Gravitation - Result Question 42
45. If $v_e$ is escape velocity and $v_0$ is orbital velocity of a satellite for orbit close to the earth’s surface, then these are related by : [2012M]
(a) $v_0=\sqrt{2} v_e$
(b) $v_0=v_e$
(c) $v_e=\sqrt{2 v_0}$
(d) $v_e=\sqrt{2} v_0$
Show Answer
Answer:
Correct Answer: 45. (d)
Solution:
(d) $v_e=\sqrt{\frac{2 G M}{R}} \Rightarrow v_0=\sqrt{\frac{G M}{R}}$
$v_e=\sqrt{2} v_0$
The orbital velocity of a satellite at a height $h$ above the surface of earth,
$v_0=\sqrt{\frac{G M}{(R+h)}}=\sqrt{\frac{g R^{2}}{(R+h)}}(\because G M=g R^{2})$
Here, $M=$ mass of earth,
$R=$ radius of earth,
$g=$ acceleration due to If the satellite is very close to the surrface of earth gravity earth on surface of then $h=0$
$\therefore \quad v_0=\sqrt{\frac{g R^{2}}{R}}=\sqrt{g R}=\sqrt{\frac{G M}{R}}$
