Question

A space station is set up in space at a distance equal to the earth’s radius from the surface of the Earth. Suppose a satellite can be launched from the space station. Let ${v_1}$ and ${v_2}$ be the escape velocities of the satellite on the earth’s surface and space station, respectively. Then
A. ${v_2} = {v_1}$
B. ${v_2} < {v_1}$
C. ${v_2} > {v_1}$
D. (a), (b) and (c) are valid depending on the mass of satellite

Answer 1

Hint: In this question, we are going to apply the concept of escape velocity and it is the minimum velocity required to leave the earth’s gravitational field. We can solve this question by calculating and comparing the two velocities.

Complete step by step answer:
It is given in the question that ${v_1}$ and ${v_2}$ are the two escape velocities.
And the distance of a space station from the surface of the earth is $R.$
Where, $R$ is the radius of the Earth.
We know that the escape velocity is given by
${V_e} = \sqrt {2\dfrac{{GM}}{R}} $ . . . (1)
Where$G \to $Gravity
$M \to $Mass of the earth
$R \to $Radius of the earth
and let ${v_1}$ and ${v_2}$ be escape velocities of the satellite on the earth’s space and space station respectively.
Therefore, from equation (1), we have
${v_1}\propto \sqrt M $
Thus, we can say that the escape velocity depends on the mass of earth and the mass of the satellite. Since, $G$ and $R$ are constants. It is obvious that the mass of earth is much greater than the mass of a space station.
$\therefore {v_1} > {v_2}$
Hence the correct option is (B) ${v_2} < {v_1}$.

Note: For an object, which is to be escaped from an orbit, should have the value of total energy equal to zero. Therefore, for space station,
Total Energy (TE)$ = \dfrac{1}{2}mv_2^2 + \left( {\dfrac{{ - GMm}}{{2R}}} \right) = 0$
$ \Rightarrow {V_2}\sqrt {\dfrac{{GM}}{R}} $, Where R is the radius of the earth. And earth’s surface, ${V_1} = \sqrt {\dfrac{{2GM}}{R}} $.