecraft with different b, all coming in at the same approach velocity and parallel to the black dashed line through the center of the planet. The thinner, dot-dashed lines represent what the spacecrafts' trajectories would be if the planet had no gravity. If b isn't large enough, as with the red and brownish-tan (formerly orange!) trajectories, they impact the planet. The thick, dashed red and brownish-tan lines show how those two trajectories would continue if they didn't run into anything, as would be the case if the planet were the same mass but a lot smaller in size (i.e., a lot denser).
The green and blue lines are for trajectories whose b are large enough that they miss the planet. They point out a couple of characteristics of these hyperbolic orbits: 1) the larger the b, the farther from the planet is the closest-approach distance; and 2) the larger the b, the smaller is the angle the planet "bends" the trajectory. Again, this is for a fixed approach velocity. The red trajectory is bent by almost 180°, while the blue one is bent only ~135-140°.
This bend is key to making gravity assists so useful.
answered Jan 28 at 21:18
Tom SpilkerTom Spilker
9,7062152
$endgroup$
To add to the answers @Hobbes & @Steve Linton posted, the mission designers indeed knew Jupiter's gravity field quite well from the orbits of Jupiter's moons. But before the Voyagers arrived they got additional measurements from the close flybys of two other spacecraft, Pioneers 10 and 11.
@Steve Linton correctly describes the effect of the "sideways" part of a spacecraft's velocity with respect to the planet it's approaching. If the sideways component is large enough, the probe will miss the planet. To increase that sideways component, on approach you aim the spacecraft [via trajectory adjustment maneuvers, where you use the spacecraft's rocket engine(s)] to precisely tune the components of velocity, one straight toward the planet and one sideways. That combination sets the "aim point", the point to the side of the planet where the spacecraft would go if the planet had no gravity. Trajectory designers call that miss distance "b", a part of the b-plane aim point.
This diagram helps to see the effect of increasing b if the approach velocity is held constant. For typical interplanetary trajectories the approach velocity doesn't depend on the chosen b. [Sorry for the pixellation—this is a very old diagram I pulled out of my presentation archives]
The brownish-tan (in the original file it was orange!) circle represents the planet, and the different colors of curves represent the trajectories followed by spacecraft with different b, all coming in at the same approach velocity and parallel to the black dashed line through the center of the planet. The thinner, dot-dashed lines represent what the spacecrafts' trajectories would be if the planet had no gravity. If b isn't large enough, as with the red and brownish-tan (formerly orange!) trajectories, they impact the planet. The thick, dashed red and brownish-tan lines show how those two trajectories would continue if they didn't run into anything, as would be the case if the planet were the same mass but a lot smaller in size (i.e., a lot denser).
The green and blue lines are for trajectories whose b are large enough that they miss the planet. They point out a couple of characteristics of these hyperbolic orbits: 1) the larger the b, the farther from the planet is the closest-approach distance; and 2) the larger the b, the smaller is the angle the planet "bends" the trajectory. Again, this is for a fixed approach velocity. The red trajectory is bent by almost 180°, while the blue one is bent only ~135-140°.
This bend is key to making gravity assists so useful.
answered Jan 28 at 21:18
Tom SpilkerTom Spilker
9,7062152
answered Jan 28 at 21:18
Tom SpilkerTom Spilker
9,7062152
answered Jan 28 at 21:18
Tom SpilkerTom Spilker
9,7062152
answered Jan 28 at 21:18
Tom SpilkerTom Spilker
9,7062152
9,7062152
add a comment |
add a comment |
$begingroup$
Let's try and understand how gravity works in space. This is kind of key idea to understanding lots of issues in space travel and astronomy.
So imagine a space probe, or rock, which is heading in from deep space, aimed almost, but not quite towards a planet. We can break that motion into two parts -- the part towards the planet and the "sideways" part. The planet's gravity accelerates the part towards the planet, but it does nothing to the sideways part, so the space probe is still not heading quite towards the planet. This keeps on happening, so the probe is moving increasingly fast, still almost but not quite heading at the planet.
After a while the probe whizzes past the side of the planet, going very fast. It can't hit it because it has never lost the "sideways" part of its motion (called angular momentum, more or less). It is still being pulled towards the planet but still moving around it, due to all the speed it has picked up. Then that speed starts to carry it away from the planet again (possibly with its course having been bent around to a greater or lesser extent). Now the gravity is slowing it down. However it is slowing it down just exactly as much as it speeded it up coming in so if it came in from far away at a certain speed, it will eventually escape equally far away at the same speed.
If someone can add some diagrams to this, that would be great.
answered Jan 28 at 20:08
Steve LintonSteve Linton
7,32711742
$endgroup$
$begingroup$
Lucidly explained, certainly. It does leave me wondering whether you've been quite clear as to how the spacecraft is speeded up, though, as you seem to have it departing the planet at the same speed it had when it arrived?
$endgroup$
– Ed999
Jan 29 at 14:18
1
$begingroup$
@Ed999: That is correct (in the planet's frame of reference). And that is a crucial insight in how and when gravity assists can work. You use them to gain speed in the Sun's frame of reference.
$endgroup$
– MSalters
Jan 29 at 14:56
2
$begingroup$
I think the OP's basic misunderstanding is that the planet is moving. Gravity assist doesn't work otherwise. The point to get across is that the probe spends a longer time falling into the planet (because its closing from behind a moving planet) and a shorter time escaping from the planet (because it slips out the side).
$endgroup$
– Oscar Bravo
Jan 29 at 15:32
4
$begingroup$
@Ed999 I was just trying to explain not being "completely attracted". Explaining gravity assist is a whole other problem. Very briefly MSalters and Oscar Bravo have it right. It's a bit like bouncing a tennis ball of the front of an oncoming train. Relative to the train, the ball bounced back at more of less the same speed it arrived at, but relative to the ground it's now moving much faster.
$endgroup$
– Steve Linton
Jan 29 at 19:40
$begingroup$
@SteveLinton I confess that I don't understand the o/p's use of the term completely attracted (unless he thinks a planet and a black hole are identical in mass?), but I like the analogy of the tennis ball and the train, which really does get the point across.
$endgroup$
– Ed999
Jan 31 at 2:54
add a comment |
$begingroup$
Let's try and understand how gravity works in space. This is kind of key idea to understanding lots of issues in space travel and astronomy.
So imagine a space probe, or rock, which is heading in from deep space, aimed almost, but not quite towards a planet. We can break that motion into two parts -- the part towards the planet and the "sideways" part. The planet's gravity accelerates the part towards the planet, but it does nothing to the sideways part, so the space probe is still not heading quite towards the planet. This keeps on happening, so the probe is moving increasingly fast, still almost but not quite heading at the planet.
After a while the probe whizzes past the side of the planet, going very fast. It can't hit it because it has never lost the "sideways" part of its motion (called angular momentum, more or less). It is still being pulled towards the planet but still moving around it, due to all the speed it has picked up. Then that speed starts to carry it away from the planet again (possibly with its course having been bent around to a greater or lesser extent). Now the gravity is slowing it down. However it is slowing it down just exactly as much as it speeded it up coming in so if it came in from far away at a certain speed, it will eventually escape equally far away at the same speed.
If someone can add some diagrams to this, that would be great.
answered Jan 28 at 20:08
Steve LintonSteve Linton
7,32711742
$endgroup$
$begingroup$
Lucidly explained, certainly. It does leave me wondering whether you've been quite clear as to how the spacecraft is speeded up, though, as you seem to have it departing the planet at the same speed it had when it arrived?
$endgroup$
– Ed999
Jan 29 at 14:18
1
$begingroup$
@Ed999: That is correct (in the planet's frame of reference). And that is a crucial insight in how and when gravity assists can work. You use them to gain speed in the Sun's frame of reference.
$endgroup$
– MSalters
Jan 29 at 14:56
2
$begingroup$
I think the OP's basic misunderstanding is that the planet is moving. Gravity assist doesn't work otherwise. The point to get across is that the probe spends a longer time falling into the planet (because its closing from behind a moving planet) and a shorter time escaping from the planet (because it slips out the side).
$endgroup$
– Oscar Bravo
Jan 29 at 15:32
4
$begingroup$
@Ed999 I was just trying to explain not being "completely attracted". Explaining gravity assist is a whole other problem. Very briefly MSalters and Oscar Bravo have it right. It's a bit like bouncing a tennis ball of the front of an oncoming train. Relative to the train, the ball bounced back at more of less the same speed it arrived at, but relative to the ground it's now moving much faster.
$endgroup$
– Steve Linton
Jan 29 at 19:40
$begingroup$
@SteveLinton I confess that I don't understand the o/p's use of the term completely attracted (unless he thinks a planet and a black hole are identical in mass?), but I like the analogy of the tennis ball and the train, which really does get the point across.
$endgroup$
– Ed999
Jan 31 at 2:54
add a comment |
$begingroup$
Let's try and understand how gravity works in space. This is kind of key idea to understanding lots of issues in space travel and astronomy.
So imagine a space probe, or rock, which is heading in from deep space, aimed almost, but not quite towards a planet. We can break that motion into two parts -- the part towards the planet and the "sideways" part. The planet's gravity accelerates the part towards the planet, but it does nothing to the sideways part, so the space probe is still not heading quite towards the planet. This keeps on happening, so the probe is moving increasingly fast, still almost but not quite heading at the planet.
After a while the probe whizzes past the side of the planet, going very fast. It can't hit it because it has never lost the "sideways" part of its motion (called angular momentum, more or less). It is still being pulled towards the planet but still moving around it, due to all the speed it has picked up. Then that speed starts to carry it away from the planet again (possibly with its course having been bent around to a greater or lesser extent). Now the gravity is slowing it down. However it is slowing it down just exactly as much as it speeded it up coming in so if it came in from far away at a certain speed, it will eventually escape equally far away at the same speed.
If someone can add some diagrams to this, that would be great.
answered Jan 28 at 20:08
Steve LintonSteve Linton
7,32711742
$endgroup$
Let's try and understand how gravity works in space. This is kind of key idea to understanding lots of issues in space travel and astronomy.
So imagine a space probe, or rock, which is heading in from deep space, aimed almost, but not quite towards a planet. We can break that motion into two parts -- the part towards the planet and the "sideways" part. The planet's gravity accelerates the part towards the planet, but it does nothing to the sideways part, so the space probe is still not heading quite towards the planet. This keeps on happening, so the probe is moving increasingly fast, still almost but not quite heading at the planet.
After a while the probe whizzes past the side of the planet, going very fast. It can't hit it because it has never lost the "sideways" part of its motion (called angular momentum, more or less). It is still being pulled towards the planet but still moving around it, due to all the speed it has picked up. Then that speed starts to carry it away from the planet again (possibly with its course having been bent around to a greater or lesser extent). Now the gravity is slowing it down. However it is slowing it down just exactly as much as it speeded it up coming in so if it came in from far away at a certain speed, it will eventually escape equally far away at the same speed.
If someone can add some diagrams to this, that would be great.
answered Jan 28 at 20:08
Steve LintonSteve Linton
7,32711742
answered Jan 28 at 20:08
Steve LintonSteve Linton
7,32711742
answered Jan 28 at 20:08
Steve LintonSteve Linton
7,32711742
answered Jan 28 at 20:08
Steve LintonSteve Linton
7,32711742
7,32711742
$begingroup$
Lucidly explained, certainly. It does leave me wondering whether you've been quite clear as to how the spacecraft is speeded up, though, as you seem to have it departing the planet at the same speed it had when it arrived?
$endgroup$
– Ed999
Jan 29 at 14:18
1
$begingroup$
@Ed999: That is correct (in the planet's frame of reference). And that is a crucial insight in how and when gravity assists can work. You use them to gain speed in the Sun's frame of reference.
$endgroup$
– MSalters
Jan 29 at 14:56
2
$begingroup$
I think the OP's basic misunderstanding is that the planet is moving. Gravity assist doesn't work otherwise. The point to get across is that the probe spends a longer time falling into the planet (because its closing from behind a moving planet) and a shorter time escaping from the planet (because it slips out the side).
$endgroup$
– Oscar Bravo
Jan 29 at 15:32
4
$begingroup$
@Ed999 I was just trying to explain not being "completely attracted". Explaining gravity assist is a whole other problem. Very briefly MSalters and Oscar Bravo have it right. It's a bit like bouncing a tennis ball of the front of an oncoming train. Relative to the train, the ball bounced back at more of less the same speed it arrived at, but relative to the ground it's now moving much faster.
$endgroup$
– Steve Linton
Jan 29 at 19:40
$begingroup$
@SteveLinton I confess that I don't understand the o/p's use of the term completely attracted (unless he thinks a planet and a black hole are identical in mass?), but I like the analogy of the tennis ball and the train, which really does get the point across.
$endgroup$
– Ed999
Jan 31 at 2:54
add a comment |
$begingroup$
Lucidly explained, certainly. It does leave me wondering whether you've been quite clear as to how the spacecraft is speeded up, though, as you seem to have it departing the planet at the same speed it had when it arrived?
$endgroup$
– Ed999
Jan 29 at 14:18
1
$begingroup$
@Ed999: That is correct (in the planet's frame of reference). And that is a crucial insight in how and when gravity assists can work. You use them to gain speed in the Sun's frame of reference.
$endgroup$
– MSalters
Jan 29 at 14:56
2
$begingroup$
I think the OP's basic misunderstanding is that the planet is moving. Gravity assist doesn't work otherwise. The point to get across is that the probe spends a longer time falling into the planet (because its closing from behind a moving planet) and a shorter time escaping from the planet (because it slips out the side).
$endgroup$
– Oscar Bravo
Jan 29 at 15:32
4
$begingroup$
@Ed999 I was just trying to explain not being "completely attracted". Explaining gravity assist is a whole other problem. Very briefly MSalters and Oscar Bravo have it right. It's a bit like bouncing a tennis ball of the front of an oncoming train. Relative to the train, the ball bounced back at more of less the same speed it arrived at, but relative to the ground it's now moving much faster.
$endgroup$
– Steve Linton
Jan 29 at 19:40
$begingroup$
@SteveLinton I confess that I don't understand the o/p's use of the term completely attracted (unless he thinks a planet and a black hole are identical in mass?), but I like the analogy of the tennis ball and the train, which really does get the point across.
$endgroup$
– Ed999
Jan 31 at 2:54
$begingroup$
Lucidly explained, certainly. It does leave me wondering whether you've been quite clear as to how the spacecraft is speeded up, though, as you seem to have it departing the planet at the same speed it had when it arrived?
$endgroup$
– Ed999
Jan 29 at 14:18
$begingroup$
Lucidly explained, certainly. It does leave me wondering whether you've been quite clear as to how the spacecraft is speeded up, though, as you seem to have it departing the planet at the same speed it had when it arrived?
$endgroup$
– Ed999
Jan 29 at 14:18
1
1
$begingroup$
@Ed999: That is correct (in the planet's frame of reference). And that is a crucial insight in how and when gravity assists can work. You use them to gain speed in the Sun's frame of reference.
$endgroup$
– MSalters
Jan 29 at 14:56
$begingroup$
@Ed999: That is correct (in the planet's frame of reference). And that is a crucial insight in how and when gravity assists can work. You use them to gain speed in the Sun's frame of reference.
$endgroup$
– MSalters
Jan 29 at 14:56
2
2
$begingroup$
I think the OP's basic misunderstanding is that the planet is moving. Gravity assist doesn't work otherwise. The point to get across is that the probe spends a longer time falling into the planet (because its closing from behind a moving planet) and a shorter time escaping from the planet (because it slips out the side).
$endgroup$
– Oscar Bravo
Jan 29 at 15:32
$begingroup$
I think the OP's basic misunderstanding is that the planet is moving. Gravity assist doesn't work otherwise. The point to get across is that the probe spends a longer time falling into the planet (because its closing from behind a moving planet) and a shorter time escaping from the planet (because it slips out the side).
$endgroup$
– Oscar Bravo
Jan 29 at 15:32
4
4
$begingroup$
@Ed999 I was just trying to explain not being "completely attracted". Explaining gravity assist is a whole other problem. Very briefly MSalters and Oscar Bravo have it right. It's a bit like bouncing a tennis ball of the front of an oncoming train. Relative to the train, the ball bounced back at more of less the same speed it arrived at, but relative to the ground it's now moving much faster.
$endgroup$
– Steve Linton
Jan 29 at 19:40
$begingroup$
@Ed999 I was just trying to explain not being "completely attracted". Explaining gravity assist is a whole other problem. Very briefly MSalters and Oscar Bravo have it right. It's a bit like bouncing a tennis ball of the front of an oncoming train. Relative to the train, the ball bounced back at more of less the same speed it arrived at, but relative to the ground it's now moving much faster.
$endgroup$
– Steve Linton
Jan 29 at 19:40
$begingroup$
@SteveLinton I confess that I don't understand the o/p's use of the term completely attracted (unless he thinks a planet and a black hole are identical in mass?), but I like the analogy of the tennis ball and the train, which really does get the point across.
$endgroup$
– Ed999
Jan 31 at 2:54
$begingroup$
@SteveLinton I confess that I don't understand the o/p's use of the term completely attracted (unless he thinks a planet and a black hole are identical in mass?), but I like the analogy of the tennis ball and the train, which really does get the point across.
$endgroup$
– Ed999
Jan 31 at 2:54
add a comment |
$begingroup$
At the heart of this question is how the gravity assist 'works', at least an intuitive understanding of it.
When a space probe uses gravity assist from a planet, it gets some energy from it, which (yes) incidentally slows the planet down slightly.
How does it do it?
Imagine a tennis ball in flight towards an ideal moving tennis racket. The ball rebounds from the racket with (ideally, more or less) no energy dissipated in ball and strings, and so leaves the racket at the same velocity with respect to the racket that it arrived with. As the racket was moving, this added the racket velocity to the ball velocity.
Now you can't just bounce a space probe off a planet. But you can swing it round the back in a hyperbolic orbit. It comes in with a certain velocity with respect to the planet, and leaves with the same relative velocity, just like the ball and racket. If you fling the probe near the trailing edge of the planet in its orbit, 'behind' the planet, then the probe will be accelerated in the direction of its orbit. However, if you fling the probe ahead of the planet, the probe will be slowed down, and the planet speeded up just a tad. Which is what the Parker Solar Probe is going to do with Venus 7 times to drop into a lower orbit around the sun.
So I like to think of a gravity assist as 'bouncing' the probe off the planet. But you can only add speed if you end up travelling in the same orbital direction as the planet. So you have to plan your mission to suit.
answered Jan 29 at 14:45
Neil_UKNeil_UK
1512
$endgroup$
add a comment |
$begingroup$
At the heart of this question is how the gravity assist 'works', at least an intuitive understanding of it.
When a space probe uses gravity assist from a planet, it gets some energy from it, which (yes) incidentally slows the planet down slightly.
How does it do it?
Imagine a tennis ball in flight towards an ideal moving tennis racket. The ball rebounds from the racket with (ideally, more or less) no energy dissipated in ball and strings, and so leaves the racket at the same velocity with respect to the racket that it arrived with. As the racket was moving, this added the racket velocity to the ball velocity.
Now you can't just bounce a space probe off a planet. But you can swing it round the back in a hyperbolic orbit. It comes in with a certain velocity with respect to the planet, and leaves with the same relative velocity, just like the ball and racket. If you fling the probe near the trailing edge of the planet in its orbit, 'behind' the planet, then the probe will be accelerated in the direction of its orbit. However, if you fling the probe ahead of the planet, the probe will be slowed down, and the planet speeded up just a tad. Which is what the Parker Solar Probe is going to do with Venus 7 times to drop into a lower orbit around the sun.
So I like to think of a gravity assist as 'bouncing' the probe off the planet. But you can only add speed if you end up travelling in the same orbital direction as the planet. So you have to plan your mission to suit.
answered Jan 29 at 14:45
Neil_UKNeil_UK
1512
$endgroup$
add a comment |
$begingroup$
At the heart of this question is how the gravity assist 'works', at least an intuitive understanding of it.
When a space probe uses gravity assist from a planet, it gets some energy from it, which (yes) incidentally slows the planet down slightly.
How does it do it?
Imagine a tennis ball in flight towards an ideal moving tennis racket. The ball rebounds from the racket with (ideally, more or less) no energy dissipated in ball and strings, and so leaves the racket at the same velocity with respect to the racket that it arrived with. As the racket was moving, this added the racket velocity to the ball velocity.
Now you can't just bounce a space probe off a planet. But you can swing it round the back in a hyperbolic orbit. It comes in with a certain velocity with respect to the planet, and leaves with the same relative velocity, just like the ball and racket. If you fling the probe near the trailing edge of the planet in its orbit, 'behind' the planet, then the probe will be accelerated in the direction of its orbit. However, if you fling the probe ahead of the planet, the probe will be slowed down, and the planet speeded up just a tad. Which is what the Parker Solar Probe is going to do with Venus 7 times to drop into a lower orbit around the sun.
So I like to think of a gravity assist as 'bouncing' the probe off the planet. But you can only add speed if you end up travelling in the same orbital direction as the planet. So you have to plan your mission to suit.
answered Jan 29 at 14:45
Neil_UKNeil_UK
1512
$endgroup$
At the heart of this question is how the gravity assist 'works', at least an intuitive understanding of it.
When a space probe uses gravity assist from a planet, it gets some energy from it, which (yes) incidentally slows the planet down slightly.
How does it do it?
Imagine a tennis ball in flight towards an ideal moving tennis racket. The ball rebounds from the racket with (ideally, more or less) no energy dissipated in ball and strings, and so leaves the racket at the same velocity with respect to the racket that it arrived with. As the racket was moving, this added the racket velocity to the ball velocity.
Now you can't just bounce a space probe off a planet. But you can swing it round the back in a hyperbolic orbit. It comes in with a certain velocity with respect to the planet, and leaves with the same relative velocity, just like the ball and racket. If you fling the probe near the trailing edge of the planet in its orbit, 'behind' the planet, then the probe will be accelerated in the direction of its orbit. However, if you fling the probe ahead of the planet, the probe will be slowed down, and the planet speeded up just a tad. Which is what the Parker Solar Probe is going to do with Venus 7 times to drop into a lower orbit around the sun.
So I like to think of a gravity assist as 'bouncing' the probe off the planet. But you can only add speed if you end up travelling in the same orbital direction as the planet. So you have to plan your mission to suit.
answered Jan 29 at 14:45
Neil_UKNeil_UK
1512
answered Jan 29 at 14:45
Neil_UKNeil_UK
1512
answered Jan 29 at 14:45
Neil_UKNeil_UK
1512
answered Jan 29 at 14:45
Neil_UKNeil_UK
1512
1512
add a comment |
add a comment |
$begingroup$
Let's focus on the "why weren't they completely attracted by their gravitational field" question here.
You've probably seen Newton's cannonball diagram before:
In this diagram, the distance from the object to the planet is fixed, and it's the speed that changes, and gives different results:
- too slow (A or B, below orbital speed): it will indeed be pulled towards the planet
- between orbital speed and the escape velocity (C and D), it will enter orbit around the planet
- beyond escape velocity (E), it will escape the planet's gravity.
However, orbital speed and escape velocity depend on the distance from the planet. So given the same speed, you can get the same results by varying the distance from the planet:
- too close: A or B, fall on the planet
- C or D, enter orbit
- far enough: E, the trajectory gets modified ("bent"), but the probe can continue its journey
So it's all a matter of getting the probe to approach Jupiter at the right combination of speed and distance (which is complicated by the fact that said distance is influenced by gravity on the approach, and that the planet is moving) to get the result you want: crash into the planet, enter orbit, or just perform a "fly-by" and continue somewhere else.
In the latter case, you want to take advantage of the operation to change to a specific direction and/or gain speed, which complicates things a bit further (in terms of picking the exact initial speed/trajectory to achieve the desired result), but it doesn't change the fact that you just need to be the right combination of fast enough and far enough to avoid crashing on the planet.
Of course, to compute all this, you indeed need to know the mass of the planet, but as others have written, there have been ways to compute this for quite a while.
answered Jan 29 at 17:11
jcaronjcaron
69348
$endgroup$
$begingroup$
Excepting of course that you cannot shoot anything to orbit, as the illustration seems to imply. Everything you shoot will either land again, or escape, never orbit.
$endgroup$
– Stian Yttervik
Jan 30 at 13:06
3
$begingroup$
@StianYttervik I'm not sure I understand what you mean? You do understand the context of Newton's cannonball? It's shot horizontally, from a high altitude where there is no atmosphere, so as long as you shoot the object at the right speed for that altitude, it will definitely orbit the planet. What did I miss?
$endgroup$
– jcaron
Jan 30 at 13:57
1
$begingroup$
Well, of course you'll also need to move the cannon out of the way after shooting it, otherwise the ball will hit it after completing its orbit. (If the planet is rotating, that'll usually do the job, especially if you shoot the ball with enough velocity to put it into an elliptical orbit with an apoapsis above the cannon's altitude, as in trajectory D in the image. Of course, even so, the ball may still hit the cannon eventually, after further rotation brings the cannon back to the ball's periapsis just as the ball is passing it.)
$endgroup$
– Ilmari Karonen
Jan 30 at 15:42
add a comment |
$begingroup$
Let's focus on the "why weren't they completely attracted by their gravitational field" question here.
You've probably seen Newton's cannonball diagram before:
In this diagram, the distance from the object to the planet is fixed, and it's the speed that changes, and gives different results:
- too slow (A or B, below orbital speed): it will indeed be pulled towards the planet
- between orbital speed and the escape velocity (C and D), it will enter orbit around the planet
- beyond escape velocity (E), it will escape the planet's gravity.
However, orbital speed and escape velocity depend on the distance from the planet. So given the same speed, you can get the same results by varying the distance from the planet:
- too close: A or B, fall on the planet
- C or D, enter orbit
- far enough: E, the trajectory gets modified ("bent"), but the probe can continue its journey
So it's all a matter of getting the probe to approach Jupiter at the right combination of speed and distance (which is complicated by the fact that said distance is influenced by gravity on the approach, and that the planet is moving) to get the result you want: crash into the planet, enter orbit, or just perform a "fly-by" and continue somewhere else.
In the latter case, you want to take advantage of the operation to change to a specific direction and/or gain speed, which complicates things a bit further (in terms of picking the exact initial speed/trajectory to achieve the desired result), but it doesn't change the fact that you just need to be the right combination of fast enough and far enough to avoid crashing on the planet.
Of course, to compute all this, you indeed need to know the mass of the planet, but as others have written, there have been ways to compute this for quite a while.
answered Jan 29 at 17:11
jcaronjcaron
69348
$endgroup$
$begingroup$
Excepting of course that you cannot shoot anything to orbit, as the illustration seems to imply. Everything you shoot will either land again, or escape, never orbit.
$endgroup$
– Stian Yttervik
Jan 30 at 13:06
3
$begingroup$
@StianYttervik I'm not sure I understand what you mean? You do understand the context of Newton's cannonball? It's shot horizontally, from a high altitude where there is no atmosphere, so as long as you shoot the object at the right speed for that altitude, it will definitely orbit the planet. What did I miss?
$endgroup$
– jcaron
Jan 30 at 13:57
1
$begingroup$
Well, of course you'll also need to move the cannon out of the way after shooting it, otherwise the ball will hit it after completing its orbit. (If the planet is rotating, that'll usually do the job, especially if you shoot the ball with enough velocity to put it into an elliptical orbit with an apoapsis above the cannon's altitude, as in trajectory D in the image. Of course, even so, the ball may still hit the cannon eventually, after further rotation brings the cannon back to the ball's periapsis just as the ball is passing it.)
$endgroup$
– Ilmari Karonen
Jan 30 at 15:42
add a comment |
$begingroup$
Let's focus on the "why weren't they completely attracted by their gravitational field" question here.
You've probably seen Newton's cannonball diagram before:
In this diagram, the distance from the object to the planet is fixed, and it's the speed that changes, and gives different results:
- too slow (A or B, below orbital speed): it will indeed be pulled towards the planet
- between orbital speed and the escape velocity (C and D), it will enter orbit around the planet
- beyond escape velocity (E), it will escape the planet's gravity.
However, orbital speed and escape velocity depend on the distance from the planet. So given the same speed, you can get the same results by varying the distance from the planet:
- too close: A or B, fall on the planet
- C or D, enter orbit
- far enough: E, the trajectory gets modified ("bent"), but the probe can continue its journey
So it's all a matter of getting the probe to approach Jupiter at the right combination of speed and distance (which is complicated by the fact that said distance is influenced by gravity on the approach, and that the planet is moving) to get the result you want: crash into the planet, enter orbit, or just perform a "fly-by" and continue somewhere else.
In the latter case, you want to take advantage of the operation to change to a specific direction and/or gain speed, which complicates things a bit further (in terms of picking the exact initial speed/trajectory to achieve the desired result), but it doesn't change the fact that you just need to be the right combination of fast enough and far enough to avoid crashing on the planet.
Of course, to compute all this, you indeed need to know the mass of the planet, but as others have written, there have been ways to compute this for quite a while.
answered Jan 29 at 17:11
jcaronjcaron
69348
$endgroup$
Let's focus on the "why weren't they completely attracted by their gravitational field" question here.
You've probably seen Newton's cannonball diagram before:
In this diagram, the distance from the object to the planet is fixed, and it's the speed that changes, and gives different results:
- too slow (A or B, below orbital speed): it will indeed be pulled towards the planet
- between orbital speed and the escape velocity (C and D), it will enter orbit around the planet
- beyond escape velocity (E), it will escape the planet's gravity.
However, orbital speed and escape velocity depend on the distance from the planet. So given the same speed, you can get the same results by varying the distance from the planet:
- too close: A or B, fall on the planet
- C or D, enter orbit
- far enough: E, the trajectory gets modified ("bent"), but the probe can continue its journey
So it's all a matter of getting the probe to approach Jupiter at the right combination of speed and distance (which is complicated by the fact that said distance is influenced by gravity on the approach, and that the planet is moving) to get the result you want: crash into the planet, enter orbit, or just perform a "fly-by" and continue somewhere else.
In the latter case, you want to take advantage of the operation to change to a specific direction and/or gain speed, which complicates things a bit further (in terms of picking the exact initial speed/trajectory to achieve the desired result), but it doesn't change the fact that you just need to be the right combination of fast enough and far enough to avoid crashing on the planet.
Of course, to compute all this, you indeed need to know the mass of the planet, but as others have written, there have been ways to compute this for quite a while.
answered Jan 29 at 17:11
jcaronjcaron
69348
edited Jan 30 at 13:58
answered Jan 29 at 17:11
jcaronjcaron
69348
answered Jan 29 at 17:11
jcaronjcaron
69348
answered Jan 29 at 17:11
jcaronjcaron
69348
69348
$begingroup$
Excepting of course that you cannot shoot anything to orbit, as the illustration seems to imply. Everything you shoot will either land again, or escape, never orbit.
$endgroup$
– Stian Yttervik
Jan 30 at 13:06
3
$begingroup$
@StianYttervik I'm not sure I understand what you mean? You do understand the context of Newton's cannonball? It's shot horizontally, from a high altitude where there is no atmosphere, so as long as you shoot the object at the right speed for that altitude, it will definitely orbit the planet. What did I miss?
$endgroup$
– jcaron
Jan 30 at 13:57
1
$begingroup$
Well, of course you'll also need to move the cannon out of the way after shooting it, otherwise the ball will hit it after completing its orbit. (If the planet is rotating, that'll usually do the job, especially if you shoot the ball with enough velocity to put it into an elliptical orbit with an apoapsis above the cannon's altitude, as in trajectory D in the image. Of course, even so, the ball may still hit the cannon eventually, after further rotation brings the cannon back to the ball's periapsis just as the ball is passing it.)
$endgroup$
– Ilmari Karonen
Jan 30 at 15:42
add a comment |
$begingroup$
Excepting of course that you cannot shoot anything to orbit, as the illustration seems to imply. Everything you shoot will either land again, or escape, never orbit.
$endgroup$
– Stian Yttervik
Jan 30 at 13:06
3
$begingroup$
@StianYttervik I'm not sure I understand what you mean? You do understand the context of Newton's cannonball? It's shot horizontally, from a high altitude where there is no atmosphere, so as long as you shoot the object at the right speed for that altitude, it will definitely orbit the planet. What did I miss?
$endgroup$
– jcaron
Jan 30 at 13:57
1
$begingroup$
Well, of course you'll also need to move the cannon out of the way after shooting it, otherwise the ball will hit it after completing its orbit. (If the planet is rotating, that'll usually do the job, especially if you shoot the ball with enough velocity to put it into an elliptical orbit with an apoapsis above the cannon's altitude, as in trajectory D in the image. Of course, even so, the ball may still hit the cannon eventually, after further rotation brings the cannon back to the ball's periapsis just as the ball is passing it.)
$endgroup$
– Ilmari Karonen
Jan 30 at 15:42
$begingroup$
Excepting of course that you cannot shoot anything to orbit, as the illustration seems to imply. Everything you shoot will either land again, or escape, never orbit.
$endgroup$
– Stian Yttervik
Jan 30 at 13:06
$begingroup$
Excepting of course that you cannot shoot anything to orbit, as the illustration seems to imply. Everything you shoot will either land again, or escape, never orbit.
$endgroup$
– Stian Yttervik
Jan 30 at 13:06
3
3
$begingroup$
@StianYttervik I'm not sure I understand what you mean? You do understand the context of Newton's cannonball? It's shot horizontally, from a high altitude where there is no atmosphere, so as long as you shoot the object at the right speed for that altitude, it will definitely orbit the planet. What did I miss?
$endgroup$
– jcaron
Jan 30 at 13:57
$begingroup$
@StianYttervik I'm not sure I understand what you mean? You do understand the context of Newton's cannonball? It's shot horizontally, from a high altitude where there is no atmosphere, so as long as you shoot the object at the right speed for that altitude, it will definitely orbit the planet. What did I miss?
$endgroup$
– jcaron
Jan 30 at 13:57
1
1
$begingroup$
Well, of course you'll also need to move the cannon out of the way after shooting it, otherwise the ball will hit it after completing its orbit. (If the planet is rotating, that'll usually do the job, especially if you shoot the ball with enough velocity to put it into an elliptical orbit with an apoapsis above the cannon's altitude, as in trajectory D in the image. Of course, even so, the ball may still hit the cannon eventually, after further rotation brings the cannon back to the ball's periapsis just as the ball is passing it.)
$endgroup$
– Ilmari Karonen
Jan 30 at 15:42
$begingroup$
Well, of course you'll also need to move the cannon out of the way after shooting it, otherwise the ball will hit it after completing its orbit. (If the planet is rotating, that'll usually do the job, especially if you shoot the ball with enough velocity to put it into an elliptical orbit with an apoapsis above the cannon's altitude, as in trajectory D in the image. Of course, even so, the ball may still hit the cannon eventually, after further rotation brings the cannon back to the ball's periapsis just as the ball is passing it.)
$endgroup$
– Ilmari Karonen
Jan 30 at 15:42
add a comment |
$begingroup$
Newton discovered gravity (and invented mathematical laws to describe it) a long time before Pioneer 10 flew!
It has been possible to calculate the mass of the planets with a fair degree of accuracy for a very long time, based on their effects on each other, as well as on their moons, and on passing NASA space probes. There were very well worked out laws of planetary motion in the pre-spaceage, which gave good values for the masses of the planets.
What seems to be missing from the other answers here is simply this bit of the explanation: a spacecraft steals energy from the planet by this type of slingshot orbit, since some of the planet's rotational motion is transfered to the vehicle (as motion): in effect, the planet is slowed down slightly, and the spacecraft is speeded up slightly.
You couldn't even begin to measure the effect on the planet's rotation: it's too minute to be measurable by our current techniques. But, in the math, the effect does occur, even though in practice we can't observe it.
The question seems to imply the spacecraft is slowed down in space, on its approach to a planet, but there is nothing in the vacuum of space to cause such an effect. And it will accelerate as it dives toward the planet, due to the planet's gravity pulling on it. If it misses the planet's atmosphere (which extends only a very short distance into space), the slingshot effect will actually increase the craft's speed, stealing rotational energy as momentum.
I'd like to recommend some further reading: the novel 2010: Odyssey 2 by Arthur C Clark contains a description of this type of slingshot orbit (slightly modified, but highly dramatic) which gives a scientifically accurate account of one, set at Jupiter.
answered Jan 29 at 14:53
Ed999Ed999
1412
$endgroup$
$begingroup$
Of course we knew the mass fairly well before Pioneer 10, and had reasonable estimates for J2 & J4, but after Pioneers 10 & 11 we knew the mass exquisitely, had better estimates for J2 and J4, and also measured J3 and J6. See Anderson & Null's papers agupubs.onlinelibrary.wiley.com/doi/pdf/10.1029/JA079i025p03661 (unfortunately behind a paywall) and adsabs.harvard.edu/full/1976AJ.....81.1153N (free!) . Those measurements were important for reducing the Voyagers' post-flyby TCM ∆Vs for the trajectory to Saturn. This decreased the propellant load, allowing a lower launch mass.
$endgroup$
– Tom Spilker
Jan 29 at 19:42
add a comment |
$begingroup$
Newton discovered gravity (and invented mathematical laws to describe it) a long time before Pioneer 10 flew!
It has been possible to calculate the mass of the planets with a fair degree of accuracy for a very long time, based on their effects on each other, as well as on their moons, and on passing NASA space probes. There were very well worked out laws of planetary motion in the pre-spaceage, which gave good values for the masses of the planets.
What seems to be missing from the other answers here is simply this bit of the explanation: a spacecraft steals energy from the planet by this type of slingshot orbit, since some of the planet's rotational motion is transfered to the vehicle (as motion): in effect, the planet is slowed down slightly, and the spacecraft is speeded up slightly.
You couldn't even begin to measure the effect on the planet's rotation: it's too minute to be measurable by our current techniques. But, in the math, the effect does occur, even though in practice we can't observe it.
The question seems to imply the spacecraft is slowed down in space, on its approach to a planet, but there is nothing in the vacuum of space to cause such an effect. And it will accelerate as it dives toward the planet, due to the planet's gravity pulling on it. If it misses the planet's atmosphere (which extends only a very short distance into space), the slingshot effect will actually increase the craft's speed, stealing rotational energy as momentum.
I'd like to recommend some further reading: the novel 2010: Odyssey 2 by Arthur C Clark contains a description of this type of slingshot orbit (slightly modified, but highly dramatic) which gives a scientifically accurate account of one, set at Jupiter.
answered Jan 29 at 14:53
Ed999Ed999
1412
$endgroup$
$begingroup$
Of course we knew the mass fairly well before Pioneer 10, and had reasonable estimates for J2 & J4, but after Pioneers 10 & 11 we knew the mass exquisitely, had better estimates for J2 and J4, and also measured J3 and J6. See Anderson & Null's papers agupubs.onlinelibrary.wiley.com/doi/pdf/10.1029/JA079i025p03661 (unfortunately behind a paywall) and adsabs.harvard.edu/full/1976AJ.....81.1153N (free!) . Those measurements were important for reducing the Voyagers' post-flyby TCM ∆Vs for the trajectory to Saturn. This decreased the propellant load, allowing a lower launch mass.
$endgroup$
– Tom Spilker
Jan 29 at 19:42
add a comment |
$begingroup$
Newton discovered gravity (and invented mathematical laws to describe it) a long time before Pioneer 10 flew!
It has been possible to calculate the mass of the planets with a fair degree of accuracy for a very long time, based on their effects on each other, as well as on their moons, and on passing NASA space probes. There were very well worked out laws of planetary motion in the pre-spaceage, which gave good values for the masses of the planets.
What seems to be missing from the other answers here is simply this bit of the explanation: a spacecraft steals energy from the planet by this type of slingshot orbit, since some of the planet's rotational motion is transfered to the vehicle (as motion): in effect, the planet is slowed down slightly, and the spacecraft is speeded up slightly.
You couldn't even begin to measure the effect on the planet's rotation: it's too minute to be measurable by our current techniques. But, in the math, the effect does occur, even though in practice we can't observe it.
The question seems to imply the spacecraft is slowed down in space, on its approach to a planet, but there is nothing in the vacuum of space to cause such an effect. And it will accelerate as it dives toward the planet, due to the planet's gravity pulling on it. If it misses the planet's atmosphere (which extends only a very short distance into space), the slingshot effect will actually increase the craft's speed, stealing rotational energy as momentum.
I'd like to recommend some further reading: the novel 2010: Odyssey 2 by Arthur C Clark contains a description of this type of slingshot orbit (slightly modified, but highly dramatic) which gives a scientifically accurate account of one, set at Jupiter.
answered Jan 29 at 14:53
Ed999Ed999
1412
$endgroup$
Newton discovered gravity (and invented mathematical laws to describe it) a long time before Pioneer 10 flew!
It has been possible to calculate the mass of the planets with a fair degree of accuracy for a very long time, based on their effects on each other, as well as on their moons, and on passing NASA space probes. There were very well worked out laws of planetary motion in the pre-spaceage, which gave good values for the masses of the planets.
What seems to be missing from the other answers here is simply this bit of the explanation: a spacecraft steals energy from the planet by this type of slingshot orbit, since some of the planet's rotational motion is transfered to the vehicle (as motion): in effect, the planet is slowed down slightly, and the spacecraft is speeded up slightly.
You couldn't even begin to measure the effect on the planet's rotation: it's too minute to be measurable by our current techniques. But, in the math, the effect does occur, even though in practice we can't observe it.
The question seems to imply the spacecraft is slowed down in space, on its approach to a planet, but there is nothing in the vacuum of space to cause such an effect. And it will accelerate as it dives toward the planet, due to the planet's gravity pulling on it. If it misses the planet's atmosphere (which extends only a very short distance into space), the slingshot effect will actually increase the craft's speed, stealing rotational energy as momentum.
I'd like to recommend some further reading: the novel 2010: Odyssey 2 by Arthur C Clark contains a description of this type of slingshot orbit (slightly modified, but highly dramatic) which gives a scientifically accurate account of one, set at Jupiter.
answered Jan 29 at 14:53
Ed999Ed999
1412
answered Jan 29 at 14:53
Ed999Ed999
1412
answered Jan 29 at 14:53
Ed999Ed999
1412
answered Jan 29 at 14:53
Ed999Ed999
1412
1412
$begingroup$
Of course we knew the mass fairly well before Pioneer 10, and had reasonable estimates for J2 & J4, but after Pioneers 10 & 11 we knew the mass exquisitely, had better estimates for J2 and J4, and also measured J3 and J6. See Anderson & Null's papers agupubs.onlinelibrary.wiley.com/doi/pdf/10.1029/JA079i025p03661 (unfortunately behind a paywall) and adsabs.harvard.edu/full/1976AJ.....81.1153N (free!) . Those measurements were important for reducing the Voyagers' post-flyby TCM ∆Vs for the trajectory to Saturn. This decreased the propellant load, allowing a lower launch mass.
$endgroup$
– Tom Spilker
Jan 29 at 19:42
add a comment |
$begingroup$
Of course we knew the mass fairly well before Pioneer 10, and had reasonable estimates for J2 & J4, but after Pioneers 10 & 11 we knew the mass exquisitely, had better estimates for J2 and J4, and also measured J3 and J6. See Anderson & Null's papers agupubs.onlinelibrary.wiley.com/doi/pdf/10.1029/JA079i025p03661 (unfortunately behind a paywall) and adsabs.harvard.edu/full/1976AJ.....81.1153N (free!) . Those measurements were important for reducing the Voyagers' post-flyby TCM ∆Vs for the trajectory to Saturn. This decreased the propellant load, allowing a lower launch mass.
$endgroup$
– Tom Spilker
Jan 29 at 19:42
$begingroup$
Of course we knew the mass fairly well before Pioneer 10, and had reasonable estimates for J2 & J4, but after Pioneers 10 & 11 we knew the mass exquisitely, had better estimates for J2 and J4, and also measured J3 and J6. See Anderson & Null's papers agupubs.onlinelibrary.wiley.com/doi/pdf/10.1029/JA079i025p03661 (unfortunately behind a paywall) and adsabs.harvard.edu/full/1976AJ.....81.1153N (free!) . Those measurements were important for reducing the Voyagers' post-flyby TCM ∆Vs for the trajectory to Saturn. This decreased the propellant load, allowing a lower launch mass.
$endgroup$
– Tom Spilker
Jan 29 at 19:42
$begingroup$
Of course we knew the mass fairly well before Pioneer 10, and had reasonable estimates for J2 & J4, but after Pioneers 10 & 11 we knew the mass exquisitely, had better estimates for J2 and J4, and also measured J3 and J6. See Anderson & Null's papers agupubs.onlinelibrary.wiley.com/doi/pdf/10.1029/JA079i025p03661 (unfortunately behind a paywall) and adsabs.harvard.edu/full/1976AJ.....81.1153N (free!) . Those measurements were important for reducing the Voyagers' post-flyby TCM ∆Vs for the trajectory to Saturn. This decreased the propellant load, allowing a lower launch mass.
$endgroup$
– Tom Spilker
Jan 29 at 19:42
add a comment |
$begingroup$
If you want a complete explanation of how the masses of the other planets have been calculated, I think that should be a separate question. But for this question, it doesn't really matter. Increasing the mass of the planet increases the escape velocity, but it also increases the gravitational force, which means that the probe would be traveling faster when it approaches the planet.
Escape velocity is the velocity one needs to reach a point infinitely far away from the planet. That is, the amount of energy to move an object from the surface of the planet to infinitely far away is equal to the kinetic energy of that object moving at escape velocity. Since energy is conserved, an object that goes from being on the surface with the kinetic energy of the escape velocity to being infinitely far away must have gravitational potential energy equal to the kinetic energy of escape velocity (keep in mind that the kinetic energy of escape velocity depends on the mass of the object, so there's an "of that object" in all of this).
Gravitational force falls off quickly with distance, so once you get to interplanetary distances, you can consider that distance to pretty much be "infinity" for planetary gravitational potential energy purposes. Thus, pretty much any probe launched from Earth will, when it reaches another planet, be traveling faster than that planet's escape velocity (it started out with the gravitational potential energy of "infinitely" far away, and that potential energy was converted to the kinetic energy of escape velocity). None of this depends on the mass of the other planet. The mass will affect the shape of the probe's path, but not that it has escape velocity.
As an analogy, suppose you have a dip in the ground, and the ground is the same height on both sides of the dip. And let's say this ground has no friction. If you roll a marble across the ground from outside the dip, it will come out the other side. It doesn't matter how deep the dip is; making the dip deeper means the marble needs more energy to climb out of it, but it also means that it gained more energy when it was falling into the dip. Similarly, if you throw something at a planet, then unless it's a direct hit, it will continue traveling at least as far away as it started (barring interactions with other bodies, of course).
answered Jan 29 at 17:17
AcccumulationAcccumulation
20314
$endgroup$
add a comment |
$begingroup$
If you want a complete explanation of how the masses of the other planets have been calculated, I think that should be a separate question. But for this question, it doesn't really matter. Increasing the mass of the planet increases the escape velocity, but it also increases the gravitational force, which means that the probe would be traveling faster when it approaches the planet.
Escape velocity is the velocity one needs to reach a point infinitely far away from the planet. That is, the amount of energy to move an object from the surface of the planet to infinitely far away is equal to the kinetic energy of that object moving at escape velocity. Since energy is conserved, an object that goes from being on the surface with the kinetic energy of the escape velocity to being infinitely far away must have gravitational potential energy equal to the kinetic energy of escape velocity (keep in mind that the kinetic energy of escape velocity depends on the mass of the object, so there's an "of that object" in all of this).
Gravitational force falls off quickly with distance, so once you get to interplanetary distances, you can consider that distance to pretty much be "infinity" for planetary gravitational potential energy purposes. Thus, pretty much any probe launched from Earth will, when it reaches another planet, be traveling faster than that planet's escape velocity (it started out with the gravitational potential energy of "infinitely" far away, and that potential energy was converted to the kinetic energy of escape velocity). None of this depends on the mass of the other planet. The mass will affect the shape of the probe's path, but not that it has escape velocity.
As an analogy, suppose you have a dip in the ground, and the ground is the same height on both sides of the dip. And let's say this ground has no friction. If you roll a marble across the ground from outside the dip, it will come out the other side. It doesn't matter how deep the dip is; making the dip deeper means the marble needs more energy to climb out of it, but it also means that it gained more energy when it was falling into the dip. Similarly, if you throw something at a planet, then unless it's a direct hit, it will continue traveling at least as far away as it started (barring interactions with other bodies, of course).
answered Jan 29 at 17:17
AcccumulationAcccumulation
20314
$endgroup$
add a comment |
$begingroup$
If you want a complete explanation of how the masses of the other planets have been calculated, I think that should be a separate question. But for this question, it doesn't really matter. Increasing the mass of the planet increases the escape velocity, but it also increases the gravitational force, which means that the probe would be traveling faster when it approaches the planet.
Escape velocity is the velocity one needs to reach a point infinitely far away from the planet. That is, the amount of energy to move an object from the surface of the planet to infinitely far away is equal to the kinetic energy of that object moving at escape velocity. Since energy is conserved, an object that goes from being on the surface with the kinetic energy of the escape velocity to being infinitely far away must have gravitational potential energy equal to the kinetic energy of escape velocity (keep in mind that the kinetic energy of escape velocity depends on the mass of the object, so there's an "of that object" in all of this).
Gravitational force falls off quickly with distance, so once you get to interplanetary distances, you can consider that distance to pretty much be "infinity" for planetary gravitational potential energy purposes. Thus, pretty much any probe launched from Earth will, when it reaches another planet, be traveling faster than that planet's escape velocity (it started out with the gravitational potential energy of "infinitely" far away, and that potential energy was converted to the kinetic energy of escape velocity). None of this depends on the mass of the other planet. The mass will affect the shape of the probe's path, but not that it has escape velocity.
As an analogy, suppose you have a dip in the ground, and the ground is the same height on both sides of the dip. And let's say this ground has no friction. If you roll a marble across the ground from outside the dip, it will come out the other side. It doesn't matter how deep the dip is; making the dip deeper means the marble needs more energy to climb out of it, but it also means that it gained more energy when it was falling into the dip. Similarly, if you throw something at a planet, then unless it's a direct hit, it will continue traveling at least as far away as it started (barring interactions with other bodies, of course).
answered Jan 29 at 17:17
AcccumulationAcccumulation
20314
$endgroup$
If you want a complete explanation of how the masses of the other planets have been calculated, I think that should be a separate question. But for this question, it doesn't really matter. Increasing the mass of the planet increases the escape velocity, but it also increases the gravitational force, which means that the probe would be traveling faster when it approaches the planet.
Escape velocity is the velocity one needs to reach a point infinitely far away from the planet. That is, the amount of energy to move an object from the surface of the planet to infinitely far away is equal to the kinetic energy of that object moving at escape velocity. Since energy is conserved, an object that goes from being on the surface with the kinetic energy of the escape velocity to being infinitely far away must have gravitational potential energy equal to the kinetic energy of escape velocity (keep in mind that the kinetic energy of escape velocity depends on the mass of the object, so there's an "of that object" in all of this).
Gravitational force falls off quickly with distance, so once you get to interplanetary distances, you can consider that distance to pretty much be "infinity" for planetary gravitational potential energy purposes. Thus, pretty much any probe launched from Earth will, when it reaches another planet, be traveling faster than that planet's escape velocity (it started out with the gravitational potential energy of "infinitely" far away, and that potential energy was converted to the kinetic energy of escape velocity). None of this depends on the mass of the other planet. The mass will affect the shape of the probe's path, but not that it has escape velocity.
As an analogy, suppose you have a dip in the ground, and the ground is the same height on both sides of the dip. And let's say this ground has no friction. If you roll a marble across the ground from outside the dip, it will come out the other side. It doesn't matter how deep the dip is; making the dip deeper means the marble needs more energy to climb out of it, but it also means that it gained more energy when it was falling into the dip. Similarly, if you throw something at a planet, then unless it's a direct hit, it will continue traveling at least as far away as it started (barring interactions with other bodies, of course).
answered Jan 29 at 17:17
AcccumulationAcccumulation
20314
answered Jan 29 at 17:17
AcccumulationAcccumulation
20314
answered Jan 29 at 17:17
AcccumulationAcccumulation
20314
answered Jan 29 at 17:17
AcccumulationAcccumulation
20314
20314
add a comment |
add a comment |
$begingroup$
Your problem appears to be too much science fiction and not enough science.
In science fiction, gravity sucks things towards the planet, sort of like a magnet. When far away you don't feel it; then you get close enough, and you are pulled down to the surface. And it holds things near the planet like glue or an elastic band.
In this science fiction model, stuff in space floats because it is far away. You go to space, you float.
In the real world, the science model, in space you fall just like you do in atmosphere. Orbit isn't far away, it is fast away -- things orbit the planet because they are going extremely fast.
Things in (low) orbit are amazingly close to the planet, from a distance they are almost skimming the surface. They are just moving so fast that the gravity of the planet pulling it down isn't fast enough; they "miss" the ground on each orbit.
Gravity doesn't stick or yank; gravity accelerates. When you go down a gravity well, you gain speed. This speed is mostly towards the planet, but it takes relative precision to actually hit the planet; even a modest velocity "horizontal" to the gravity well will make you miss.
And when you are going away from the planet, gravity doesn't "stick", it decelerates. The amount it decelerates on the way out is the same as it accelerated on the way in, at least in its frame of reference.
So you can set a trajectory that skims the surface of a planet (without an atmosphere, or outside it), and you'll only lose the speed you gained while falling in.
The slingshot maneuver uses the fact that the planet's "stationary" frame of reference isn't the same as the sun. So you approach the planet from "behind" in its orbital track, then leave roughly perpendicular to its orbital track.
If the planet was moving 13 km/s and you approached it from the side at 20 km/s.
In the planet's perspective you are moving sqrt(20^2 + 13^2) = 23.9 km/s.
You then leave the planet say, parallel to the orbital track in the same direction. You leave at 23.9 km/s.
But in the sun's frame of reference, you approached the planet at 20km/s and left at (13+23.9) = 36.9 km/s, a massive speed boost.
answered Jan 29 at 20:55
YakkYakk
25016
$endgroup$
add a comment |
$begingroup$
Your problem appears to be too much science fiction and not enough science.
In science fiction, gravity sucks things towards the planet, sort of like a magnet. When far away you don't feel it; then you get close enough, and you are pulled down to the surface. And it holds things near the planet like glue or an elastic band.
In this science fiction model, stuff in space floats because it is far away. You go to space, you float.
In the real world, the science model, in space you fall just like you do in atmosphere. Orbit isn't far away, it is fast away -- things orbit the planet because they are going extremely fast.
Things in (low) orbit are amazingly close to the planet, from a distance they are almost skimming the surface. They are just moving so fast that the gravity of the planet pulling it down isn't fast enough; they "miss" the ground on each orbit.
Gravity doesn't stick or yank; gravity accelerates. When you go down a gravity well, you gain speed. This speed is mostly towards the planet, but it takes relative precision to actually hit the planet; even a modest velocity "horizontal" to the gravity well will make you miss.
And when you are going away from the planet, gravity doesn't "stick", it decelerates. The amount it decelerates on the way out is the same as it accelerated on the way in, at least in its frame of reference.
So you can set a trajectory that skims the surface of a planet (without an atmosphere, or outside it), and you'll only lose the speed you gained while falling in.
The slingshot maneuver uses the fact that the planet's "stationary" frame of reference isn't the same as the sun. So you approach the planet from "behind" in its orbital track, then leave roughly perpendicular to its orbital track.
If the planet was moving 13 km/s and you approached it from the side at 20 km/s.
In the planet's perspective you are moving sqrt(20^2 + 13^2) = 23.9 km/s.
You then leave the planet say, parallel to the orbital track in the same direction. You leave at 23.9 km/s.
But in the sun's frame of reference, you approached the planet at 20km/s and left at (13+23.9) = 36.9 km/s, a massive speed boost.
answered Jan 29 at 20:55
YakkYakk
25016
$endgroup$
add a comment |
$begingroup$
Your problem appears to be too much science fiction and not enough science.
In science fiction, gravity sucks things towards the planet, sort of like a magnet. When far away you don't feel it; then you get close enough, and you are pulled down to the surface. And it holds things near the planet like glue or an elastic band.
In this science fiction model, stuff in space floats because it is far away. You go to space, you float.
In the real world, the science model, in space you fall just like you do in atmosphere. Orbit isn't far away, it is fast away -- things orbit the planet because they are going extremely fast.
Things in (low) orbit are amazingly close to the planet, from a distance they are almost skimming the surface. They are just moving so fast that the gravity of the planet pulling it down isn't fast enough; they "miss" the ground on each orbit.
Gravity doesn't stick or yank; gravity accelerates. When you go down a gravity well, you gain speed. This speed is mostly towards the planet, but it takes relative precision to actually hit the planet; even a modest velocity "horizontal" to the gravity well will make you miss.
And when you are going away from the planet, gravity doesn't "stick", it decelerates. The amount it decelerates on the way out is the same as it accelerated on the way in, at least in its frame of reference.
So you can set a trajectory that skims the surface of a planet (without an atmosphere, or outside it), and you'll only lose the speed you gained while falling in.
The slingshot maneuver uses the fact that the planet's "stationary" frame of reference isn't the same as the sun. So you approach the planet from "behind" in its orbital track, then leave roughly perpendicular to its orbital track.
If the planet was moving 13 km/s and you approached it from the side at 20 km/s.
In the planet's perspective you are moving sqrt(20^2 + 13^2) = 23.9 km/s.
You then leave the planet say, parallel to the orbital track in the same direction. You leave at 23.9 km/s.
But in the sun's frame of reference, you approached the planet at 20km/s and left at (13+23.9) = 36.9 km/s, a massive speed boost.
answered Jan 29 at 20:55
YakkYakk
25016
$endgroup$
Your problem appears to be too much science fiction and not enough science.
In science fiction, gravity sucks things towards the planet, sort of like a magnet. When far away you don't feel it; then you get close enough, and you are pulled down to the surface. And it holds things near the planet like glue or an elastic band.
In this science fiction model, stuff in space floats because it is far away. You go to space, you float.
In the real world, the science model, in space you fall just like you do in atmosphere. Orbit isn't far away, it is fast away -- things orbit the planet because they are going extremely fast.
Things in (low) orbit are amazingly close to the planet, from a distance they are almost skimming the surface. They are just moving so fast that the gravity of the planet pulling it down isn't fast enough; they "miss" the ground on each orbit.
Gravity doesn't stick or yank; gravity accelerates. When you go down a gravity well, you gain speed. This speed is mostly towards the planet, but it takes relative precision to actually hit the planet; even a modest velocity "horizontal" to the gravity well will make you miss.
And when you are going away from the planet, gravity doesn't "stick", it decelerates. The amount it decelerates on the way out is the same as it accelerated on the way in, at least in its frame of reference.
So you can set a trajectory that skims the surface of a planet (without an atmosphere, or outside it), and you'll only lose the speed you gained while falling in.
The slingshot maneuver uses the fact that the planet's "stationary" frame of reference isn't the same as the sun. So you approach the planet from "behind" in its orbital track, then leave roughly perpendicular to its orbital track.
If the planet was moving 13 km/s and you approached it from the side at 20 km/s.
In the planet's perspective you are moving sqrt(20^2 + 13^2) = 23.9 km/s.
You then leave the planet say, parallel to the orbital track in the same direction. You leave at 23.9 km/s.
But in the sun's frame of reference, you approached the planet at 20km/s and left at (13+23.9) = 36.9 km/s, a massive speed boost.
answered Jan 29 at 20:55
YakkYakk
25016
answered Jan 29 at 20:55
YakkYakk
25016
answered Jan 29 at 20:55
YakkYakk
25016
answered Jan 29 at 20:55
YakkYakk
25016
25016
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63
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I kind of feel like we should have some kind of a canonical "how does gravity work?" question that we could link to whenever these kinds of questions based on a fundamental misunderstanding of orbital mechanics come up. Unfortunately, searching for that phrase only turns up questions about zero-gravity cat litter boxes and other random stuff like that. :P
$endgroup$
– Ilmari Karonen
Jan 28 at 17:41
8
$begingroup$
That said, searching for "gravity assist" instead does turn up this question, which I highly recommend the OP should read.
$endgroup$
– Ilmari Karonen
Jan 28 at 17:43
4
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I have a question that branches off this one... How do you kind of crash into something????
$endgroup$
– Happy Koala
Jan 28 at 18:58
4
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Never mind these unhelpful/unkind comments above, it's a good question! You've taken the time to find out that escape velocity is important and added the equation (I converted it to MathJax for you). Your question has certainly attracted several great answers below. Welcome to Space!
$endgroup$
– uhoh
Jan 29 at 0:48
4
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@Baldrickk: In fact, we do have space.stackexchange.com/questions/22834/… and, more generally, space.stackexchange.com/questions/32776/…. But while playing KSP is a good way to learn basic orbital mechanics, it does require an investment of (at least) several hours before paying off. As much as it might improve question quality here, I don't think it would be fair or reasonable to tell new users to "come back after you've landed a kerbal on the Mun". :)
$endgroup$
– Ilmari Karonen
Jan 30 at 15:14