The minimum energy required to launch a satellite of mass m from the surface of a planet of mass M and radius R in a circular orbit at an altitude of 3 R is { G = gravitational constant }

Question

TestHub · Accepted Answer

Initial energy of the satellite on the surface of the planet is:

$E_{i} = - \frac{GM m}{R}$

The radius of the circular orbit at an altitude of $3 R$ is $r = R + 3 R = 4 R$ .

Final total energy of the satellite in the circular orbit is:

$E_{f} = - \frac{GM m}{2 r} = - \frac{GM m}{2 ( 4 R )} = - \frac{GM m}{8 R}$

The minimum energy required to launch the satellite is the difference between the final and initial energies:

$Δ E = E_{f} - E_{i}$

$Δ E = - \frac{GM m}{8 R} - (- \frac{GM m}{R})$

$Δ E = \frac{GM m}{R} - \frac{GM m}{8 R} = \frac{7 GM m}{8 R}$

Answer: $\frac{7 GM m}{8 R}$

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