MCQ
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
  • A
    $\frac{1}{2} m V^2$
  • $m V^2$
  • C
    $\frac{3}{2} m V^2$
  • D
    $2 m V^2$

Answer

Correct option: B.
$m V^2$
b
At height $r$ from center of earth. orbital velocity

$=\sqrt{\frac{\overline{G M}}{r}}$

$\therefore$ By energy conservation

KE of ' $m ^{\prime}+\left(-\frac{ GMm }{ r }\right)=0+0$

(At infinity, $PE = KE = o$ )

$\Rightarrow KE$ of ' m ' $=\frac{ GMm }{ r }=\left(\sqrt{\frac{ GM }{ r }}\right)^2 m= mv ^2$

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