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(B) Let the radius of the circular orbit be $r$. The orbital speed of the satellite is given by $V = \sqrt{\frac{GM}{r}}$,which implies $V^2 = \frac{G

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$m V^2$

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(B) Let the radius of the circular orbit be $r$. The orbital speed of the satellite is given by $V = \sqrt{\frac{GM}{r}}$,which implies $V^2 = \frac{GM}{r}$.For an object to just escape the gravitational pull of the Earth,its total mechanical energy at the point of ejection must be equal to its total mechanical energy at infinity,which is $0$.Let $K$ be the kinetic energy of the object of mass $m$ at the time of ejection.The potential energy of the object at distance $r$ from the center of the Earth is $U = -\frac{GMm}{r}$.According to the law of conservation of energy: $K + U = 0$.$K - \frac{GMm}{r} = 0$.$K = \frac{GMm}{r}$.Substituting $\frac{GM}{r} = V^2$,we get $K = m V^2$.

$A$ satellite is moving with a constant speed '$V$' in a circular orbit about 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 its ejection,the kinetic energy of the object is

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