# Work caused by gravity

1. Oct 10, 2008

### Symstar

1. The problem statement, all variables and given/known data
A 2500 kg space vehicle, initially at rest, falls vertically from a height of 3100 km above the Earth's surface.

2. Relevant equations
$$\frac{m_1m_2}{r^2}G=F_g$$
$$W=\int_a^b Fcos\theta dl$$

3. The attempt at a solution
$$W=\int_a^b F_g dl$$
$$r=r_{Earth} + r_{ship}$$
$$W=\int_a^b \frac{m_1m_2}{r^2}G$$
$$W=m_1m_2G\int_a^b r^{-2} dl$$
$$W=m_1m_2G\bigg[-\tfrac{1}{r}\Big|_a^b\bigg]$$
$$W=m_1m_2G\bigg[-\frac{1}{r_{Earth} + r_{ship}}\Big|_a^b\bigg]$$
$$W=m_1m_2G\bigg[-\frac{1}{r_{Earth} + a}+\frac{1}{r_{Earth} + b}\bigg]$$
$$a=3.1*10^6m; b=0$$
$$W=(2500)(5.98*10^{24})(6.67*10^{-11})\bigg[-\frac{1}{(6.38*10^6) + (3.1*10^6)}+\frac{1}{(6.38*10^6) + 0}\bigg]$$
$$W=5.1*10^{10}$$

I believe this work is correct, can someone verify that my integral is correct please?

2. Oct 11, 2008

### D H

Staff Emeritus
Looks good!