Potential Energy EX 9

Since the radius of the orbit is $r=1.5R_{E}$, we have to use the general formula for the change in gravitational potential energy of the system satellite-Earth. The initial distance between the satellite and the Earth is equal to the radius of the Earth, or $r_{i}=R_{E}=6.37\times 10^{6}m$. When the satellite orbits at the altitude $7\times 10^{5}m$, the final distance between the Earth and the satellite becomes $r_{f}=(7\times 10^{5})+(6.37\times 10^{6})=7.07\times 10^{6}m$. Now we can compute the change in gravitational potential energy:

$\Delta U={-Gm_{1}m_{2} \over r_{f}}-\left({-Gm_{1}m_{2} \over r_{i}}\right)=Gm_{1}m_{2}\left({\frac {1}{r_{i}}}-{\frac {1}{r_{f}}}\right)$

Since $m_{1}=M_{E}$ and $m_{2}=m_{S}$, we substitute for $r_{i}$ and $r_{f}$ and write

$\Delta U=GM_{E}m_{S}\left({\frac {1}{6.37\times 10^{6}}}-{\frac {1}{7.07\times 10^{6}}}\right)$

$\Delta U=(6.67\times 10^{-11})(6\times 10^{24})(1000)\left({\frac {1}{6.37\times 10^{6}}}-{\frac {1}{7.07\times 10^{6}}}\right)$

$\Delta U=6.23\times 10^{9}J$