Question

If 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 earth, the height of the satellite above the surface of the earth is
A. $2R$
B. $\dfrac{R}{2}$
C. $R$
D. $\dfrac{R}{4}$

Answer 1

Hint: The gravitational pull should be equal to the needed centrifugal force on the satellite. Assume the height of the satellite. Find the gravitational pull of earth at that height using Newton’s theorem. Find the centrifugal force needed to maintain that height. Equate these two forces.

Formula Used:
The centrifugal force needed to maintain circular motion is given by,
$F=\dfrac{m{{v}^{2}}}{r}$

Where,
$v$ is the speed at any particular moment
$r$ is the radius of the circle
$m$ is the mass of the particle

The gravitational pull at a given point outside earth is given by,
$F=G\dfrac{mM}{{{r}^{2}}}$

Escape velocity of Earth is given by,
${{v}_{e}}=\sqrt{\dfrac{2GM}{R}}$

Where,
$m$ is the mass of the particle
$M$ is the mass of Earth
$r$ is the distance from the centre of earth
$G$ is the gravitational constant.

Complete step by step answer:
Let’s assume that the satellite is at a distance r from the centre of the earth.

Escape velocity of earth is,
${{v}_{e}}=\sqrt{\dfrac{2GM}{R}}$

Hence the speed of the satellite is,
${{v}_{o}}=\dfrac{{{v}_{e}}}{2}=\dfrac{1}{2}\sqrt{\dfrac{2GM}{R}}=\sqrt{\dfrac{GM}{2R}}$

So, we can write equation (1) as,
$F=\dfrac{m(\dfrac{GM}{2R})}{r}=\dfrac{GMm}{2Rr}$

This force should be equal to the gravitational pull at that location.
The gravitational pull at that height is given by,
$F=G\dfrac{mM}{{{r}^{2}}}$

Hence, we can write,
$\dfrac{GMm}{{{r}^{2}}}=\dfrac{GMm}{2Rr}$
$\Rightarrow r=2R$

So, the distance of the satellite from the centre of the earth is,
$(2R-R)=R$

Hence, the height of the satellite is, $R$

The correct option is - (C).

Note: All the satellites follow the same principle that we have mentioned in the solution. It uses the gravitational pull to maintain the orbit at a fixed location. As there is no atmospheric drag on the high satellites, it maintains a fixed height without using any energy for propulsion for years.