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:

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

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

$\mathrm{\Delta }U=G{M}_E{m}_S(\frac{1}{6.37\times 10^6}-\frac{1}{7.07\times 10^6})$

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

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