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 height of satellite above earth's surface is (R is radius of earth)

Answers (1)

@ABHISHEK

escape speed of a body=V_e=\sqrt{2gR}

and orbital speed of a body=V_0=\sqrt{ \frac{GM}{r}}

but here V_0= \frac{1}{4}V_e

so \sqrt{ \frac{GM}{r}}=\frac{1}{4}\sqrt{2gR}=\sqrt{ \frac{gR}{8}}=\sqrt{ \frac{GM}{8R}}

So r=8R

r=R+h

so h=7R

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