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

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Answer 1

$Energy at earth’s surface = potential energy+kinetic energy = - \frac{G M m}{R} + K . E$
$Energy at earth’s surface = - \frac{G M m}{R} + 0 = - \frac{G M m}{R}$
$Energy at altitude of 2R = E = P . E + K . E .$
$E = - \frac{G M m}{3 R} + \frac{1}{2} m (\frac{G M}{3 R})^{\frac{1}{2} \times 2}$
$[A s, o r b i t a l v e l o c i t y = (\frac{G M}{R})^{\frac{1}{2}}]$
$By taking the difference of energies, we get$
$E = - \frac{G M m}{6 R} - (- \frac{G M m}{R})$
$E = \frac{5 G M m}{6 R} = \frac{5}{6} m g R$ $A s, G M = g R^{2}$

Answer 2

Energy at earth’s surface = potential energy+kinetic energy = - \frac{G M m}{R} + K . E

Energy at earth’s surface = - \frac{G M m}{R} + 0 = - \frac{G M m}{R}

Energy at altitude of 2R = E = P . E + K . E .

E = - \frac{G M m}{3 R} + \frac{1}{2} m (\frac{G M}{3 R})^{\frac{1}{2} \times 2}

[A s, o r b i t a l v e l o c i t y = (\frac{G M}{R})^{\frac{1}{2}}]

By taking the difference of energies, we get

E = - \frac{G M m}{6 R} - (- \frac{G M m}{R})

E = \frac{5 G M m}{6 R} = \frac{5}{6} m g R

A s, G M = g R^{2}