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) (1/2) mv2 (B) mv2 (C) (1/3) mv2 (D) (2/3) mv2

Tardigrade Admin · Accepted Answer

Escape speed, ve=√2 × orbital speed =√2v ∴ Kinetic energy of the object at the time of ejection =(1/2) mv2e=(1/2) m(√2v)2=mv2

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

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

$m v^{2}$

$\frac{1}{3} m v^{2}$

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

Escape speed,
$v_{e} = 2 \times$ orbital speed $= 2 v$
$∴$ Kinetic energy of the object at the time of
ejection $= \frac{1}{2} m v_{e}^{2} = \frac{1}{2} m (2 v)^{2} = m v^{2}$

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

21​mv2

mv2

31​mv2

32​mv2