Question: An earth satellite of mass 200 kg lost energy slowly through atmospheric resistance and fell from an orbit of radius ${\displaystyle 8.0x10^{6}}$ m to ${\displaystyle 7.8x10^{6}}$ m. Calculate the changes in the potential, kinetic and total energies of the satellite during this transition period.

Solution: ${\displaystyle \Delta PE\ =-G{\frac {Mm}{r_{2}}}\ -(-G{\frac {Mm}{r_{1}}}\ )}$ ${\displaystyle \Delta PE\ =-6.67x10^{-11}{\frac {(6.0x10^{24})(200)}{7.8x10^{6}}}\ -(-6.67x10^{-11}{\frac {(6.0x10^{24})(200)}{8.0x10^{6}}}\ )}$ ${\displaystyle \Delta PE\ =-2.57x10^{8}J}$

Assuming the orbits are in circular motion as it decays from ${\displaystyle r_{1}=8.0x10^{6}}$ to ${\displaystyle r_{2}=7.8x10^{6}}$

equation ${\displaystyle G{\frac {Mm}{r^{2}}}\ ={\frac {mv^{2}}{r}}\ }$ can be assumed to be valid to approximate this motion

Thus, ${\displaystyle \Delta KE\ ={\frac {1}{2}}\ mv_{2}^{2}-{\frac {1}{2}}\ mv_{1}^{2}}$

${\displaystyle \Delta KE\ ={\frac {1}{2}}\ mG{\frac {M}{r_{2}}}\ -}$ ${\displaystyle {\frac {1}{2}}\ mG{\frac {M}{r_{1}}}\ }$

${\displaystyle \Delta KE\ ={\frac {1}{2}}\ (200)(6.67x10^{-11}){\frac {6.0x10^{24}}{7.8x10^{6}}}\ -}$ ${\displaystyle {\frac {1}{2}}\ (200)(6.67x10^{-11}){\frac {6.0x10^{24}}{8.0x10^{6}}}\ }$

${\displaystyle \Delta KE\ =1.28x10^{8}J}$

since ${\displaystyle TE=PE+KE}$

imply ${\displaystyle \Delta TE\ =}$ ${\displaystyle \Delta PE\ +}$ ${\displaystyle \Delta KE\ }$

${\displaystyle \Delta TE\ =-2.57x10^{8}+1.28x10^{8}}$

${\displaystyle \Delta TE\ =-1.28x10^{8}J}$

${\displaystyle \Delta TE\ =-1.28x10^{8}J}$