First slide

Motion of planets and satellites

Question

The minimum energy required to launch a m kg satellite from the Earth's surface in a circular orbit at altitude 2R, where R is the radius of earth is

Difficult

A

$\frac{5}{3} mgR$
B

$\frac{4}{3} mgR$
C

$\frac{5}{6} mgR$
D

$\frac{5}{4} mgR$

Solution

$Total energy, E = \frac{1}{2} {mv}^{2} - \frac{GmM}{r}$
$\begin{array}{l} = \frac{GmM}{2 r} - \frac{GMm}{r} = - G \frac{mM}{2 r} \\ r = 2 R = R = 3 R \\ E = \frac{GmM}{6 R} \end{array}$
$Potential energy = - \frac{GMm}{R}$
$Minimum energy required = \frac{1}{6} \frac{GMm}{R} - (\frac{- GMm}{R}) = \frac{5}{6} \frac{GMm}{R}$
$= \frac{5}{6} mgR$

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