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 $r_f=0.9(1.5R_E)$.

$\mathrm{\Delta }U=\frac{-G{m}_1{m}_2}{r_f}-\left(\frac{-G{m}_1{m}_2}{r_i}\right)=G{m}_1{m}_2(\frac{1}{r_i}-\frac{1}{r_f})$

Since $m_1=M_E$ and $m_2=m_S$, we substitute for $r_i$ and $r_f$ and write

$\mathrm{\Delta }U=G{M}_E{m}_S(\frac{1}{1.5R_E}-\frac{1}{0.9(1.5R_E)})=\frac{G{M}_E{m}_S}{1.5R_E}(\frac{1}{1}-\frac{1}{0.9})=\frac{-0.1G{M}_E{m}_S}{0.9(1.5R_E)}$

Substituting the values, we compute

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