Question

The orbital velocity of an artificial satellite a circular orbit above the earth’s surface is v. For a satellite orbiting at an altitude of half of the earth’s radius, the orbital velocity is
A. $\left( \dfrac{3}{2} \right)v$
B. $\sqrt{\left( \dfrac{3}{2} \right)}v$
C. $\sqrt{\left( \dfrac{2}{3} \right)}v$
D. $\left( \dfrac{2}{3} \right)v$

Answer 1

Hint:Use the formula for the gravitational force and write the expressions for the force exerted by the earth both the satellites. Then use the formula for the centripetal acceleration and Newton’s second to find an expression for the orbital velocity.

Formula used:
$F=\dfrac{GMm}{{{r}^{2}}}$
$a=\dfrac{{{v}^{2}}}{r}$
$g=\dfrac{GM}{{{r}^{2}}}$

Complete step by step answer:
Let us first understand some points about the gravitational force of attraction between two bodies.According to the law of gravitation, every object in the universe attracts another object with a force called gravitational force. Similarly, the earth exerts a gravitational force of attraction (directed towards it) on the artificial satellite. It is due to this attractive force that the satellite revolves around the earth. The earth appears to be at rest since its mass is very large compared to that of the satellite.

The magnitude of the gravitational force is given as $F=\dfrac{GMm}{{{r}^{2}}}$ …. (i),
where G is the universal gravitational constant, M is the mass of earth, m is the mass of the satellite and r is the distance between the two objects (i.e. earth and satellite).
It is given that in this case the satellites are revolving in circular orbits. Therefore, they will have a centripetal acceleration. The centripetal acceleration of an object in a uniform circular is given as $a=\dfrac{{{v}^{2}}}{r}$,
where v is the speed of the object and r is the radius of the circular path.

From Newton’s second law of motion we know that $F=ma$.
$\Rightarrow F=ma=\dfrac{m{{v}^{2}}}{r}$ …. (ii).
Now, equate (i) and (ii).
$\Rightarrow \dfrac{GMm}{{{r}^{2}}}=\dfrac{m{{v}^{2}}}{r}$.
$\Rightarrow v=\sqrt{\dfrac{GM}{r}}$ …… (*)
It is given that when $r=R$, the orbital velocity of the satellite v.
$\Rightarrow v=\sqrt{\dfrac{GM}{R}}$ …. (iii)
For a satellite orbiting at an altitude of half of the earth’s radius, $r=R+\dfrac{R}{2}=\dfrac{3R}{2}$.

Let the velocity of the satellite in this orbit be v’.Therefore,
$\Rightarrow v'=\sqrt{\dfrac{GM}{\dfrac{3R}{2}}}=\sqrt{\dfrac{2GM}{3R}}$ …. (iv).
Now, divide (iv) by (iii).
$ \dfrac{v'}{v}=\dfrac{\sqrt{\dfrac{2GM}{3R}}}{\sqrt{\dfrac{GM}{R}}}=\sqrt{\dfrac{2}{3}}$.
$\therefore v'=\sqrt{\dfrac{2}{3}}v$.
Therefore, for a satellite orbiting at an altitude of half of the earth’s radius, the orbital velocity is equal to $\sqrt{\dfrac{2}{3}}v$.

Hence, the correct option is C.

Note: Equation (*) can be written as $v=\sqrt{\dfrac{GM}{{{r}^{2}}}r}$.And the acceleration of the object due the gravitational force orbit in a circular orbit of radius r is equal to $g=\dfrac{GM}{{{r}^{2}}}$. Therefore, the orbital velocity of the satellite can be written as $v=\sqrt{gr}$. The value of g is equal to $9.8m{{s}^{-2}}$ only at the surface of the earth.