A 3000 kg space vehicle, initially at rest, falls vertically from a height of 3500 km above the Earth's surface.
Determine how much work is done by the force of gravity in bringing the vehicle to the Earth's surface.
Gravitational potential energy: mgh
Law of Gravitation: GMm/r2
Newton's Second Law: [tex]\Sigma[/tex]F=ma
The Attempt at a Solution
The work done should be the change in potential gravitational energy, correct?
There's zero gravitational potential energy when the spacecraft is on earth (y=0). When the spacecraft's at the height above earth from which it will fall, its gravitational potential energy is:
Where g is the gravitational acceleration. This can be found with:
Where Fg = GMearthmcraft/h2
Putting it altogether:
W=mcraft*[(GMearthmcraft/h2)/mcraft]*h - 0
Am I doing this correctly? Also, since we haven't technically covered potential energy, how would I set up an integral to find out how much work was done, since the force of gravity (and hence its acceleration) varies with r, the distance to the center of the earth?
Thank you for your help.