Art Vandelay
Illuminator
- Joined
- May 8, 2004
- Messages
- 4,787
Well, I suppose you could, but I don't think the moderators would appreciate a detailed discussion of the matterneutrino_cannon said:"You can't power something with... thingy! I guess it will make charted engineership much more exciting!"
You can do it be writing out the equation for gravitational acceleration versus distance, then integrating from the current radius to the radius of the earth. It comes out to 1/(earth radius)-1/(moon radius) times some constant. I'm just too lazy to figure out what it is. Actually, I believe that for a circular orbit, the ratio between the KE and PE energy is a particular number that I don't remeber.The moon is 7.36*10^22 kilograms, and orbits at ~1028 m/s. That comes to a kinetic energy of 3.89*10^28 joules. I'm not sure how to calculate the moon's potential energy because the dropoff in earth's gravity at that distance is significant, so multiplying the mass time acceleration times distance would give an innacurate answer.