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Relative Position

An artificial...

Question

An artificial satellite is moving around earth in a circular orbit with speed equal to one fourth the escape speed of a body. From the surface of earth the hight of satellite above earth's surface is (R = radius of earth)

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Solution

We know that,
Escape velocity = $\sqrt{2 g R}$
Given, Orbital speed of satellite = $\frac{\sqrt{2 g R}}{4}$
By force balance we have,
Gravitational force = Centripetal force
$G \frac{M m}{x^{2}} = \frac{m v^{2}}{x}$
$⟹ x = \frac{G M}{v^{2}} = \frac{G M}{2 g R} \times 16 = \frac{8 G M}{\frac{G M}{R^{2}}} \times R = 8 R$
Height of satellite above earth's surface is $8 R - R = 7 R$ .

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