Gravitation Ques 3
- Two satellites $A$ and $B$ have masses $m$ and $2 m$ respectively. $A$ is in a circular orbit of radius $R$ and $B$ is in a circular orbit of radius $2 R$ around the earth. The ratio of their kinetic energies, $T_A / T_B$ is
(2019 Main, 12 Jan II)
(a) $\frac{1}{2}$
(b) 2
(c) $\sqrt{\frac{1}{2}}$
(d) 1
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Answer:
Correct Answer: 3.( d )
Solution:
- Orbital speed of a satellite in a circular orbit is
$ v_0=\sqrt{\left(\frac{G M}{r_0}\right)} $
where $r_0$ is the radius of the circular orbit. So, kinetic energies of satellites $A$ and $B$ are
$ \begin{aligned} & T_A=\frac{1}{2} m_A v_{O A}^2=\frac{G M m}{2 R} \\ & T_B=\frac{1}{2} m_B v_{O B}^2=\frac{G M(2 m)}{2(2 R)}=\frac{G M m}{2 R} \end{aligned} $
So, ratio of their kinetic energies is $ \frac{T_A}{T_B}=1 $
BETA
