Question

An artificial satellite is moving in a circular orbit around the earth with a speed equal to half the magnitude of the escape velocity from the earth. The height $(h)$ of the satellite above the earth’s surface is (Take radius of earth as $R_e$).
A. $h=R_e ^2$
B. $h=R_e$
C. $h=2R_e$
D. $h=4R_e$

Answer 1

Hint: The velocity of satellite has been given as half of that of escape velocity, so we will use the formulae for the escape velocity and the velocity of satellite orbiting around earth, we will find the relation between the height of satellite and radius of earth.

Formulae used:
Escape velocity, $v_e =\sqrt{2gR_e}$
Orbital velocity of satellite, $v_s =\dfrac{\sqrt{GM_e}}{R_e +h}$
Acceleration due to gravity, $g=\dfrac{GM_e}{R_e ^2}$

Complete step by step solution:
We have been given that the speed of the artificial satellite$(v_s)$ is half of the magnitude of escape velocity of earth $(v_e)$ and its height from the earth’s surface is $h$.
Since, the expression for escape velocity from the earth is given by
$v_e = \sqrt{2gR_e}$ ………. (i)
where $R_e$ is the radius of earth and $g$ is the acceleration due to gravity.
Now, the orbital velocity of a satellite at a height $h$ above the earth’s surface is given by,
$v_s =\dfrac{\sqrt{GM_e}}{R_e +h}$ ………. (ii)
From the gravity equation, we can write $GM_e =gR_e ^2$, where $G$ is the gravitational constant.
$\therefore v_s =\dfrac{\sqrt{gR_e ^2}}{R_e +h}$ ………. (ii)
According to the question, $v_s =\dfrac{v_e}{2}\implies \dfrac{v_e}{v_s}=2$
Substituting equations (i) and (ii), we get
$\dfrac{\sqrt{2gR_e}}{\left(\dfrac{\sqrt{gR_e ^2}}{R_e +h}\right)}=2$
$\implies \dfrac{\sqrt{2(R_e +h)}}{R_e}=2$
On squaring both sides, we get
$\implies \dfrac{2(R_e +h)}{R_e}=4\implies R_e +h=2R_e$
Thus, $h=R_e$
Hence, option b is the correct answer.

Note: The height of the satellite is measured from the surface of earth and not from the centre and that is why the radius has been added in the expressions for velocity, If, we would have considered earth a point object, the height would have been from the centre and radius of earth can be subtracted later.