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(B) For a satellite in a circular orbit of radius $r$,the orbital speed $v$ is given by $v = \sqrt{\frac{GM_e}{r}}$,which implies $v^2 = \frac{GM_e}{r

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$mv^2$

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(B) For a satellite in a circular orbit of radius $r$,the orbital speed $v$ is given by $v = \sqrt{\frac{GM_e}{r}}$,which implies $v^2 = \frac{GM_e}{r}$.The potential energy of the object of mass $m$ at distance $r$ is $U = -\frac{GM_em}{r}$.For the object to just escape the gravitational pull of the Earth,its total mechanical energy must be zero at infinity,meaning its total energy at the point of ejection must also be zero.Let $K$ be the kinetic energy of the object at the time of ejection.Total energy $E = K + U = 0$.$K - \frac{GM_em}{r} = 0$.$K = \frac{GM_em}{r}$.Substituting $\frac{GM_e}{r} = v^2$,we get $K = mv^2$.

$A$ satellite is moving with a constant speed $v$ in a 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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