Potential Energy EX 5

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 radius is $r_{i}=1.5R_{E}$. The final radius is 10% smaller, therefore $r_{f}=0.9(1.5R_{E})$.

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

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

$\Delta U=GM_{E}m_{S}\left({\frac {1}{1.5R_{E}}}-{\frac {1}{0.9(1.5R_{E})}}\right)={GM_{E}m_{S} \over 1.5R_{E}}\left({\frac {1}{1}}-{\frac {1}{0.9}}\right)={-0.1GM_{E}m_{S} \over 0.9(1.5R_{E})}$

Substituting the values, we compute

$\Delta U={-0.1(6.67\times 10^{-11})(6\times 10^{24})(500) \over 0.9(1.5(6.37\times 10^{6}))}=-2.327\times 10^{9}J$