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\[(a)\;\frac{v_e^2}{v_0} \quad (b)\; v_0 \quad (c)\;\sqrt {v_e^2-v_0^2} \quad (d)\;\sqrt {v_e^2-2 v_0^2}\]

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Let r be the radius of satellite then,

$v_0^2 =\large\frac{GM}{r}$

Applying conservation of Medical energy between points A (the satellite position when its suddenly stops ) to point B on the surface of earth where it strikes with velocity v

$KE$=change in potential energy

$\large\frac{1}{2}$$mv^2 =\large\frac{-GMm}{r}-\bigg(\large\frac{-GMm}{R}\bigg)$

$v^2=\large\frac{2GM}{R}-\frac{2GM}{r}$

$\qquad= v_e^2 -2 v_0^2$

$v=\sqrt {v_e^2-2v_0^2}$

Hence d is the correct answer.

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