Is Earth an inertial reference frame?












6














Is earth considered as inertial frame? i was confused because we learned about coriolis effect. We know that earth spins therefore coriolis effect should take place . But does it have minimal effect for motion of balls etc when they move with respect to the ground?










share|cite|improve this question
























  • you can calculate it's size, e.g, for a BMG 50 cal (853 m/s muzzle velocity) shot at its max effective range (1,800m) and tell us.
    – JEB
    Dec 2 '18 at 0:05










  • This question seems synonymous with "Does the earth rotate (with respect to the rest of the universe)?"
    – Geremia
    Dec 2 '18 at 4:31






  • 2




    That the Earth's surface is not an inertial frame is nicely demonstrated by the Foucault pendulum.
    – Ruslan
    Dec 2 '18 at 15:51






  • 1




    It depends on the physics problem you look at whether the Earth can be considered an inertial frame of reference or not.
    – jjack
    Dec 2 '18 at 18:14










  • To emphasize @jjack's point, it's notoriously hard work to get a Foucault pendulum to work properly because the non-inertial effects, while present, are very small.
    – dmckee
    Dec 2 '18 at 19:28
















6














Is earth considered as inertial frame? i was confused because we learned about coriolis effect. We know that earth spins therefore coriolis effect should take place . But does it have minimal effect for motion of balls etc when they move with respect to the ground?










share|cite|improve this question
























  • you can calculate it's size, e.g, for a BMG 50 cal (853 m/s muzzle velocity) shot at its max effective range (1,800m) and tell us.
    – JEB
    Dec 2 '18 at 0:05










  • This question seems synonymous with "Does the earth rotate (with respect to the rest of the universe)?"
    – Geremia
    Dec 2 '18 at 4:31






  • 2




    That the Earth's surface is not an inertial frame is nicely demonstrated by the Foucault pendulum.
    – Ruslan
    Dec 2 '18 at 15:51






  • 1




    It depends on the physics problem you look at whether the Earth can be considered an inertial frame of reference or not.
    – jjack
    Dec 2 '18 at 18:14










  • To emphasize @jjack's point, it's notoriously hard work to get a Foucault pendulum to work properly because the non-inertial effects, while present, are very small.
    – dmckee
    Dec 2 '18 at 19:28














6












6








6


1





Is earth considered as inertial frame? i was confused because we learned about coriolis effect. We know that earth spins therefore coriolis effect should take place . But does it have minimal effect for motion of balls etc when they move with respect to the ground?










share|cite|improve this question















Is earth considered as inertial frame? i was confused because we learned about coriolis effect. We know that earth spins therefore coriolis effect should take place . But does it have minimal effect for motion of balls etc when they move with respect to the ground?







newtonian-mechanics reference-frames inertial-frames earth coriolis-effect






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 2 '18 at 2:25









Qmechanic

101k121831151




101k121831151










asked Dec 1 '18 at 23:46









ado sar

30910




30910












  • you can calculate it's size, e.g, for a BMG 50 cal (853 m/s muzzle velocity) shot at its max effective range (1,800m) and tell us.
    – JEB
    Dec 2 '18 at 0:05










  • This question seems synonymous with "Does the earth rotate (with respect to the rest of the universe)?"
    – Geremia
    Dec 2 '18 at 4:31






  • 2




    That the Earth's surface is not an inertial frame is nicely demonstrated by the Foucault pendulum.
    – Ruslan
    Dec 2 '18 at 15:51






  • 1




    It depends on the physics problem you look at whether the Earth can be considered an inertial frame of reference or not.
    – jjack
    Dec 2 '18 at 18:14










  • To emphasize @jjack's point, it's notoriously hard work to get a Foucault pendulum to work properly because the non-inertial effects, while present, are very small.
    – dmckee
    Dec 2 '18 at 19:28


















  • you can calculate it's size, e.g, for a BMG 50 cal (853 m/s muzzle velocity) shot at its max effective range (1,800m) and tell us.
    – JEB
    Dec 2 '18 at 0:05










  • This question seems synonymous with "Does the earth rotate (with respect to the rest of the universe)?"
    – Geremia
    Dec 2 '18 at 4:31






  • 2




    That the Earth's surface is not an inertial frame is nicely demonstrated by the Foucault pendulum.
    – Ruslan
    Dec 2 '18 at 15:51






  • 1




    It depends on the physics problem you look at whether the Earth can be considered an inertial frame of reference or not.
    – jjack
    Dec 2 '18 at 18:14










  • To emphasize @jjack's point, it's notoriously hard work to get a Foucault pendulum to work properly because the non-inertial effects, while present, are very small.
    – dmckee
    Dec 2 '18 at 19:28
















you can calculate it's size, e.g, for a BMG 50 cal (853 m/s muzzle velocity) shot at its max effective range (1,800m) and tell us.
– JEB
Dec 2 '18 at 0:05




you can calculate it's size, e.g, for a BMG 50 cal (853 m/s muzzle velocity) shot at its max effective range (1,800m) and tell us.
– JEB
Dec 2 '18 at 0:05












This question seems synonymous with "Does the earth rotate (with respect to the rest of the universe)?"
– Geremia
Dec 2 '18 at 4:31




This question seems synonymous with "Does the earth rotate (with respect to the rest of the universe)?"
– Geremia
Dec 2 '18 at 4:31




2




2




That the Earth's surface is not an inertial frame is nicely demonstrated by the Foucault pendulum.
– Ruslan
Dec 2 '18 at 15:51




That the Earth's surface is not an inertial frame is nicely demonstrated by the Foucault pendulum.
– Ruslan
Dec 2 '18 at 15:51




1




1




It depends on the physics problem you look at whether the Earth can be considered an inertial frame of reference or not.
– jjack
Dec 2 '18 at 18:14




It depends on the physics problem you look at whether the Earth can be considered an inertial frame of reference or not.
– jjack
Dec 2 '18 at 18:14












To emphasize @jjack's point, it's notoriously hard work to get a Foucault pendulum to work properly because the non-inertial effects, while present, are very small.
– dmckee
Dec 2 '18 at 19:28




To emphasize @jjack's point, it's notoriously hard work to get a Foucault pendulum to work properly because the non-inertial effects, while present, are very small.
– dmckee
Dec 2 '18 at 19:28










4 Answers
4






active

oldest

votes


















12














The surface of the Earth is not, rigorously speaking, an inertial frame of reference. Objects at rest relative to Earth's surface are actually subject to a series of inertial effects, like the ficticious forces (Coriolis, centrifugal etc.) because of Earth's rotation, precession and other kinds of acceleration.



When solving physics problems, however, we usually take the Earth frame as being inertial. This is because the inertial effects are minuscule for most of our day-to-day experiences and experiments. For example, objects in the Equator are the ones subject to the strongest centrifugal force and it is only about $3 times10^{-3}$ or $0.3%$ of their weight.



So for the most part, if an experiment is short enough and happens in a small enough region, the surface of Earth can indeed be approximated to an inertial frame of reference since the effects on the experiment's results are very, very tiny.



This of course has exceptions, as cited in njspeer's answer.



If however by "Earth" you mean the reference frame in Earth's center, it is an inertial frame according to General Relativity (GR), since observers in free fall are inertial in GR. The Earth actually does have some proper acceleration due to external forces such as radiation pressure, but these are also minuscule effects.






share|cite|improve this answer



















  • 2




    The earth is in free-fall, so its acceleration due to the sun's gravity (or the Milky Way's) does not cause the earth-centered frame to be non-inertial. (Except for even more miniscule effects, e.g. solar radiation pressure.)
    – David
    Dec 2 '18 at 8:23












  • I wasn't really sure about this. I mean, it would be considered inertial in GR, but I was using a more classical idea. I'm editing the post to address this issue. Thank you.
    – João Vítor G. Lima
    Dec 2 '18 at 12:52



















9














Because the earth is rotating, it is never strictly an inertial reference frame. However, because the effects are small in many situations, it can often be approximated as one. When to use Coriolis forces will have to be determined on a case-by-case basis. E.g. ballistic problems that cover large distances will most certainly require Coriolis-force corrections, and pendulums that swings for a long time would also require Coriolis-force corrections. For a block sliding down an inclined plane, or a spring on a mass, or a vibrating string, you should not need to take it into consideration.






share|cite|improve this answer































    3














    Mach would say that non-inertial effects are due to the relative motion between the earth and the rest of the universe.



    See:
    • Assis, André K. T. Relational Mechanics and Implementation of Mach’s Principle with Weber’s Gravitational Force. Montréal: Apeiron, 2014.






    share|cite|improve this answer































      0














      I know it's being a little pedantic, but I would say "earth" is a thing, not a reference frame. You could define an inertial reference frame that contains the earth.



      But suppose you mean a reference frame defined with reference to the earth: Z is perpendicular to the ground where you are standing, X and Y are parallel to the ground and perpendicular to each other.



      If you stand at rest in this reference frame, you may notice you feel acceleration (gravity). It is therefore not an inertial reference frame.






      share|cite|improve this answer





















        Your Answer





        StackExchange.ifUsing("editor", function () {
        return StackExchange.using("mathjaxEditing", function () {
        StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
        StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
        });
        });
        }, "mathjax-editing");

        StackExchange.ready(function() {
        var channelOptions = {
        tags: "".split(" "),
        id: "151"
        };
        initTagRenderer("".split(" "), "".split(" "), channelOptions);

        StackExchange.using("externalEditor", function() {
        // Have to fire editor after snippets, if snippets enabled
        if (StackExchange.settings.snippets.snippetsEnabled) {
        StackExchange.using("snippets", function() {
        createEditor();
        });
        }
        else {
        createEditor();
        }
        });

        function createEditor() {
        StackExchange.prepareEditor({
        heartbeatType: 'answer',
        autoActivateHeartbeat: false,
        convertImagesToLinks: false,
        noModals: true,
        showLowRepImageUploadWarning: true,
        reputationToPostImages: null,
        bindNavPrevention: true,
        postfix: "",
        imageUploader: {
        brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
        contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
        allowUrls: true
        },
        noCode: true, onDemand: true,
        discardSelector: ".discard-answer"
        ,immediatelyShowMarkdownHelp:true
        });


        }
        });














        draft saved

        draft discarded


















        StackExchange.ready(
        function () {
        StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f444580%2fis-earth-an-inertial-reference-frame%23new-answer', 'question_page');
        }
        );

        Post as a guest















        Required, but never shown

























        4 Answers
        4






        active

        oldest

        votes








        4 Answers
        4






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        12














        The surface of the Earth is not, rigorously speaking, an inertial frame of reference. Objects at rest relative to Earth's surface are actually subject to a series of inertial effects, like the ficticious forces (Coriolis, centrifugal etc.) because of Earth's rotation, precession and other kinds of acceleration.



        When solving physics problems, however, we usually take the Earth frame as being inertial. This is because the inertial effects are minuscule for most of our day-to-day experiences and experiments. For example, objects in the Equator are the ones subject to the strongest centrifugal force and it is only about $3 times10^{-3}$ or $0.3%$ of their weight.



        So for the most part, if an experiment is short enough and happens in a small enough region, the surface of Earth can indeed be approximated to an inertial frame of reference since the effects on the experiment's results are very, very tiny.



        This of course has exceptions, as cited in njspeer's answer.



        If however by "Earth" you mean the reference frame in Earth's center, it is an inertial frame according to General Relativity (GR), since observers in free fall are inertial in GR. The Earth actually does have some proper acceleration due to external forces such as radiation pressure, but these are also minuscule effects.






        share|cite|improve this answer



















        • 2




          The earth is in free-fall, so its acceleration due to the sun's gravity (or the Milky Way's) does not cause the earth-centered frame to be non-inertial. (Except for even more miniscule effects, e.g. solar radiation pressure.)
          – David
          Dec 2 '18 at 8:23












        • I wasn't really sure about this. I mean, it would be considered inertial in GR, but I was using a more classical idea. I'm editing the post to address this issue. Thank you.
          – João Vítor G. Lima
          Dec 2 '18 at 12:52
















        12














        The surface of the Earth is not, rigorously speaking, an inertial frame of reference. Objects at rest relative to Earth's surface are actually subject to a series of inertial effects, like the ficticious forces (Coriolis, centrifugal etc.) because of Earth's rotation, precession and other kinds of acceleration.



        When solving physics problems, however, we usually take the Earth frame as being inertial. This is because the inertial effects are minuscule for most of our day-to-day experiences and experiments. For example, objects in the Equator are the ones subject to the strongest centrifugal force and it is only about $3 times10^{-3}$ or $0.3%$ of their weight.



        So for the most part, if an experiment is short enough and happens in a small enough region, the surface of Earth can indeed be approximated to an inertial frame of reference since the effects on the experiment's results are very, very tiny.



        This of course has exceptions, as cited in njspeer's answer.



        If however by "Earth" you mean the reference frame in Earth's center, it is an inertial frame according to General Relativity (GR), since observers in free fall are inertial in GR. The Earth actually does have some proper acceleration due to external forces such as radiation pressure, but these are also minuscule effects.






        share|cite|improve this answer



















        • 2




          The earth is in free-fall, so its acceleration due to the sun's gravity (or the Milky Way's) does not cause the earth-centered frame to be non-inertial. (Except for even more miniscule effects, e.g. solar radiation pressure.)
          – David
          Dec 2 '18 at 8:23












        • I wasn't really sure about this. I mean, it would be considered inertial in GR, but I was using a more classical idea. I'm editing the post to address this issue. Thank you.
          – João Vítor G. Lima
          Dec 2 '18 at 12:52














        12












        12








        12






        The surface of the Earth is not, rigorously speaking, an inertial frame of reference. Objects at rest relative to Earth's surface are actually subject to a series of inertial effects, like the ficticious forces (Coriolis, centrifugal etc.) because of Earth's rotation, precession and other kinds of acceleration.



        When solving physics problems, however, we usually take the Earth frame as being inertial. This is because the inertial effects are minuscule for most of our day-to-day experiences and experiments. For example, objects in the Equator are the ones subject to the strongest centrifugal force and it is only about $3 times10^{-3}$ or $0.3%$ of their weight.



        So for the most part, if an experiment is short enough and happens in a small enough region, the surface of Earth can indeed be approximated to an inertial frame of reference since the effects on the experiment's results are very, very tiny.



        This of course has exceptions, as cited in njspeer's answer.



        If however by "Earth" you mean the reference frame in Earth's center, it is an inertial frame according to General Relativity (GR), since observers in free fall are inertial in GR. The Earth actually does have some proper acceleration due to external forces such as radiation pressure, but these are also minuscule effects.






        share|cite|improve this answer














        The surface of the Earth is not, rigorously speaking, an inertial frame of reference. Objects at rest relative to Earth's surface are actually subject to a series of inertial effects, like the ficticious forces (Coriolis, centrifugal etc.) because of Earth's rotation, precession and other kinds of acceleration.



        When solving physics problems, however, we usually take the Earth frame as being inertial. This is because the inertial effects are minuscule for most of our day-to-day experiences and experiments. For example, objects in the Equator are the ones subject to the strongest centrifugal force and it is only about $3 times10^{-3}$ or $0.3%$ of their weight.



        So for the most part, if an experiment is short enough and happens in a small enough region, the surface of Earth can indeed be approximated to an inertial frame of reference since the effects on the experiment's results are very, very tiny.



        This of course has exceptions, as cited in njspeer's answer.



        If however by "Earth" you mean the reference frame in Earth's center, it is an inertial frame according to General Relativity (GR), since observers in free fall are inertial in GR. The Earth actually does have some proper acceleration due to external forces such as radiation pressure, but these are also minuscule effects.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 2 '18 at 12:54

























        answered Dec 2 '18 at 0:10









        João Vítor G. Lima

        861319




        861319








        • 2




          The earth is in free-fall, so its acceleration due to the sun's gravity (or the Milky Way's) does not cause the earth-centered frame to be non-inertial. (Except for even more miniscule effects, e.g. solar radiation pressure.)
          – David
          Dec 2 '18 at 8:23












        • I wasn't really sure about this. I mean, it would be considered inertial in GR, but I was using a more classical idea. I'm editing the post to address this issue. Thank you.
          – João Vítor G. Lima
          Dec 2 '18 at 12:52














        • 2




          The earth is in free-fall, so its acceleration due to the sun's gravity (or the Milky Way's) does not cause the earth-centered frame to be non-inertial. (Except for even more miniscule effects, e.g. solar radiation pressure.)
          – David
          Dec 2 '18 at 8:23












        • I wasn't really sure about this. I mean, it would be considered inertial in GR, but I was using a more classical idea. I'm editing the post to address this issue. Thank you.
          – João Vítor G. Lima
          Dec 2 '18 at 12:52








        2




        2




        The earth is in free-fall, so its acceleration due to the sun's gravity (or the Milky Way's) does not cause the earth-centered frame to be non-inertial. (Except for even more miniscule effects, e.g. solar radiation pressure.)
        – David
        Dec 2 '18 at 8:23






        The earth is in free-fall, so its acceleration due to the sun's gravity (or the Milky Way's) does not cause the earth-centered frame to be non-inertial. (Except for even more miniscule effects, e.g. solar radiation pressure.)
        – David
        Dec 2 '18 at 8:23














        I wasn't really sure about this. I mean, it would be considered inertial in GR, but I was using a more classical idea. I'm editing the post to address this issue. Thank you.
        – João Vítor G. Lima
        Dec 2 '18 at 12:52




        I wasn't really sure about this. I mean, it would be considered inertial in GR, but I was using a more classical idea. I'm editing the post to address this issue. Thank you.
        – João Vítor G. Lima
        Dec 2 '18 at 12:52











        9














        Because the earth is rotating, it is never strictly an inertial reference frame. However, because the effects are small in many situations, it can often be approximated as one. When to use Coriolis forces will have to be determined on a case-by-case basis. E.g. ballistic problems that cover large distances will most certainly require Coriolis-force corrections, and pendulums that swings for a long time would also require Coriolis-force corrections. For a block sliding down an inclined plane, or a spring on a mass, or a vibrating string, you should not need to take it into consideration.






        share|cite|improve this answer




























          9














          Because the earth is rotating, it is never strictly an inertial reference frame. However, because the effects are small in many situations, it can often be approximated as one. When to use Coriolis forces will have to be determined on a case-by-case basis. E.g. ballistic problems that cover large distances will most certainly require Coriolis-force corrections, and pendulums that swings for a long time would also require Coriolis-force corrections. For a block sliding down an inclined plane, or a spring on a mass, or a vibrating string, you should not need to take it into consideration.






          share|cite|improve this answer


























            9












            9








            9






            Because the earth is rotating, it is never strictly an inertial reference frame. However, because the effects are small in many situations, it can often be approximated as one. When to use Coriolis forces will have to be determined on a case-by-case basis. E.g. ballistic problems that cover large distances will most certainly require Coriolis-force corrections, and pendulums that swings for a long time would also require Coriolis-force corrections. For a block sliding down an inclined plane, or a spring on a mass, or a vibrating string, you should not need to take it into consideration.






            share|cite|improve this answer














            Because the earth is rotating, it is never strictly an inertial reference frame. However, because the effects are small in many situations, it can often be approximated as one. When to use Coriolis forces will have to be determined on a case-by-case basis. E.g. ballistic problems that cover large distances will most certainly require Coriolis-force corrections, and pendulums that swings for a long time would also require Coriolis-force corrections. For a block sliding down an inclined plane, or a spring on a mass, or a vibrating string, you should not need to take it into consideration.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 2 '18 at 3:20

























            answered Dec 2 '18 at 0:01









            njspeer

            4374




            4374























                3














                Mach would say that non-inertial effects are due to the relative motion between the earth and the rest of the universe.



                See:
                • Assis, André K. T. Relational Mechanics and Implementation of Mach’s Principle with Weber’s Gravitational Force. Montréal: Apeiron, 2014.






                share|cite|improve this answer




























                  3














                  Mach would say that non-inertial effects are due to the relative motion between the earth and the rest of the universe.



                  See:
                  • Assis, André K. T. Relational Mechanics and Implementation of Mach’s Principle with Weber’s Gravitational Force. Montréal: Apeiron, 2014.






                  share|cite|improve this answer


























                    3












                    3








                    3






                    Mach would say that non-inertial effects are due to the relative motion between the earth and the rest of the universe.



                    See:
                    • Assis, André K. T. Relational Mechanics and Implementation of Mach’s Principle with Weber’s Gravitational Force. Montréal: Apeiron, 2014.






                    share|cite|improve this answer














                    Mach would say that non-inertial effects are due to the relative motion between the earth and the rest of the universe.



                    See:
                    • Assis, André K. T. Relational Mechanics and Implementation of Mach’s Principle with Weber’s Gravitational Force. Montréal: Apeiron, 2014.







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Dec 4 '18 at 16:46

























                    answered Dec 2 '18 at 4:35









                    Geremia

                    1,1802927




                    1,1802927























                        0














                        I know it's being a little pedantic, but I would say "earth" is a thing, not a reference frame. You could define an inertial reference frame that contains the earth.



                        But suppose you mean a reference frame defined with reference to the earth: Z is perpendicular to the ground where you are standing, X and Y are parallel to the ground and perpendicular to each other.



                        If you stand at rest in this reference frame, you may notice you feel acceleration (gravity). It is therefore not an inertial reference frame.






                        share|cite|improve this answer


























                          0














                          I know it's being a little pedantic, but I would say "earth" is a thing, not a reference frame. You could define an inertial reference frame that contains the earth.



                          But suppose you mean a reference frame defined with reference to the earth: Z is perpendicular to the ground where you are standing, X and Y are parallel to the ground and perpendicular to each other.



                          If you stand at rest in this reference frame, you may notice you feel acceleration (gravity). It is therefore not an inertial reference frame.






                          share|cite|improve this answer
























                            0












                            0








                            0






                            I know it's being a little pedantic, but I would say "earth" is a thing, not a reference frame. You could define an inertial reference frame that contains the earth.



                            But suppose you mean a reference frame defined with reference to the earth: Z is perpendicular to the ground where you are standing, X and Y are parallel to the ground and perpendicular to each other.



                            If you stand at rest in this reference frame, you may notice you feel acceleration (gravity). It is therefore not an inertial reference frame.






                            share|cite|improve this answer












                            I know it's being a little pedantic, but I would say "earth" is a thing, not a reference frame. You could define an inertial reference frame that contains the earth.



                            But suppose you mean a reference frame defined with reference to the earth: Z is perpendicular to the ground where you are standing, X and Y are parallel to the ground and perpendicular to each other.



                            If you stand at rest in this reference frame, you may notice you feel acceleration (gravity). It is therefore not an inertial reference frame.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Dec 2 '18 at 15:51









                            Owen

                            27729




                            27729






























                                draft saved

                                draft discarded




















































                                Thanks for contributing an answer to Physics Stack Exchange!


                                • Please be sure to answer the question. Provide details and share your research!

                                But avoid



                                • Asking for help, clarification, or responding to other answers.

                                • Making statements based on opinion; back them up with references or personal experience.


                                Use MathJax to format equations. MathJax reference.


                                To learn more, see our tips on writing great answers.





                                Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


                                Please pay close attention to the following guidance:


                                • Please be sure to answer the question. Provide details and share your research!

                                But avoid



                                • Asking for help, clarification, or responding to other answers.

                                • Making statements based on opinion; back them up with references or personal experience.


                                To learn more, see our tips on writing great answers.




                                draft saved


                                draft discarded














                                StackExchange.ready(
                                function () {
                                StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f444580%2fis-earth-an-inertial-reference-frame%23new-answer', 'question_page');
                                }
                                );

                                Post as a guest















                                Required, but never shown





















































                                Required, but never shown














                                Required, but never shown












                                Required, but never shown







                                Required, but never shown

































                                Required, but never shown














                                Required, but never shown












                                Required, but never shown







                                Required, but never shown







                                Popular posts from this blog

                                Biblatex bibliography style without URLs when DOI exists (in Overleaf with Zotero bibliography)

                                ComboBox Display Member on multiple fields

                                Is it possible to collect Nectar points via Trainline?