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 :

Question

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) (3/2) m v2 (B)  m v2 (C) 2 m v2 (D)  (1/2)m v2

Tardigrade Admin · Accepted Answer

At height r from center of earth. orbital velocity =√( GM / r ) ∴ By energy conservation KE of 'm prime+(-( GMm / r ))=0+0 (At infinity, PE = KE =0 ) ' m prime=(G M m/r)=(√(G M/r))2 m=m v2

Q. 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 :

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

$m v^{2}$

$2 m v^{2}$

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

At height $r$ from center of earth. orbital velocity
$= \frac{GM}{r}$
$∴$ By energy conservation
$K E$ of $^{'} m^{'} + (- \frac{GM m}{r}) = 0 + 0$
(At infinity, $PE = K E = 0$ )
$^{'} m^{'} = \frac{GM m}{r} = (\frac{GM}{r})^{2} m = m v^{2}$

Q. 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 :

23​mv2

mv2

2mv2

21​mv2