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 2 R is

Question

TestHub · Accepted Answer

The initial total energy of the satellite on the surface of the planet is:

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

The satellite is launched into a circular orbit at an altitude $h = 2 R$ . The orbital radius is:

$r = R + h = R + 2 R = 3 R$

The final total energy of the satellite in this circular orbit is:

$E_{f} = - \frac{GM m}{2 r} = - \frac{GM m}{2 ( 3 R )} = - \frac{GM m}{6 R}$

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

$Δ E = E_{f} - E_{i}$

$Δ E = - \frac{GM m}{6 R} - (- \frac{GM m}{R})$

$Δ E = \frac{GM m}{R} - \frac{GM m}{6 R} = \frac{5 GM m}{6 R}$

Answer: $\frac{5 GMm}{6 R}$

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