Given the names of two planets, give the distance

CJam, 54 51 44 bytes

2{"X84VT:Z/3KD'Y->>6\ Ta "3/r26b93%=70be4}*-

Try it online in the CJam interpreter.

Idea

We use a simple hashing function to identify all eight planets. By considering each name as the array of its code points, converting them from base 26 to integer and taking the result modulo 93 then modulo 8, Mercury, Venus, Earth, etc. map to 2, 4, 0, 1, 3, 5, 6 and 7.

Now, we choose a point that lies 320,000 km behind Neptune and calculate the distances of all eight planets to that point. After dropping four trailing zeroes and reordering the planets so that they fit the 8 indexes from above, we obtain the array

[435172 427338 444341 372299 439312 307672 162777 32]

which, if we encode each integer in base 70, yields the following:

[
   [1 18 56 52] [1 17 14 58] [1 20 47 51] [1 5 68 39]
   [1 19 45 62] [  62 55 22] [  33 15 27] [       32]
]

Remembering that two adjacent digits (A B) can be replaced with ((A-1) (B+70)), we can modify the array from above so that all integers can be encoded as printable ASCII characters:

["X84" "VT:" "Z/3" "KD'" "Y->" ">6\\" " Ta" " "]

Code

2{                         e# Do twice:   
  "X84VT:Z/3KD'Y->>6\ Ta " e#   Push that string.
  3/                       e#   Chop it into chunks of length 3.
  r                        e#   Read a token from STDIN.
  26b                      e#   Convert from base 26 to integer.
  93%                      e#   Take the result modulo 93.
  =                        e#   Retrieve the chunk at that index.
  70b                      e#   Convert from base 70 to integer.
  e4                       e#   Multiply by 10,000.
}*                         e#
-                          e# Subtract the two results.

Python 2, 149 147 142 138 128 123 119 Bytes

Just uses a simple lookup to figure out which distances to use :) This defines an anonymous function, so to use it you'll need to give it a name.

Thanks to Sp3000 for ideas that saved a bunch of bytes!

lambda*x:int.__sub__(*[[0,5029,9169,17003,72042,136669,281564,444309]['MeVeEaMaJuSaUr'.find(k[:2])/2]for k in x])*~9999

Indented properly and ungolfed slightly for readability:

def f(*x):
 d=0,5029,9169,17003,72042,136669,281564,444309
 a,b=[d['MeVeEaMaJuSaUr'.find(k[:2])/2]for k in x]
 print(b-a)*10000

Call like so:

f("Mercury","Mars")    -> 170030000
f("Neptune","Jupiter") -> -3722670000L

Prolog, 190 174 151 bytes

Thanks to Fatalize for guidance.

g(A,X):-sub_atom(A,2,2,_,B),member(B:X,[rc:0,nu:5029,rt:9169,rs:17003,pi:72042,tu:136669,an:281564,pt:444309]).
s(A,B,R):-g(A,X),g(B,Y),R is(Y-X)*10^4.

---

$ gprolog --consult-file src.pro 
| ?- s('Mercury','Mars',R).   
R = 170030000 ? 
yes
| ?- s('Neptune','Jupiter',R).
R = -3722670000 ? 
yes
| ?- s('Earth','Earth',R).    
R = 0 ? 
yes