# 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)

@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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