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What would be the projection velocity of a satellite so that interstellar (interspace with no gravitation) velocity of the satellite becomes the orbital velocity when it is in the orbit at height of 10R from earth's surface, where R is the radius of the earth?

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$\sqrt{\frac{23 gR}{11}}$

$\sqrt{\frac{25 gR}{11}}$

$\frac{{v_{a}}^{2}}{{gR}^{2}} - R$

$\frac{{v_{a}}^{2}}{2 {gR}^{2}} - R$

answer is A.

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Detailed Solution

Interstellar velocity $v^{'} = \sqrt{\frac{GM}{r}} = R \sqrt{\frac{g}{(R + h)}}$

= $\sqrt{v^{2} - {v_{e}}^{2}}$

where v = projection velocity

$\frac{R^{2} g}{(R + h)} = v^{2} - 2 gR v^{2} = \frac{R^{2} g}{(R + 10 R)} + 2 g R Solving v^{2} = \frac{23 gR}{11} V = \sqrt{\frac{23 g R}{11}}$

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