A satellite is moving with a constant speed v in circular orbit around the earth. An object of mass ‘m’ is ejected from the satellite such that it just escapes from the gravitational pull of the earth. At the time of ejection, the kinetic energy of the object is ..

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$\frac{1}{2} m v^{2}$

$2 m v^{2}$

$\frac{3}{2} m v^{2}$

$m v^{2}$

answer is B.

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Detailed Solution

At height r from center of earth. Orbital velocity $v = \sqrt{\frac{G M}{r}} . . . . . . . (1)$
$∴$ By energy conservation
$KE of' m' + (- \frac{G M m}{r}) = 0 + 0$
(At infinity, PE = KE = 0)
$\Rightarrow KE of' m' = \frac{G M m}{r} = {(\sqrt{\frac{G M}{r}})}^{2} m = m v^{2}$

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