Question

If the moon describes a circular path of radius ‘r’ around the earth with uniform angular speed \[\omega \], what will the time period of revolution of the moon be?
A. \[2\pi \sqrt {\dfrac{{{r^2}}}{{g{R^2}}}} \]
B. \[2\pi \sqrt {\dfrac{{g{R^2}}}{{{r^2}}}} \]
C. \[2\pi \sqrt {\dfrac{{g{R^3}}}{{{r^3}}}} \]
D. \[2\pi \sqrt {\dfrac{{{r^3}}}{{g{R^2}}}} \]

Answer 1

Hint: In this question it has been given that the moon describes a circular motion around earth. Hence the moon experiences 2 forces: centripetal force and gravitational force. For the moon to keep moving in the orbit we will equate centripetal force to gravitational force as our first step, from there we find \[\omega \] and then we will find the time period.

Formula Used:
Gravitational force=\[{F_G} = \dfrac{{GMm}}{{{r^2}}}\]
Where G=gravitational constant, M=mass of planet, r=radius of planet
Centripetal force =\[{F_C} = m{\omega ^2}r\]
 time period T=\[T = \dfrac{{2\pi }}{\omega }\]
and acceleration due to gravity, \[g = \dfrac{{GM}}{{{R^2}}}.................(1)\]

Complete answer:
In the given question the moon is moving around the earth and to allow this motion we need to know that centripetal force will be equal to gravitational force to keep the moon revolving. If this condition is not satisfied the moon would not orbit around the earth.
Hence:
\[{F_c} = {F_g}\]
\[m{\omega ^2}r = \dfrac{{GMm}}{{{r^2}}}\]
\[{\omega ^2} = \dfrac{{GMm}}{{m{r^3}}}\]
\[\omega = \sqrt {\dfrac{{GMm}}{{m{r^3}}}} \]
\[\omega = \sqrt {\dfrac{{GM}}{{{r^3}}}} ...............(2)\]
And from equation 1 we see that \[GM = g{R^2}\] hence we will substitute this value in equation 2
\[\omega = \sqrt {\dfrac{{g{R^2}}}{{{r^3}}}} \]
We have found the value of angular frequency \[\omega \], hence now we will find time period T.
\[T = \dfrac{{2\pi }}{\omega }\]
\[T = \dfrac{{2\pi }}{{\sqrt {\dfrac{{g{R^2}}}{{{r^3}}}} }}\]
\[T = 2\pi \sqrt {\dfrac{{{r^3}}}{{g{R^2}}}} \]
Hence the period of revolution of the moon will be \[T = 2\pi \sqrt {\dfrac{{{r^3}}}{{g{R^2}}}} \]
Hence the correct answer to this question is option D.

Note: We can also solve this question taking frequency into consideration, as \[\omega = 2\pi f\] where \[\omega \] is angular frequency and f is frequency. After finding \[\omega \] we will find frequency. The time period is given by \[T = \dfrac{1}{f}\] and then we will solve the time period.