Question

The orbital velocity of an artificial satellite in a circular orbit just above the centre's surface is v. For a satellite orbiting at an altitude of half of the earth's radius, the orbital velocity is :

Answer 1

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Answer 2

A “satellite orbiting” at an altitude at “half of the earth’s radius” is $\bold{\sqrt{\frac{2}{3}} V}$

Explanation:

The orbital velocity at height h above the earth surface is,

$V_{0}=\sqrt{\frac{G M}{R_{e}+h}}$

$V_{0}=\sqrt{\frac{g R_{E}^{2}}{R_{E}+h}}$

Orbital velocity just above the earth surface is (V)

$V=\sqrt{\frac{g R_{E}^{2}}{R_{E}+h}}$

Orbital velocity at height of half of earth radius

$V^{\prime}=\sqrt{\frac{g R_{E}^{2}}{R_{E}+\frac{R E}{2}}}$

$V^{\prime}=\sqrt{\frac{g R_{E}^{2}}{\frac{3}{2} R_{E}}}$

$V^{\prime}=\sqrt{\frac{2}{3} g R_{E}}$

$V=\sqrt{g R_{E}}$

Where,

$R_{E}= \text{Radius of earth}$

Orbiting at an altitude at “half of the earth’s radius” is,

$V^{\prime}=\sqrt{\frac{2}{3}} V$