PowerShell, 3 bytes, score 3637

2e4

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Very naive, boring, implementation; just returns 20000 no matter the input. Experimentation with things like special-casing the sun or using floating-point values instead of 2 all resulted in worse scores because the length of code increased enough to offset any size-comparison gains.

Jelly, 34 bytes, score = 37

OḌ“⁸|5/!‘%ƒị“RNFLOJMjs⁽u[USJ‘1.1*

Input is in uppercase, output is the power of 1.1 with the least error.

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How it works

OḌ“⁸|5/!‘%ƒị“RNFLOJMjs⁽u[USJ‘1.1*  Main link. Argument: s (string)



O                                   Ordinal; map the char in s to their code points.

                                        "ERIS" -> [69,82,73,83]

 Ḍ                                  Undecimal; treat the result as an array of digits

                                    in base 10 and convert it to integer.

                                        [69,82,73,83] -> 69000+8200+730+83 = 78013

  “⁸|5/!‘                           Literal; yield [136, 124, 53, 47, 33].

         %ƒ                         Fold the array by modulus, using the computed

                                    integer as initial value.

                                        78013 -> 78013%136%124%53%47%33 = 32

            “RNFLOJMjs⁽u[USJ‘      Literal; yield [82, 78, 70, 76, 79, 74, 77, ...

                                    106, 115, 141, 92, 117, 91, 85, 83, 74].

           ị                        Retrieve the element from the array to the right,

                                    at the index to the left.

                                    Indexing is 1-based and modular.

                                        32 = 16 (mod 16) -> 'J' = 74

                              1.1*  Raise 1.1 to the computed power.

                                        74 = 1.1**74 = 1156.268519450066

Java (JDK), 90 bytes, score = 97

s->("ýCĄ (ᬺ!˂Fɍ".charAt(s.substring(2).chars().sum()%96%49%25)-7)*100

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• This entry uses both undisplayable and multi-byte Unicode characters (but Java accepts them nevertheless). Check the TIO for the accurate code.


• The input must be title-case.


• This code rounds the values to the best multiple-of-100 (sometimes up, sometimes down) so that the last two digits can be skipped when encoded, and the value can then be approximated by multiplying by 100.


• This entry uses various hashes to fit a 25 codepoints string (shortest string I could find).

Credits

• -48 score (-45 bytes) thanks to Kevin Cruijssen by encoding the radiuses (divided by 100) directly in a String instead of hardcoding them in an explicit int array..

Wolfram Language 114 103 97 88 86 82 bytes. score = 114 103 97 89 87 83 points

(#&@@EntityValue[Interpreter["AstronomicalObject"]@#,"Radius"]/._String->507)1.61&

At least 6 points saved thanks to Dennis, several more thanks to lirtosiast, and 6 more thanks to user202729.

Although Mathematica can fetch solar system data (as well as much additional astronomical data), some minor tweaks are needed, as explained below.

Interpreter[#,"AstronomicalObject"]& will return the entity (i.e. the machine computable object) associated with the term represented by #.

EntityValue[AstronomicalObject,"Radius"] returns the radius, in miles, of the entity. In the case of "Haumea", the value, 816.27 (i.e. 507*1.61), is returned.

Multiplication of the radius by 1.61 converts from miles to km. Decimal values, rather than integers, account for much less than 1% error, even in the most extreme case.

[[1]] returns the magnitude without the unit, km. This was later changed to #&@@, yielding the same result.

Python 3, score 95, 95 bytes

lambda n:ord("ؙҢ򪀖ਏ𑄗ാᣣ४ঈ挒ឤ?̰ҋ??ۉՉ怮ܞ੊̔"[int(n,35)%87%52%24-1])

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Python 3, score 133, 133 bytes

lambda n:int(f'00e0{10**18+10**6}10x1h2411j4?00??811i1207wazxmwuvko?mw??xc1ze1ldyujz6zysi4?ob??k9lym6w'[int(n,35)%87%52%24-1::23],36)

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Powershell, 150 bytes, score 163

($args|% t*y|?{'Su680J68S57U25N24Ea6V6Ma3.3G2.6Ti2.5Me2.4C2.4I1.8M1.7Eu1.5T1.3P1.2E1.1H.8Titani.8'-cmatch"$(($y+=$_))([d.]+)"}|%{1kb*$Matches.1})[-1]

Test script:

$f = {

($args|% t*y|?{'Su680J68S57U25N24Ea6V6Ma3.3G2.6Ti2.5Me2.4C2.4I1.8M1.7Eu1.5T1.3P1.2E1.1H.8Titani.8'-cmatch"$(($y+=$_))([d.]+)"}|%{1kb*$Matches.1})[-1]

}



$penalty = @(

    ,("Sun"      , 696342)

    ,("Jupiter"  ,  69911)

    ,("Saturn"   ,  58232)

    ,("Uranus"   ,  25362)

    ,("Neptune"  ,  24622)

    ,("Earth"    ,   6371)

    ,("Venus"    ,   6052)

    ,("Mars"     ,   3390)

    ,("Ganymede" ,   2634)

    ,("Titan"    ,   2575)

    ,("Mercury"  ,   2440)

    ,("Callisto" ,   2410)

    ,("Io"       ,   1822)

    ,("Moon"     ,   1737)

    ,("Europa"   ,   1561)

    ,("Triton"   ,   1353)

    ,("Pluto"    ,   1186)

    ,("Eris"     ,   1163)

    ,("Haumea"   ,    816)

    ,("Titania"  ,    788)

) | % {

    $s,$expected = $_

    $result = &$f $s

    $ratio = [Math]::Max($result/$expected, $expected/$result)

    $ratio*$ratio

}

$scriptLength = $f.ToString().Length - 2  # -4 if CRLF mode

$penaltyMax = ($penalty|Measure-Object -Maximum).Maximum

$score = $scriptLength * $penaltyMax

"$score = $scriptLength * $penaltyMax"

Output:

162.113324228916 = 150 * 1.08075549485944

Explanation:

• Names contain letters only, radiuses contain digits and dots. So we can write all the data in a data string and perform a regexp search.


• The script searches for all substrings from left to right and takes the last result found.


• The input must be title-case to reduce the data string.


• The end of line mode is LF only.

Example:

Titania         Triton         Titan

--------------  -------------  -------------

T       -> 1.3  T      -> 1.3  T      -> 1.3

Ti      -> 2.5  Tr     ->      Ti     -> 2.5

Tit     ->      Tri    ->      Tit    ->

Tita    ->      Trit   ->      Tita   ->

Titan   ->      Triton ->      Titan  ->

Titani  -> .8

Titania ->



Result is .8    Result is 1.3  Result is 2.5

Powershell, 178 bytes, score 178

($args|% t*y|?{'Su696342J69911S58232U25362N24622Ea6371V6052Ma3390G2634Ti2575Me2440C2410I1822M1737Eu1561T1353P1186E1163H816Titani788'-cmatch"$(($y+=$_))(d+)"}|%{+$Matches.1})[-1]

05AB1E, score 100 66 60 (100 61 56 bytes)

•1∞²îc|I‰∍T‡sÇ3¡ò½в…»Ë•§•1ë£ñƒq£û¿’…•S£y¦¦ÇO96%49%25%èт*

Port of @OlivierGrégoire's Java answer, so if you like this first answer, make sure to upvote him as well!

Input in titlecase.

Verify all test cases.

05AB1E, score 100 (100 bytes)

•*Òâ%ÌÜS…Ùb‹Úi{e!]ɸ·vÌBUSηHã£āðxyµŠ•§•3«8¹ØмS7Ç•S£.•WùηƵ@,Sº,ûεβʒóÃX¹Θäáá’Ý)”Ωož∞-z.A±D•3ôI2£Iθ«kè

Input in full lowercase. Outputs the exact radius, so no penalty is added.

Verify all test cases.

Explanation:

•*Òâ%ÌÜS…Ùb‹Úi{e!]ɸ·vÌBUSηHã£āðxyµŠ•

                   # Compressed integer 696342699115823225362246226371605233902634257524402410182217371561135311861163816788

 §                 # Casted to string (bug, should have been implicitly..)

  •3«8¹ØмS7Ç•      # Compressed integer 65555444444444444433

   S               # Converted to a list of digits: [6,5,5,5,5,4,4,4,4,4,4,4,4,4,4,4,4,4,3,3]

    £              # The first integer is split into parts of that size: ["696342","69911","58232","25362","24622","6371","6052","3390","2634","2575","2440","2410","1822","1737","1561","1353","1186","1163","816","788"]

     .•WùηƵ@,Sº,ûεβʒóÃX¹Θäáá’Ý)”Ωož∞-z.A±D•

                   # Compressed string "sunjursanursneeeahvesmasgaetinmeycaoioomoneuatrnploershaatia"

      3ô           # Split into parts of size 3: ["sun","jur","san","urs","nee","eah","ves","mas","gae","tin","mey","cao","ioo","mon","eua","trn","plo","ers","haa","tia"]

        I2£        # The first two characters of the input

           Iθ      # The last character of the input

             «     # Merged together

              k    # Get the index of this string in the list of strings

               è   # And use that index to index into the list of integers

                   # (and output the result implicitly)

See this 05AB1E tip of mine (sections How to compress large integers? and How to compress strings not part of the dictionary?) to understand how the compression used works.

I did create a 70-bytes alternative which would map sun to 600,000; [jupiter,saturn] to 60,000; [uranus,neptune] to 30,000; [earth,venus] to 6,000; [mars,ganymede,titan,mercury,callisto] to 3,000; [io,moon,europa,triton,pluto,eris] to 1,500; and [haumea;titania] to 750. Unfortunately that got a score of 117. I will see if I can get below 100 with an alternative approach later.

Mathematica, 57 bytes, score = 62 58

^{-4 bytes/score thanks to lirtosiast!}

#&@@WolframAlpha[#<>" size km","Result"]]/._Missing->816&

Just does a Wolfram Alpha lookup for the mean radius.

Python 2, 155 bytes, score = 155

lambda p:int('G11KK54222111111XXNM8MCO37WQ53YXHE93V8BIF2IMH1WU9KPU2MLN    HGR'['uSJJaSrUNNrEnVsMeGtTMMoCoInMuErTuPsEaHTT'.find(p[7%len(p)]+p[0])/2::20],35)

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Surprisingly well for this lazy solution... will look into improving as well. ;-)

Japt, 86 bytes, score = 94

g5 ¥'i?788:[7*A³7*L6*LG²G²IIÉHÄDÑDÑCÑCÑGÄÄGÄGECC8]g`suj«a¨Ì¼và@ã/eÖô¶e©rp¤r`bU¯2)z)*L

Try it for all inputs, Calculate the score, or Check the highest error

Very similar to Olivier's original answer. Input is all lowercase.

After various improvements to the output values, the current highest error is Venus at just over 4%.

Explanation now that things are a bit more stable:

¤¥`Éa`?                             :If the fifth character of the input is 'i':

       788                          : Output 788.

          :                         :Otherwise:

           [...]                    : From the array representing radii

                g                   : Get the value at the index:

                 `...`              :  In the string representing names

                      b             :  Find the first index where this string appears:

                       U¯2)         :   The first two characters of the input

                           z)       :  And divide it by two

                             *L     : Multiply that value by 100

The string for the names is sujusaurneeavemagatimecaiomoeutrplerha compressed using Japt's built-in compression. The numbers representing the radii are calculated like so:

                          My value | Actual value

                          ---------+-------------

7 * 10 ^ 3 = 7000 * 100 =   700000 | 696342

7 * 100    = 700  * 100 =    70000 |  69911

6 * 100    = 600  * 100 =    60000 |  58232

16 * 16    = 256  * 100 =    25600 |  25362

16 * 16    = 256  * 100 =    25600 |  24622

64         = 64   * 100 =     6400 |   6371

64 - 1     = 63   * 100 =     6300 |   6052

32 + 1     = 33   * 100 =     3300 |   3390

13 * 2     = 26   * 100 =     2600 |   2634

13 * 2     = 26   * 100 =     2600 |   2575

12 * 2     = 24   * 100 =     2400 |   2440

12 * 2     = 24   * 100 =     2400 |   2410

16 + 1 + 1 = 18   * 100 =     1800 |   1822

16 + 1     = 17   * 100 =     1700 |   1737

16         = 16   * 100 =     1600 |   1561

14         = 14   * 100 =     1400 |   1353

12         = 12   * 100 =     1200 |   1186

12         = 12   * 100 =     1200 |   1163

8          = 8    * 100 =      800 |    816

788                     =      788 |    788

Japt, 77 76 75 bytes, score = 75

First pass at this; I wanted to try a 0 penalty solution to give myself a baseline to work off. Will come back to it tomorrow to see what improvements can be made, hopefully still for 0 penalty.

Input is case-insensitive.

n35 %87%52 g"..."ò)mc

Try it or test all inputs

The "..." represents a string containing many unprintables. The codepoints are:

32,32,15,61,11,86,696,342,25,75,699,11,33,90,63,71,24,10,24,40,253,62,60,52,32,32,8,16,11,63,32,32,32,32,58,232,17,37,135,3,246,22,18,22,26,34,7,88

To offer a quick explanation: the string gets split into chunks of 2 characters. We then index into that array using part of ovs' formula plus some index-wrapping and then map the 2 characters to their codepoints.

• Saved a byte/point thanks to ETH
  

54 bytes, score = 58

A port of Olivier's solution.

"ýCĄ (ᬺ!˂Fɍ"cU¤¬xc %96%49)-7 *L

Test all inputs

PowerShell, 203 bytes, score 203

param($a)if($a-eq'Titan'){2575;exit}(696342,69911,58232,25362,24622,6371,6052,3390,2634,2440,2410,1822,1737,1561,1353,1186,1163,816,788)["SuJuSaUrNeEaVeMaGaMeCaIoMoEuTrPlErHaTi".indexOf(-join$a[0..1])/2]

Try it online!

Very similar to Olivier's answer, now that I see it, but developed independently.

T-SQL, 203 bytes, score = 217

SELECT IIF(v='Titan',26,SUBSTRING(value,3,4))*100

FROM i,STRING_SPLIT('Ca24,Ea64,Er12,Eu16,Ga26,Ha8,Io18,Ju699,Ma34,Me24,Mo17,Ne246,Pl12,Sa582,Su6963,Ti8,Tr14,Ur254,Ve61',',')

WHERE LEFT(v,2)=LEFT(value,2)

Line breaks are for readability only.

Input is taken via pre-existing table i with varchar column v, per our IO standards.

Joins the input table to an in-memory table on the first two characters, and returns the remaining digits x100.

Treats "Titan" as a special case using IIF.

Ruby, 105 bytes, score 109

->n{7E5/('!)"0 r&zZ&1#}3Mfh-~~d@'[0,j=" =1&%)AM<I>2,-B#($D  7@".index((n[1,9].sum%50+34).chr)].sum-j*32)}

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If we divide 700000 by the radii, we get a sequence which increases reasonably linearly (though rather erratically). The increments in the table below can be approximated by the ASCII values of characters. The problem with this approach is it requires the input to be decoded to a value which orders the different names by size.

A minor issue is that the difference between Eris and Haumea is quite large. Three characters ~~d are required to encode this increment in ASCII only format. The planet-to-index string has two "ghost planet"spaces in it to pad the index.

700000/r    increment from previous

0.994774    

9.960407    8.965633

11.95806    1.997657

27.45612    15.49805

28.28129    0.825178

109.2987    81.0174

115.0598    5.761118

205.4106    90.3508

264.3667    58.95612

270.4241    6.057335

285.3861    14.96199

288.9386    3.552524

382.1855    93.24692

400.8877    18.70223

446.0871    45.19939

514.6652    68.57806

587.1349    72.46972

598.7463    11.61144

853.3603    254.6139

883.6827    30.32245

Jelly, 28 bytes, score = 31

“__ʋ7ṗ“RUu⁽NMsOSJj[FL‘ḥ1.1*

This uses a configurable hashing built-in that I added to Jelly at @lirtosiast's suggestion.

Input is in titlecase, output is the power of 1.1 with the least error.

Try it online!

How it works

This answer consists of merely two parts.

• First, “__ʋ7ṗ“RUu⁽NMsOSJj[FL‘ḥ uses the new built-in to map each of the 20 possible inputs to 15 different integers.


• Then, 1.1* elevates 1.1 to the computed power.

“__ʋ7ṗ“RUu⁽NMsOSJj[FL‘ is a literal; every non-quote character is replaced by it's 0-based index in Jelly's code page, yielding $[95, 95, 169, 55, 242], [82, 85, 117, 141, 78, 77, 92, 115, 79, 83, 74, 106, 91, 70, 76]$.

The hashing built-in ḥ first maps $[95, 95, 169, 55, 242]$ to an integer by incrementing each number, then treating the result as the bijective base-250 digits integer and adding $1$. This yields $376510639244$.

By halving and flooring this integer until the result is $0$, we get the sequence $[376510639244, 188255319622, 94127659811, 47063829905, dots, 5, 2, 1, 0]$, which has the forward differences $[188255319622, 94127659811, 47063829906, dots, 3, 1, 1]$.

Next, we generate 64 64-bit integers by applying SHAKE256-4096 to the string representation of the internal representation of ḥ's right argument, then chopping the resulting 4096 bits into 64 64-bit chunks.

ḥ now computes the dot product of the 39 differences and the first 39 generated 64-bit integers, modulo $2^{64}$. This yields an integer in $[0, 2^{64})$.

The list $[82, 85, 117, 141, 78, 77, 92, 115, 79, 83, 74, 106, 91, 70, 76]$ has length 15, so we multiply the generated integer by 15 and take the 64 higher bits of the result. This yields an integer in $[0, 15)$, which we use to index into the list.

To find the appropriate hash configuration, I've used a brute-forcer in C that is part of the Jelly repo.

Charcoal, 101 bytes, score = 101

Ｉ⍘§⪪“_″FJ⁼⦄bl≕)Ｔ‹#⊙xO-nη⁻À↓ζ↥ς§%Ｈ8H“ρj✳Hρl× S↶…|ＵＤ⎇ＬkfZ”³⌕⪪”@/rjmq_↙§Ｅ▶νＦ↨oº⁷÷Ｋ⁻eDH:_Tbk¦�”²⁺§θ⁰§θχγ

Try it online! Link is to verbose version of code. Explanation:

⁺§θ⁰§θχ

Take the 1st and 11th character (cyclically) of the input string and concatenate them.

⌕⪪”@/rjmq_↙§Ｅ▶νＦ↨oº⁷÷Ｋ⁻eDH:_Tbk¦�”²

Look them up in the string SuJiSrUuNtEEVVMrGnTTMcClIIMoEpToPPEiHeTa split into pairs of characters.

§⪪“_″FJ⁼⦄bl≕)Ｔ‹#⊙xO-nη⁻À↓ζ↥ς§%Ｈ8H“ρj✳Hρl× S↶…|ＵＤ⎇ＬkfZ”³

Split the string m.w'fv&J|"l|"e1 c& _c Ca ;e ;* 9a 9C 31 2; 0I .7 ,N ,7 (X (< into groups of three characters and take the corresponding group.

Ｉ⍘ ... γ

Decode the result as a base-95 number using the printable ASCII character set as the digits. Example: Io's 11th character is I, so we look up II and find it's the 13th largest object and its size is 31 which maps to 19 * 95 + 17 = 1822.

Swift 4, 225 bytes, score = 241

Probably golfable a bunch more (maybe in the "Ga-Me-Ca" area?), but Swift is not often used (for a reason, maybe.)

func b(i:String){print(i=="Titan" ?2575:["Su":6963,"Ju":699,"Sa":582,"Ur":253,"Ne":246,"Ea":63,"Ve":60,"Ma":33,"Ga":26,"Me":24,"Ca":24,"Io":18,"Mo":17,"Eu":16,"Tr":14,"Pl":12,"Er":12,"Ha":8,"Ti":8][String(i.prefix(2))]!*100)}

and ungolfed

func size(ofAstralObject object: String) {

  let objectToRadius = // Map size/100 of all objects to the first two chars

   ["Su":6963,

    "Ju":699,

    "Sa":582,

    "Ur":253,

    "Ne":246,

    "Ea":63,

    "Ve":60,

    "Ma":33,

    "Ga":26,

    "Me":24,

    "Ca":24,

    "Io":18,

    "Mo":17,

    "Eu":16,

    "Tr":14,

    "Pl":12,

    "Er":12,

    "Ha":8,

    "Ti":8] // Ti is Titania, while Titan is treated differently



  print(object == "Titan" ? 

    2575 : // If "Titan", print the exact size

    objectToRadius[String(i.prefix(2))]!*100 // get the size from the map and multiply by 100

  )

}

Try It Online!

I tried different "key sizes" for the map, but of course 1 has many clashes and using three chars doesn't give me i=="Titan" ?2575:'s 17 chars, since there's "Io" to manage (and it'll take more than 3 chars, I think).

JavaScript (ES6), 152 bytes, score = 163

Well, it's a pretty standard solution, but I enjoyed the challenge anyway!

s=>s=='Titan'?2575:[6963,699,582,254,246,64,60,34,26,24,24,18,17,16,14,12,12,8,8]["SuJuSaUrNeEaVeMaGaMeCaIoMoEuTrPlErHaTi".match(s[0]+s[1]).index/2]*100

My Score:

Max. penalty ratio = 1.07068 for Triton

Score = ceil(152 x 1.07068) = 163

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FALSE, 152 bytes, Score = 563

[911*.]^$0[~][1+^]#$$2=$4=8=||[2 0!]?$3=[764 0!]?$5=[$$69=86=|$[6]?~[2]?0!]?$6=[$$83=85=|$[46]?~[$72=$[1]?~[2]?]?0!]?$7=[$84=$[1]?~[52]?0!]?

Lazy answer using word lengths and first letters but my excuse is that I'm using a weird language

Try it online! (copy paste the code, hit show and then run)

[911*.]          {defines a function that multiplies a number by 911 and then prints it}

^$0[~][1+^]#    {counts the length of the name as it's input, also records the first char}

$$2=$4=8=||[1 0!]?  {if the name is 2, 4, or 8 chars long print 911*2 (0! calls the function)}

$3=[764 0!]?          {if name is 3 long print 911*764}

$5=[$$69=86=|$[6]?~[2]?0!]? {5 long? print 911*6 if it starts with E or V, otherwise *2}

$6=[$$83=85=|$[46]?~[       {6 long? print 911*46 if it starts with S or U, otherwise:}

    $72=$[1]?~[2]?            {if name starts with H print 911*1 else *2

]?0!]?

$7=[$84=$[1]?~[26]?0!]?      {7 long? print 1822*1 if it starts with NT otherwise *26 (for jupiter}

My results:

Sun       : 696004.00 penalty ratio = (696342.00 / 696004.00 )² = 1.00097

Jupiter   : 47372.00  penalty ratio = (69911.00  / 47372.00  )² = 2.17795

Saturn    : 41906.00  penalty ratio = (58232.00  / 41906.00  )² = 1.93095

Uranus    : 41906.00  penalty ratio = (41906.00  / 25362.00  )² = 2.73014

Neptune   : 47372.00  penalty ratio = (47372.00  / 24622.00  )² = 3.70166

Earth     : 5466.00   penalty ratio = (6371.00   / 5466.00   )² = 1.35855

Venus     : 5466.00   penalty ratio = (6052.00   / 5466.00   )² = 1.22591

Mars      : 1822.00   penalty ratio = (3390.00   / 1822.00   )² = 3.46181

Ganymede  : 1822.00   penalty ratio = (2634.00   / 1822.00   )² = 2.08994

Titan     : 1822.00   penalty ratio = (2575.00   / 1822.00   )² = 1.99737

Mercury   : 1822.00   penalty ratio = (2440.00   / 1822.00   )² = 1.79342

Callisto  : 1822.00   penalty ratio = (2410.00   / 1822.00   )² = 1.74959

Io        : 1822.00   penalty ratio = (1822.00   / 1822.00   )² = 1.00000

Moon      : 1822.00   penalty ratio = (1822.00   / 1737.00   )² = 1.10026

Europa    : 1822.00   penalty ratio = (1822.00   / 1561.00   )² = 1.36236

Triton    : 1822.00   penalty ratio = (1822.00   / 1353.00   )² = 1.81343

Pluto     : 1822.00   penalty ratio = (1822.00   / 1186.00   )² = 2.36008

Eris      : 1822.00   penalty ratio = (1822.00   / 1163.00   )² = 2.45435

Haumea    : 911.00    penalty ratio = (911.00    / 816.00    )² = 1.24640

Titania   : 911.00    penalty ratio = (911.00    / 788.00    )² = 1.33655



Max. penalty ratio = 3.70166 for Neptune

Score = ceil(152 x 3.70166) = 563

Python 2, 89 bytes, Score = 234

lambda(p):39**4/'zzuSJJaSrUNNrEnVsMeGtTMMoCoInMuErTuPsEaHTT'.find(p[7%len(p)]+p[0])**2.18

Try it online!

Most answers posted appear to have used a "encode/decode" strategy. I wondered how well I could do by estimating the diameter of celestial bodies using a simple equation. It's been a fun exercise, but the moderate byte savings are more than made up for by the accuracy penalty.

The core of this solution is the estimating equation:

Radius = 39**4/x**2.18

where x is twice the rank order of the radius of the body.

I generate the value of x based on the input string using a modification of @Erik the Outgolfer's Python 2 solution. I saved a few bytes on his code by recasting my equations to work with [2..40] instead of [1..20].

The code for generating rank orders takes up more than 2/3 of the bytes of the whole solution. If anyone has a more compact way of generating ranks, this solution could be shortened further. Because of the accuracy penalty (around 2.6), the score would improve quite a bit.

Generating the Equation

I used statistical methods to search for simple equations to estimate the size of each body based on its rank. In part following up on the insights in @Level River St's Ruby solution and generalizing, I settled on equations of the form:

Radius = A/(Rank)**B

Working in R, I used linear models on the log of the radii to develop initial estimates, and then used non-linear optimization, seeding the optimization with the results of the linear models, to search for solutions that minimized the penalty function specified in the problem.

The estimated value of A in the above equation is seven digits, so I searched for a simple expression to save a couple of bytes. I looked for expressions of the form

x**y

for two digit x and 1 digit y (for a total of five bytes, saving two bytes, or about five points, given the penalty) that was not too different from the optimum value of A and did not inflate the penalty much, and ended up with the (otherwise inexplicable):

39**4