It's Just Rocket Science

Question

Write a program/function that finds the amount of fuel needed to escape Earth's gravity well given the exhaust velocity of the fuel and the amount of mass to transport using the Tsiolkovsky rocket equation: $$\Delta v = v_\text{exhaust}\ln\dfrac{m_\text{start}}{m_\text{end}}$$ where

• \$\Delta v\$ is the change in velocity from start to finish (in this case, escape velocity is about 11.2 km/s, so you will use that)
• \$v_\text{exhaust}\$ is the exhaust speed of the fuel (you will be given this)
• \$m_\text{start}\$ is the starting mass of the ship (i.e. including fuel)
• \$m_\text{end}\$ is the ending mass of the ship (i.e. excluding fuel) (you will also be given this)

Equivalently, you have to calculate \$m_\text{start}-m_\text{end},\$ which is the mass of the fuel.

The mass you need to get away from Earth will be given in kg*, and the exhaust speed in km/s with precision in the tenths. Return the mass of fuel in kg*. You can take input in any acceptable format, including (but not limited to):

• Mass as integer, exhaust speed as double/float
• Mass as integer, exhaust speed as integer representing tenths (for example, 24 would represent 2.4)
• Mass as string, exhaust speed as string

Output at least with precision to the integer (that is, when rounded to the nearest integer, your answer should be correct), and in any reasonable format (integer, float, string). Standard loopholes apply. Scoring is shortest answer per language wins, as is standard code-golf.

Sample implementation in Python:

import math

DELTA_V = 11.2

def fuel(mass: int, exhaust_speed: float) -> float:
    ratio = math.exp(DELTA_V / exhaust_speed) # m_start / m_end
    return mass * (ratio - 1)

Try it online!

Test cases (note that they aren't exact):

(0, 1) -> 0
(1, 2.4) -> 105.34267539816554
(100, 2.4) -> 10534.267539816554
(100, 100) -> 11.851286064504517
(1000, 4) -> 15444.646771097048
(597, 4) -> 9220.454122344938
(597, 4.2) -> 7994.973908804485

Leaderboards

Here is a Stack Snippet to generate both a regular leaderboard and an overview of winners by language.

To make sure that your answer shows up, please start your answer with a headline, using the following Markdown template:

# Language Name, N bytes

where N is the size of your submission. If you improve your score, you can keep old scores in the headline, by striking them through. For instance:

# Ruby, <s>104</s> <s>101</s> 96 bytes

If there you want to include multiple numbers in your header (e.g. because your score is the sum of two files or you want to list interpreter flag penalties separately), make sure that the actual score is the last number in the header:

# Perl, 43 + 2 (-p flag) = 45 bytes

You can also make the language name a link which will then show up in the leaderboard snippet:

# [><>](http://esolangs.org/wiki/Fish), 121 bytes

var QUESTION_ID=252264;
var OVERRIDE_USER=114332;
var ANSWER_FILTER="!t)IWYnsLAZle2tQ3KqrVveCRJfxcRLe",COMMENT_FILTER="!)Q2B_A2kjfAiU78X(md6BoYk",answers=[],answers_hash,answer_ids,answer_page=1,more_answers=!0,comment_page;function answersUrl(d){return"https://api.stackexchange.com/2.2/questions/"+QUESTION_ID+"/answers?page="+d+"&pagesize=100&order=desc&sort=creation&site=codegolf&filter="+ANSWER_FILTER}function commentUrl(d,e){return"https://api.stackexchange.com/2.2/answers/"+e.join(";")+"/comments?page="+d+"&pagesize=100&order=desc&sort=creation&site=codegolf&filter="+COMMENT_FILTER}function getAnswers(){jQuery.ajax({url:answersUrl(answer_page++),method:"get",dataType:"jsonp",crossDomain:!0,success:function(d){answers.push.apply(answers,d.items),answers_hash=[],answer_ids=[],d.items.forEach(function(e){e.comments=[];var f=+e.share_link.match(/\d+/);answer_ids.push(f),answers_hash[f]=e}),d.has_more||(more_answers=!1),comment_page=1,getComments()}})}function getComments(){jQuery.ajax({url:commentUrl(comment_page++,answer_ids),method:"get",dataType:"jsonp",crossDomain:!0,success:function(d){d.items.forEach(function(e){e.owner.user_id===OVERRIDE_USER&&answers_hash[e.post_id].comments.push(e)}),d.has_more?getComments():more_answers?getAnswers():process()}})}getAnswers();var SCORE_REG=function(){var d=String.raw`h\d`,e=String.raw`\-?\d+\.?\d*`,f=String.raw`[^\n<>]*`,g=String.raw`<s>${f}</s>|<strike>${f}</strike>|<del>${f}</del>`,h=String.raw`[^\n\d<>]*`,j=String.raw`<[^\n<>]+>`;return new RegExp(String.raw`<${d}>`+String.raw`\s*([^\n,]*[^\s,]),.*?`+String.raw`(${e})`+String.raw`(?=`+String.raw`${h}`+String.raw`(?:(?:${g}|${j})${h})*`+String.raw`</${d}>`+String.raw`)`)}(),OVERRIDE_REG=/^Override\s*header:\s*/i;function getAuthorName(d){return d.owner.display_name}function process(){var d=[];answers.forEach(function(n){var o=n.body;n.comments.forEach(function(q){OVERRIDE_REG.test(q.body)&&(o="<h1>"+q.body.replace(OVERRIDE_REG,"")+"</h1>")});var p=o.match(SCORE_REG);p&&d.push({user:getAuthorName(n),size:+p[2],language:p[1],link:n.share_link})}),d.sort(function(n,o){var p=n.size,q=o.size;return p-q});var e={},f=1,g=null,h=1;d.forEach(function(n){n.size!=g&&(h=f),g=n.size,++f;var o=jQuery("#answer-template").html();o=o.replace("{{PLACE}}",h+".").replace("{{NAME}}",n.user).replace("{{LANGUAGE}}",n.language).replace("{{SIZE}}",n.size).replace("{{LINK}}",n.link),o=jQuery(o),jQuery("#answers").append(o);var p=n.language;p=jQuery("<i>"+n.language+"</i>").text().toLowerCase(),e[p]=e[p]||{lang:n.language,user:n.user,size:n.size,link:n.link,uniq:p}});var j=[];for(var k in e)e.hasOwnProperty(k)&&j.push(e[k]);j.sort(function(n,o){return n.uniq>o.uniq?1:n.uniq<o.uniq?-1:0});for(var l=0;l<j.length;++l){var m=jQuery("#language-template").html(),k=j[l];m=m.replace("{{LANGUAGE}}",k.lang).replace("{{NAME}}",k.user).replace("{{SIZE}}",k.size).replace("{{LINK}}",k.link),m=jQuery(m),jQuery("#languages").append(m)}}

body{text-align:left!important}#answer-list{padding:10px;float:left}#language-list{padding:10px;float:left}table thead{font-weight:700}table td{padding:5px}

 <script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script> <link rel="stylesheet" type="text/css" href="https://cdn.sstatic.net/Sites/codegolf/primary.css?v=f52df912b654"> <div id="language-list"> <h2>Winners by Language</h2> <table class="language-list"> <thead> <tr><td>Language</td><td>User</td><td>Score</td></tr></thead> <tbody id="languages"> </tbody> </table> </div><div id="answer-list"> <h2>Leaderboard</h2> <table class="answer-list"> <thead> <tr><td></td><td>Author</td><td>Language</td><td>Size</td></tr></thead> <tbody id="answers"> </tbody> </table> </div><table style="display: none"> <tbody id="answer-template"> <tr><td>{{PLACE}}</td><td>{{NAME}}</td><td>{{LANGUAGE}}</td><td><a href="{{LINK}}">{{SIZE}}</a></td></tr></tbody> </table> <table style="display: none"> <tbody id="language-template"> <tr><td>{{LANGUAGE}}</td><td>{{NAME}}</td><td><a href="{{LINK}}">{{SIZE}}</a></td></tr></tbody> </table> 

* Technically, you can use any unit, as long as it's the same for input and output - it doesn't really matter.

Answer 1

Raku, 19 18 bytes

*×(e**(11.2/*)-1)

saved one byte by using the built-in constant e instead of the exp function

Try it online!

This is a "WhateverCode" expression, a short way of defining an anonymous function where the lone asterisks mark where the function arguments will go.

Answer 2

Factor, 24 bytes

[ 11.2 swap / e^ 1 - * ]

Attempt This Online!

       ! 1 2.4
11.2   ! 1 2.4 11.2
swap   ! 1 11.2 2.4
/      ! 1 4.666666666666667
e^     ! 1 106.3426753981655
1      ! 1 106.3426753981655 1
-      ! 1 105.3426753981655
*      ! 105.3426753981655

Answer 3

JavaScript (ES7), 22 bytes

Approximation

Expects (ending_mass)(exhaust_speed).

m=>s=>m*73130**(1/s)-m

Try it online!

This is:

$$m\times \left(e^{11.2}\right)^{1/s}-m$$

with \$e^{11.2}\approx 73130\$

Maximum error for the test cases: \$\approx0.027\$

---

JavaScript (ES6), 25 bytes

-1 by using tenths for the exhaust speed, as pointed out by @CommandMaster

Exact formula

Expects (ending_mass)(exhaust_speed) where the exhaust speed is expressed in tenths.

m=>s=>m*Math.exp(112/s)-m

Try it online!

Answer 4

Fig, \$11\log_{256}(96)\approx\$ 9.054 bytes

*{^mE\x11.2

Try it online!

Forgot to mention that it expects exhaust speed first and then mass.

*{^mE\x11.2  : Full program
     \x11.2  : Divides 11.2 by first input
  ^mE        : e^result
 {           : Decrement
*            : Multiply given only one argument so second input is implied

Answer 5

Python, 31 bytes

lambda m,e:m*(1+7e-9/e)**16e8-m

Attempt This Online!

Avoids the math import by using \$e^x\approx(1+x/n)^n\$ for large n.

Answer 6

Rust, 28 27 bytes

|m,v:f64|m*(11.2/v).exp()-m

Attempt This Online!

• -1 byte by @Steffan

Answer 7

Desmos, 20 bytes

f(m,v)=me^{11.2/v}-m

Try It On Desmos!

Try It On Desmos! - Prettified

Answer 8

Jelly,  9  8 bytes

-1 by taking the exhaust velocity in 10^ths - thanks Kevin Cruijssen!

112÷Æe’×

A dyadic Link that accepts the exhaust velocity on the left, in tenths, and the mass on the right and yields the fuel mass.

Try it online!

How?

112÷Æe’× - Link: number v, number m
112      - 112
   ÷     - divide by v
    Æe   - Exp(that)
      ’  - decrement
       × - multiply by m

Answer 9

I, 50 bytes

I does not have exponentiation, so e^x is approximated with the Taylor series.

*H 1/(inf. i+1p2f *.r)*(11.2/].  *p inf.i H]). +.r

Answer 10

Go, 64 bytes

import."math"
func f(m,v float64)float64{return m*Expm1(11.2/v)}

Attempt This Online!

Answer 11

MathGolf, 10 8 bytes

8D*╠e▬(*

Inputs in the order: \$v_{exhaust}, m_{end}\$, where \$v_{exhaust}\$ is taken as a float^† representing tenths (e.g. 24.0).

Try it online.

Explanation:

8D*       # Multiply 8 by 14: 112
   ╠      # Divide it by the first (implicit) input-float (exhaust speed as tenths)
    e▬    # Push e (2.718281828459045) to the power this
      (   # Decrement it by 1
       *  # Multiply it to the second (implicit) input-integer (mass)
          # (after which the entire stack is output implicitly as resulting fuel mass)

† : / is used for both integer division (if both arguments are integers) and regular division (if either or both arguments are floats). So I can't take the first input \$v_{exhaust}\$ as tenths as an integer.

Answer 12

C (gcc), 29 bytes

#define f(m,v)m*exp(11.2/v)-m

Try it online!

Update: Removed typdef and unnecessary includes. Thanks to @AZTECCO

Answer 13

Python, 27 Bytes

lambda m,e:m*73130**(1/e)-m

Try it online

Answer 14

BQN, 14 13 12 bytes

Edit: -1 byte thanks to faun locke
Edit 2: -1 byte by taking input in tenths of km/s (thanks to Kevin Cruijssen's comment)

⊣×1-˜⟜⋆112⊸÷

Try it at BQN REPL

Exhaust speed given as an integer representing tenths of km/s.

Or as a function for 1 more byte: {𝕨×1-˜⋆112÷𝕩} (try it)

        11.2⊸÷   # 11.2 divided by right arg
       ⋆         # exponent of that
   1-˜⟜          # one minus that
 ⊣×              # times left arg

Answer 15

05AB1E, 11 9 bytes

žrƵBI/m<*

Inputs in the order: \$v_{exhaust}, m_{end}\$, where \$v_{exhaust}\$ is taken as an integer representing tenths.

Try it online or verify all test cases.

Explanation:

žr         # Push constant e: 2.718281828459045
  ƵB       # Push compressed integer 112
    I/     # Divide it by the first input (exhaust speed as tenths)
      m    # Take e to the power this
       <   # Decrease it by 1
        *  # Multiply it to the second (implicit) input (mass)
           # (after which the resulting fuel mass is output implicitly)

See this 05AB1E tip of mine (section How to compress large integers?) to understand why ƵB is 112.

Answer 16

Vyxal, 9 8 bytes

⁺»?/∆e‹*

Try it Online!

Answer 17

Charcoal, 18 bytes

Ｉ×Ｎ⊖Ｘ≕math.e∕¹¹·²Ｎ

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

  Ｎ                 First input as a number
 ×                  Multiplied by
     ≕math.e        Python variable `math.e`
    Ｘ               Rasied to power
             ¹¹·²   Literal number `11.2`
            ∕       Divided by
                 Ｎ  Second input as a number
   ⊖                Decremented
Ｉ                   Cast to string
                    Implicitly print

Answer 18

Japt, 9 bytes

Takes mass as the first input and speed, represented as tenths, as the second.

nU*Me#p/V

Try it

nU*Me#p/V     :Implicit input of integers U=mass & V=speed
n             :Subtract U from
 U*           :  U multiplied by
   Me         :  Euler's Constant raised to the power of
     #p       :    112
       /V     :    Divided by V

Answer 19

Python, 45 43 bytes

lambda m,e:m*math.exp(11.2/e)-m
import math

Attempt This Online!

Uninteresting answer.

---

-2 bytes from @att

Answer 20

MATLAB, 45 bytes

function y=fuel(m,v) y=m*(exp(11.2/v)-1); end