# Work done by gravity on falling object

1. Oct 7, 2008

### Serik

1. The problem statement, all variables and given/known data

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.

2. Relevant equations

Gravitational potential energy: mgh
Law of Gravitation: GMm/r2
Newton's Second Law: $$\Sigma$$F=ma

3. 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:

mcraft*g*h

Where g is the gravitational acceleration. This can be found with:

Fg=ma
a=Fg/mcraft

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?

2. Oct 7, 2008

### LowlyPion

So long as you realize that the h you're talking about is the distance to the center of the Earth.
And the distance from the surface to the center becomes your rest potential.

3. Oct 7, 2008

### Serik

Ahhh, thanks.

In the corner of my notes, I scrawled h=Habove earth + Rearth, but failed to carry that substitution through the remainder of my calculations.

So at the surface, which I shouldn't really call y=0, the spacecraft will have gravitational potential energy because it's a distance Rearth from the center of the planet.