Question

If the density of moon is $d$, the time period of revolution of an artificial satellite in a circular orbit very close to the moon is:
$\begin{aligned} & (A)\sqrt{\dfrac{3\pi }{16dG}} \\\ & (B)\sqrt{\dfrac{16\pi }{3dG}} \\\ & (C)\sqrt{\dfrac{9\pi }{dG}} \\\ & (D)\sqrt{\dfrac{3\pi }{dG}} \\\ \end{aligned}$

Answer 1

Since the satellite is revolving very close to the moon, the radius of orbit will be equal to the radius of the moon. Now, at first we will derive the formula for the time period of revolution of an object under the gravitational effect of the moon. Then, we shall use this formula to calculate the time period of revolution of the satellite.

Complete answer:
Let the mass of the moon be given by $M$ and the radius of moon be denoted by $R$ .
Then for a satellite of mass (say $m$) orbiting the moon at a fixed distance (say $d$ ) above the surface of the moon, the radius of orbit will be equal to:
$=R+d$
$\approx R$ [This is because the satellite is very close to the surface of the moon.]
Now, the gravitational pull of the moon on the satellite is also the centripetal force that is enabling the satellite to revolve around it.
Let the speed of revolution be given by $v$ . Then we can write:
Centripetal force $=\dfrac{m{{v}^{2}}}{R}$ [Let this expression be equation number (1)]
Force of gravity $=m{{g}_{M}}$
Or, $\Rightarrow {{F}_{g}}=\dfrac{GMm}{{{R}^{2}}}$ [Let this expression be equation number (2)]
On equating equation number (1) and (2), we get:
$\begin{aligned} & \Rightarrow \dfrac{m{{v}^{2}}}{R}=\dfrac{GMm}{{{R}^{2}}} \\\ & \Rightarrow v=\sqrt{\dfrac{GM}{R}} \\\ \end{aligned}$
Now, the time period of revolution (say $T$) can be given as:
$\Rightarrow T=\dfrac{2\pi R}{v}$
Putting the values, we get:
$\begin{aligned} & \Rightarrow T=\dfrac{2\pi R}{\sqrt{\dfrac{GM}{R}}} \\\ & \Rightarrow T=\sqrt{\dfrac{4{{\pi }^{2}}{{R}^{2}}}{\dfrac{GM}{R}}} \\\ & \Rightarrow T=\sqrt{\dfrac{3\pi }{\dfrac{MG}{\dfrac{4\pi {{R}^{3}}}{3}}}} \\\ & \therefore T=\sqrt{\dfrac{3\pi }{dG}}\text{ }\left[ \text{Using, density(d)=}\dfrac{M}{\dfrac{4\pi {{R}^{3}}}{3}} \right] \\\ \end{aligned}$
Hence, the time period of revolution of the satellite in the circular orbit comes out to be $\sqrt{\dfrac{3\pi }{dG}}\text{ }$.

Hence, option (D) is the correct option.

Note:
We have worked under the assumption that the satellite is very close to the moon. If we had not considered this fact then, then the acceleration due to gravity at the surface of the moon and at some finite comparable distance would not be equal. It is because the acceleration due to gravity at moon will be inversely proportional to the square of the distance from the center of the moon.