If an artificial satellite is moving in a circular orbit around the earth with a speed equal to half the magnitude of the escape velocity from the surface of earth, the height of the satellite above the surface of the earth is :

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$\frac{R}{2}$

$\frac{R}{4}$

answer is C.

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Detailed Solution

$v_{e} = escape velocity = \sqrt{\frac{2 GM}{R}} v_{0} = orbial velocity = \sqrt{\frac{GM}{R + h}}$ Given that; $U_{0} = \frac{U_{e}}{2}$

$∴ \sqrt{\frac{GM}{R + h}} = \frac{1}{2} \sqrt{\frac{2 GM}{R}} or \frac{GM}{R + h} = \frac{1}{4} \times \frac{2 GM}{R} Solving, we get; h = R$

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