Question

The radius of a planet is $R_1$, and a satellite revolves around it in a radius $R_2$ time period of revolution is T. Find the acceleration due to gravity.

A. $\dfrac{{4{\pi ^2}{\rm{R}}_2^3}}{{{\rm{R}}_1^2{{\rm{T}}^2}}}$

B. $\dfrac{{4{\pi ^2}{\rm{R}}_2^3}}{{{{\rm{R}}_1}{{\rm{T}}^2}}}$

C. $\dfrac{{2{\pi ^2}{\rm{R}}_2^3}}{{{{\rm{R}}_1}{{\rm{T}}^2}}}$

D. $\dfrac{{4{\pi ^2}{{\rm{R}}_2}}}{{{{\rm{T}}^2}}}$

Answer 1

We know that there are satellites revolving around the earth. It is because the satellite is revolving around earth due to the gravitational attraction between earth and the satellite. The centripetal force required to maintain the satellite in its orbit is provided by the gravitational attraction of the earth.

Complete step by step answer:

Let us assume that the orbit of the satellite is circular, and the satellite is revolving around the earth in a circular path. The gravitational force between the earth (mass M) and the satellite (mass m) provide a centripetal force to the satellite to move in a circular orbit.

i.e. centripetal force = gravitational force

$\dfrac{{m{v^2}}}{{\rm{R}}} = \dfrac{{GMm}}{{{{\rm{R}}^2}}}$

$\Rightarrow \dfrac{{GM}}{R} = {v^2}$

Now, the velocity of the satellite is v = $\dfrac{{2\pi R}}{T}$

Where 2πR is the circumference of the circle, and T is the time period.

Substitute the value of velocity in the above equation, we get

$\dfrac{{GM}}{R} = {\dfrac{{4{\pi ^2}R}}{{{T^2}}}^2}$

$\Rightarrow GM = \dfrac{{4{\pi ^2}{R^3}}}{{{T^2}}}$

$\therefore {T^2} = \dfrac{{4{\pi ^2}{R^3}}}{{GM}}$

We use this relation to solve the above problem, let two radii $R_1$ (planet) and $R_2$ (satellite).

Now the time period of a satellite revolving around the planet of mass $M$ in radius $R_2$ is given as,

$GM = \dfrac{{4{\pi ^2}{R_2}^3}}{{{T^2}}}$

Also, acceleration due to gravity of the planet of mass $M$ of radius $R_1$ is given as,

$g = \dfrac{{GM}}{{R_1^2}}$

Put the value of GM in the second equation,

$g = \dfrac{{4{\pi ^2}{R_2}^3}}{{R_1^2{T^2}}}$

Hence the correct option is (A).

Note:
The acceleration experienced by a body due to the gravitational force of the earth is known as acceleration due to gravity. Here “g” is the vector quantity; it depends on the mass of the planet and radius of the planet. It is independent of mass of the body. Also, the value of g is different for different heavy bodies.