# Potential Energy and Velocity of a Spaceship

1. Jun 23, 2015

### Staff: Mentor

1. The problem statement, all variables and given/known data
You plan to take a trip to the moon. Since you do not have a traditional spaceship with rockets, you will need to leave the earth with enough speed to make it to the moon. Some information that will help during this problem:

mearth = 5.9742 x 1024 kg
rearth = 6.3781 x 106 m
mmoon = 7.36 x 1022 kg
rmoon = 1.7374 x 106 m
dearth to moon = 3.844 x 108 m (center to center)
G = 6.67428 x 10-11 N-m2/kg2

1)On your first attempt you leave the surface of the earth at v = 5534 m/s. How far from the center of the earth will you get?
2) Since that is not far enough, you consult a friend who calculates (correctly) the minimum speed needed as vmin = 11068 m/s. If you leave the surface of the earth at this speed, how fast will you be moving at the surface of the moon? Hint carefully write out an expression for the potential and kinetic energy of the ship on the surface of earth, and on the surface of moon. Be sure to include the gravitational potential energy of the earth even when the ship is at the surface of the moon!
2. Relevant equations
ΔE = 0
U1+K1 = U2 + K2

3. The attempt at a solution

Found the answer to Q1. It's 8.44x106 meters.

Q2 is where I'm having trouble. I don't know how to set it up. For Q1, ΔE=0 so ΔU + ΔK = 0 as well. And ΔU = U2-U1.

For Q2, I was thinking you would add the potential energy of the spacecraft for both the Earth and the Moon together, like: U1 = U1E + U1M. That would make ΔU = (U2E + U2M) - (U1E + U1M).

Unfortunately my answer using this method appears to be incorrect. Not sure what to do. I made sure to account for the differing distance between the centers of the Earth and Moon, and the distance between the surfaces of each and the center of the other.

2. Jun 23, 2015

### haruspex

In principle, you need to consider the PE wrt the moon in Q1 too, but that shot falls too far short for it to matter.
Please post the details of your attempt on Q2. Can't tell where or if you are going wrong otherwise.

3. Jun 23, 2015

### Staff: Mentor

Roger. I'll get on that as soon as I can.

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