Question

A satellite is revolving around earth in its equatorial plane with a period $T$. If the radius of earth suddenly shrinks to half its radius without change in the mass. Then, the new period of revolution will be
A. $8T$
B. $2\sqrt{2}T$
C. $2T$
D. $T$

Answer 1

Hint: The centrifugal force will be equal to the gravitational force between earth and satellite. Use this expression to understand the relation between the time period and the variables. Determine the change in time period if the radius of the planet changes.

Formula used:
$mR{{\omega }^{2}}=\dfrac{GmM}{{{R}^{2}}}$
 $\omega =\dfrac{2\pi }{T}$

Complete step by step solution:
Consider a satellite of mass $m$ revolving around earth of mass $M$ in a circular orbit. The gravitational force of attraction that exists between the two bodies will be equal to the centrifugal force that exists on the satellite due to its circular motion. It can be seen as,
$\begin{align}
  & mR{{\omega }^{2}}=\dfrac{GmM}{{{R}^{2}}} \\
 & \omega =\sqrt{\dfrac{GM}{{{R}^{3}}}}
\end{align}$
Here,$\omega $ is the angular velocity of the satellite and $R$ is the radius of orbit measured from the center of earth. The expression for time period of satellite is calculated as,
$\begin{align}
  & \omega =\dfrac{2\pi }{T} \\
 & T=\dfrac{2\pi }{\omega } \\
 & =2\pi \sqrt{\dfrac{{{R}^{3}}}{GM}}
\end{align}$
The above expression shows that the time period of the satellite is independent of the mass of the satellite and radius of the planet. The time period depends only on the mass of earth and the distance of satellite from the center of earth.
Now, the questions mention that the radius of earth is halved keeping the mass the same. We can see from the obtained expression for the time period that there will be no change on the time period of the satellite.

Thus, the correct option is (D).

Note:
Deduce the relation for the time period of the satellite correctly. Do not get confused between the mass of earth and mass of satellites. The radius of orbit is measured from the center of earth and should not be confused with the radius of earth.