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A satellite is revolving in a circular orbit round the earth at a height of h. If the acceleration due to gravity at that height is g. The orbital speed of the satellite.

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$\sqrt{g h}$

$\sqrt{2 g h}$

$\sqrt{g (R + h)}$

$\sqrt{g R}$

answer is C.

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Detailed Solution

$∴ \frac{m V^{2}}{r} = m g$

$\Rightarrow V = \sqrt{g r} = \sqrt{g (R + h)}$

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