$A$ $400 \; kg$ satellite is in a circular orbit of radius $2 R_{E}$ about the Earth. How much energy is required to transfer it to a circular orbit of radius $4 R_{E}$? What are the changes in the kinetic and potential energies?

(N/A) The total energy of a satellite in a circular orbit of radius $r$ is given by $E = -\frac{G M_{E} m}{2r}$.
Initial total energy at $r_{i} = 2 R_{E}$ is $E_{i} = -\frac{G M_{E} m}{4 R_{E}}$.
Final total energy at $r_{f} = 4 R_{E}$ is $E_{f} = -\frac{G M_{E} m}{8 R_{E}}$.
The energy required is $\Delta E = E_{f} - E_{i} = -\frac{G M_{E} m}{8 R_{E}} - (-\frac{G M_{E} m}{4 R_{E}}) = \frac{G M_{E} m}{8 R_{E}}$.
Using $g = \frac{G M_{E}}{R_{E}^{2}}$,we get $\Delta E = \frac{g m R_{E}}{8} = \frac{9.8 \times 400 \times 6.37 \times 10^{6}}{8} \approx 3.12 \times 10^{9} \; J$.
Kinetic energy $K = \frac{G M_{E} m}{2r}$,so $\Delta K = K_{f} - K_{i} = \frac{G M_{E} m}{8 R_{E}} - \frac{G M_{E} m}{4 R_{E}} = -\frac{G M_{E} m}{8 R_{E}} = -3.12 \times 10^{9} \; J$.
Potential energy $V = -\frac{G M_{E} m}{r}$,so $\Delta V = V_{f} - V_{i} = -\frac{G M_{E} m}{4 R_{E}} - (-\frac{G M_{E} m}{2 R_{E}}) = \frac{G M_{E} m}{4 R_{E}} = 6.24 \times 10^{9} \; J$.