# What do I need to destroy the world?

The gravitational binding energy of a planet is the amount of energy required to separate every tiny piece of it so that no piece will orbit or collide with another piece. For a uniform sphere, Wikipedia gives this formula:

$$E = \frac{3GM^2}{5R}$$

Where G is the gravitational constant (6.672e-11 m3•kg-1•s-2), M is mass, and R is the radius.

I could make this challenge about merely calculating this in Joules (kg•m2•s-2), but… that would be boring. And besides, not everyone understands Joules.

So instead, let's convert it into the following units:

• Kilotons of TNT (4.184 trillion Joules, shorthand kilotons)
• Hiroshima nuclear weapons (62.76 trillion Joules, shorthand nukes)
• Hostess™ Twinkies (560,000 Joules, shorthand twinkies)
• Kilowatt-hours (exactly 3.6 million Joules, shorthand kWh)
• Kilograms of mass-energy (1 kg mass-energy = 299,792,4582 Joules, shorthand kg)

So, take in the following input:

• mass in kilograms
• one of six distinct inputs representing the unit to use

And output the binding energy in the specified units. Append the shorthand unit to the end of the output.

# Rules

• The shorthand for Joules is J.
• Arbitrary whitespace is allowed, as long as it doesn't split the shorthand unit or the number.
• Scientific notation is allowed, in whatever format your language uses.
• You only need 4 significant figures. Floating points are highly recommended.
• Blah blah shortest answer in bytes.

# Test Cases

Mass (kg)   Radius (m)  J           kilotons    nukes       twinkies    kWh         kg
3.302E+23   2440000     1.789E+30   4.275E+17   2.850E+16   3.194E+24   4.969E+23   1.990E+13
4.869E+24   6052000     1.568E+32   3.748E+19   2.499E+18   2.800E+26   4.356E+25   1.745E+15
5.974E+24   6371000     2.242E+32   5.360E+19   3.573E+18   4.004E+26   6.229E+25   2.495E+15
6.419E+23   3390000     4.866E+30   1.163E+18   7.753E+16   8.689E+24   1.352E+24   5.414E+13
1.899E+27   69911000    2.065E+36   4.935E+23   3.290E+22   3.687E+30   5.736E+29   2.298E+19
5.685E+26   58232000    2.222E+35   5.310E+22   3.540E+21   3.968E+29   6.172E+28   2.472E+18
8.683E+25   25360000    1.190E+34   2.845E+21   1.896E+20   2.125E+28   3.306E+27   1.324E+17
1.024E+26   24620000    1.705E+34   4.075E+21   2.717E+20   3.045E+28   4.736E+27   1.897E+17
1.311E+22   1186000     5.801E+27   1.387E+15   9.244E+13   1.036E+22   1.611E+21   6.455E+10
1.989E+30   696300000   2.274E+41   5.436E+28   3.624E+27   4.062E+35   6.318E+34   2.531E+24
7.350E+22   1737000     1.245E+29   2.976E+16   1.984E+15   2.223E+23   3.458E+22   1.385E+12

Inspired by Powering the Death Star with Twinkies by Scott Manley.

• what do you mean by "default to joules"? does that mean that the output unit will not always be given in the input? Nov 27, 2017 at 2:02
• also, I love how there's actually a name for the amount of energy needed to completely annihilate a thing :P Nov 27, 2017 at 2:03
• can we take the output unit as any of 6 distinct inputs? otherwise it's a cumbersome I/O format and those are discouraged Nov 27, 2017 at 2:06
• @HyperNeutrino sure, but you still need the shorthand in the output. Nov 27, 2017 at 2:16
• @StephenLeppik "one of six distinct inputs representing the unit to use" - can those distinct inputs be 4.184, 62.76, etc?
– ngn
Nov 30, 2017 at 18:50

# JavaScript (ES6), 78 bytes

Takes input as 3 distinct variables (mass, radius, unit).

(m,r,u)=>m*m/r/[8987e9,56,6276e6,,4184e5,1e-4,360][parseInt(u,30)%7]/2498e11+u

### Test cases

The following snippet runs all test cases in the provided spreadsheet.

let f=

(m,r,u)=>m*m/r/[8987e9,56,6276e6,,4184e5,1e-4,360][parseInt(u,30)%7]/2498e11+u

console.log(
[
'3.302e+23 2440000   1.789e+30 4.275e+17 2.850e+16 3.194e+24 4.969e+23 1.990e+13',
'4.869e+24 6052000   1.568e+32 3.748e+19 2.499e+18 2.800e+26 4.356e+25 1.745e+15',
'5.974e+24 6371000   2.242e+32 5.360e+19 3.573e+18 4.004e+26 6.229e+25 2.495e+15',
'6.419e+23 3390000   4.866e+30 1.163e+18 7.753e+16 8.689e+24 1.352e+24 5.414e+13',
'1.899e+27 69911000  2.065e+36 4.935e+23 3.290e+22 3.687e+30 5.736e+29 2.298e+19',
'5.685e+26 58232000  2.222e+35 5.310e+22 3.540e+21 3.968e+29 6.172e+28 2.472e+18',
'8.683e+25 25360000  1.190e+34 2.845e+21 1.896e+20 2.125e+28 3.306e+27 1.324e+17',
'1.024e+26 24620000  1.705e+34 4.075e+21 2.717e+20 3.045e+28 4.736e+27 1.897e+17',
'1.311e+22 1186000   5.801e+27 1.387e+15 9.244e+13 1.036e+22 1.611e+21 6.455e+10',
'1.989e+30 696300000 2.274e+41 5.436e+28 3.624e+27 4.062e+35 6.318e+34 2.531e+24',
'7.350e+22 1737000   1.245e+29 2.976e+16 1.984e+15 2.223e+23 3.458e+22 1.385e+12'
]
.map((l, n) => {
[m, r, ...res] = l.match(/\S+/g).map(Number);

return 'Test case #' + -~n + ' -> ' + (
['J', 'kilotons', 'nukes', 'twinkies', 'kWh', 'kg']
.every((u, i) => +(+f(m, r, u).split(u).join).toExponential(3) === +res[i])
? 'OK' : 'FAIL'
)
})
)

### How?

This is quite similar to @ColeraSu's C answer, with a JS-friendly hash function that works on the whole unit string.

Unit       | Parsed as (*) | Base-30 -> Dec. | mod 7
-----------+---------------+-----------------+------
"J"        | "j"           | 19              | 5        (*):
"kilotons" | "kilotons"    | 451052545318    | 4          Because we're parsing as Base-30,
"nukes"    | "n"           | 23              | 2          characters [u-z] (and anything
"twinkies" | "t"           | 29              | 1          after them) are simply ignored.
"kWh"      | "k"           | 20              | 6
"kg"       | "kg"          | 616             | 0

# Jelly, 78 bytes

²×⁽LÐ÷ȷ14÷⁴÷5÷⁵ị“¤Fðẏẏż“?A⁺ẏẏż“¬ŀż“8Ƙż“¡uþ³⁾ẉṿÆ“¡’¤;⁵ị“ÞṠlT“¡ṁæ-“;ØḲ“ȧ¹“ṫD“}»¤

Try it online!

Stuff to do with strings isn't really Jelly's type of challenge...

-7 bytes thanks to user202729 (thanks for helping me not get beaten by JavaScript :D)

• Yeah, now you're tied with JS and still not on top either. Nov 27, 2017 at 14:40
• @StephenLeppik At least I'm not getting beaten :D and shush 05AB1E is inherently better at string challenges :P Nov 27, 2017 at 14:50

# 05AB1E, 56 54 bytes

Takes input as unit, radius, mass

6n56•G*•x619+•Oć•4°z)•1¡‡•S°*"Wwigu"I1èkè*2498T>°P/*¹J

Try it online!

Saved 2 bytes thanks to Erik the outgolfer (base 255 encode 4184 and 6276).

# C, 113 bytes

Hash the second character, using the trailing '\0' of "J". Inline array in C is used.

Note we need to multiply the second M last, or it will overflow to inf.

f(m,r,c)float m,r;char*c;{printf("%e%s",m/r/(float[]){1e-4,360,6276e6,8987e9,56,4184e5}[c[1]%36%7]/2498e11*m,c);}

Try it online!

• +1. Not sure if possible, but f(m,r,char*c)float m,r{…} would save some bytes. Also, the semicolon after printf is optional Nov 27, 2017 at 6:03
• @xanoetux Which compiler do you use? Both don't work on my machine though. Nov 27, 2017 at 8:25
• I'm using GCC. Untested, but I'm quite sure about the last semicolon. (will test later) Nov 27, 2017 at 14:27

# APL (Dyalog Unicode), 153 71 69 bytes

op←{⍺,⍨40032E¯15÷÷/8987E13 36E5 6276E10 56E4 4184E9 1['gWuw '⍳⊃1↓⍺]⍵⍺⍺⍵}

Try it online!

Thanks to @Adám for absolutely crushing my old answer into this one. Thanks to @ngn for 2 bytes.

This 'function' is actually an operator called op. All the values for the measurement units were transformed to scientific notation.

### How it works:

{⍺,⍨40032E¯15÷÷/8987E13 36E5 6276E10 56E4 4184E9 1['gWuw '⍳⊃1↓⍺]⍵⍺⍺⍵}
1↓⍺         ⍝ take the measurement unit without the first letter (because it is ambiguous)
⊃            ⍝ picks the first element returned by
⍳             ⍝ Dyadic ⍳ota, which finds the index of
'gWuw '                ⍝ each of these characters
[             ]         ⍝ square brackets indicate the index
8987E13 36E5 6276E10 56E4 4184E9 1                        ⍝ of the element of this vector (1-indexed) which will be used as the divisor for
÷/                                                  ⍵⍺⍺⍵     ⍝ dividing by the left operand (⍺⍺) and the right argument squared.
40032E¯15÷                                                           ⍝ then divide this number by the result
⍺,⍨                                                                     ⍝ and append the measurement unit

# Python 3, 112106108 101 Bytes

lambda m,r,u:str(40032e-19*m*m/r/{'':1e-4,'i':4184e5,'u':6276e6,'w':56,'W':360,'g':8987e9}[u[1:2]])+u

Pretty straight forward. Uses Python's e notation. Uses a dict that distinguishes unit based on the second letter and divides by the corrosponding value. u[1:2] is basically u[1], expect it returns '' instead of an error (which I only learned now). Then applies the formula with a few simplifications.

edit: down to 106 thanks to totallyhuman, back to 108 because of new nukes value and down to 101 thanks to Colera Su.

Try it online!

# Excel (Office 365), 106101 97 bytes

Assuming cell A1 is the mass, A2 is the radius, and B1 is the unit, written as the shorthand for the output.

=40032E-19*A1^2/(A2*SWITCH(MID(B1,2,1),"u",6276E6,"w",56,"W",360,"g",89876E8,"i",4184E5,1E-4))&B1

# Excel, 130 bytes

Old-fashioned Excel doesn't have the SWITCH function, so you have

=40032E-15*A1^2/(A2*IF(B1="J",1,IF(B1="nukes",6276E10,IF(B1="twinkies",56E4,IF(B1="kWh",36E5,IF(B1="kg",299792E3^2,4184E9))))))&B1
• You can save 2 bytes by substituting 299792E3^2 with 89876E12 (Still accurate enough for question requirements.) Divide all values by 1000 saves another 2: =40032E-19*A1^2/(A2*SWITCH(MID(B1,2,1),"u",6276E6,"w",56,"W",360,"g",89876E8,"i",4184E5,1E-4))&B1 Dec 1, 2017 at 13:10
• Hey, just want to throw it out there that the Switch() function is available to desktop copies of Excel, and was introduced with Office 2016. Dec 1, 2017 at 21:44

# Ruby , 83 82 bytes

->m,r,u{"#{40032E-18*m*m/r/[36E2,56,8987E9,1E-3,4184E6,6276E7][u.sum*3%71%6]} "+u}

Uses a custom hash to index a LUT for the units (i * k mod m formula). The numbers were found using brute-force.

# Excel, 94 bytes

Takes Mass from A1, Radius from B1, Unit in C1. Look up on 2*Code(Unit) + Length(Unit) modulo 11.

=4.0032E-15*A1^2/B1/CHOOSE(MOD(2*CODE(C1)+LEN(C1),11),,4184E5,,,6276E6,1E-4,89876E8,360,56)

# Jelly,  59  54 bytes

Saved 5 bytes thanks to @ChartZBelatedly

I just wanted to try another weird method to convert the units. But the rest of the code is probably badly golfed.

⁵OẒḄ%9ị56,4184ȷ5,8987ȷ9,360,0,6276ȷ6,ȷ-4
²÷÷2498ȷ11÷¢⁵

Try it online!

### How?

We build a 'signature' of a unit string out of the primality of its ASCII codes. We interpret it as a binary value and apply a modulo to get the final index in the lookup table.

⁵OẒḄ%9ị[...]    - helper link to convert the unit to a hardcoded constant
⁵               - the 3rd input        e.g. "kilotons"
O              - to ASCII codes       --> [107, 105, 108, 111, 116, 111, 110, 115]
Ẓ             - is prime?            --> [1, 0, 0, 0, 0, 0, 0, 0]
Ḅ            - binary -> decimal    --> 128
%9          - modulo 9             --> 2
ị[...]    - get value from table --> 418400000

²÷÷2498ȷ11÷¢⁵   - main link taking the mass, the radius and the unit
²               - square the mass
÷              - divide by the radius
÷2498ȷ11      - divide by 2498e11
÷¢    - divide by the unit constant, invoking the helper link as a nilad
⁵   - append the unit

The helper link gives the following indices for each unit:

unit     | ASCII codes                     | is prime?       | bin -> dec | mod 9
---------+---------------------------------+-----------------+------------+------
J        | 74                              | 0               |      0     |   0
kilotons | 107,105,108,111,116,111,110,115 | 1,0,0,0,0,0,0,0 |    128     |   2
nukes    | 110,117,107,101,115             | 0,0,1,1,0       |      6     |   6
twinkies | 116,119,105,110,107,105,101,115 | 0,0,0,0,1,0,1,0 |     10     |   1
kWh      | 107,87,104                      | 1,0,0           |      4     |   4
kg       | 107,103                         | 1,1             |      3     |   3
• The [ and ] are unnecessary. Additionally, the can be removed and the ÆP has a 1 byte alias: . Finally, if the number on the left of ȷ is 1, it can be omitted. -5 bytes Apr 4, 2021 at 15:51
• @ChartZBelatedly Thank you! Apr 4, 2021 at 20:01