Question

A satellite in a circular orbit at a height $h$ from the earth surface (radius of earth $R$; $h < < R$ ). The minimum increased in its orbital velocity required, so that the satellite could escape from the earth’s gravitational field, is close to:
(Neglect the effect of atmosphere)
A. $\sqrt {gR} $
B. $\sqrt {\dfrac{{gR}}{2}} $
C. $\sqrt {gR} \left( {\sqrt 2 - 1} \right)$
D. $\sqrt {2gR} $

Answer 1

Hint: here, we will calculate the minimum velocity required by the satellite by subtracting the orbital velocity from the escape velocity. Now, orbital velocity is the speed required by the satellite to revolve around the orbit. Also, the escape velocity is the minimum speed required by the satellite to escape the gravitational influence.

Complete step by step answer:
It is given in the question that a satellite is revolving in a circular orbit. Let this satellite is at a height $h$ from the earth’s surface such that $h < < R$ . this means that $h$ is negligible as compared to the radius $R$ of the earth. Hence, it will not accelerate the satellite due to gravity.
Now, as the satellite is orbiting in circular orbit around the earth, therefore, the velocity of the satellite will be orbital velocity and is given by
${V_o} = \sqrt {\dfrac{{GM}}{{R + h}}} $
Now, as $h < < R$ , we get
${V_o} = \sqrt {\dfrac{{GM}}{R}} $
Now, the minimum velocity required to escape the gravitational force is known as escape velocity and is given by
${V_e} = \sqrt {\dfrac{{2GM}}{R}} $ (Because $h < < R$ )
Now, the minimum velocity required by the satellite to escape the earth’s gravitational force can be calculating by finding the difference between orbital velocity and the escape velocity, which is shown below
$v = {V_e} - {V_o}$
$ \Rightarrow \,v = \sqrt {\dfrac{{2GM}}{R}} - \sqrt {\dfrac{{GM}}{R}} $
Now, $g = \dfrac{{GM}}{{{R^2}}}$ , we get
$ \Rightarrow \,v = \sqrt {2gR} - \sqrt {gR} $
$ \Rightarrow \,v = \sqrt {gR} \left( {\sqrt 2 - 1} \right)$
Therefore, the minimum velocity require by the satellite to escape the gravitational field of earth is $\sqrt {gR} \left( {\sqrt 2 - 1} \right)$ .

So, the correct answer is “Option C”.

Note:
Here, in the above question, if height $h$ will be comparable to radius of earth $R$ , then we cannot put $g = \dfrac{{GM}}{{{R^2}}}$ in the values of orbital velocity and escape velocity.
We can only put this value in the expressions where $h$ will be negligible to $R$ .