# Calculating the Total Work Done by Gravity

1. Dec 1, 2011

### Simon777

1. The problem statement, all variables and given/known data
A satellite in a circular orbit around the earth with a radius 1.011 times the mean radius of the earth is hit by an incoming meteorite. A large fragment (m = 83.0 kg) is ejected in the backwards direction so that it is stationary with respect to the earth and falls directly to the ground. Its speed just before it hits the ground is 355.0 m/s. Find the total work done by gravity on the satellite fragment. RE 6.37·10^3 km; Mass of the earth= ME 5.98·10^24 kg.

2. Relevant equations
Gravitational PE= (-GmM)/R

3. The attempt at a solution

Delta PE= PE intial - PE final

= (-(6.67x10^-11) (83kg) (5.98x10^24kg))/((1.011) (6.37x10^3)) - (-(6.67x10^-11) (83kg) (5.98x10^24kg))/(6.37x10^3)

= 5.65x10^10J

This is incorrect and I have tried for hours to get something to work so any help would be greatly appreciated.

2. Dec 1, 2011

### Staff: Mentor

Something happened to your orders of magnitude. I think you're result is about 1000x to big. Check your math.

You can save yourself a lot of digit pushing if you do some of the algebra symbolically ahead of time:

Let r = 1.011; M = Mass of Earth; R = radius of Earth; m = mass of fragment;

$$\Delta E = \left(\frac{G M m}{R} - \frac{G M m}{r R}\right) = \frac{G M m}{R}\left(1 - \frac{1}{r}\right)$$

3. Dec 1, 2011

### Simon777

That does make it easier, thank you. Using it, I still end up with 5.65x10^10J. Perhaps one of my terms is wrong. This is what I used:
M=5.98·10^24 kg
G=6.67x10^-11
m=83.0kg
R= 6.37·10^3 km
r= 1.011

4. Dec 1, 2011