Question

If mass of the earth were $2$ times the present mass, mass of the moon were half the present mass, and the moon were revolving round the earth at the same present distance, the time period of revolution of the moon would be (in day)
$\begin{aligned} & A)56 \\\ & B)28 \\\ & C)14\sqrt{2} \\\ & D)7 \\\ \end{aligned}$

Answer 1

Time period of revolution of the moon around the earth is given by Kepler’s third law of planetary motion. To solve this problem, we take the ratio of time periods of both the cases when mass of the earth is kept the same as well as when mass of the earth is twice its original mass. This ratio helps us to determine the time period of the moon around the earth, when the earth’s mass is considered to be twice its original mass.

Formula used:
$T=2\pi \sqrt{\dfrac{{{r}^{3}}}{GM}}$

Complete answer:
Kepler’s third law of planetary motion states that the square of time period of revolution of a celestial body$(A)$ revolving around another celestial body$(B)$ is proportional to the cube of orbital radius of celestial body$(A)$, which revolves around the other celestial body$(B)$. Mathematically, Kepler’s third law of planetary motion is given by
$T=2\pi \sqrt{\dfrac{{{r}^{3}}}{GM}}$
where
$T$ is the time period of revolution of a celestial body$(A)$ around another celestial body$(B)$
$r$ is the orbital radius of the celestial body$(A)$
$G$ is the gravitational constant
$M$ is the mass of the celestial body $(B)$ around which the other celestial body$(A)$ revolves
Let this be equation 1.
Considering the revolution of the moon around the earth, using equation 1, we have
${{T}_{m}}=2\pi \sqrt{\dfrac{{{r}_{m}}^{3}}{G{{M}_{e}}}}$
where
${{T}_{m}}$ is the time period of revolution of the moon around the earth
${{r}_{m}}$ is the orbital radius of the moon
$G$ is the gravitational constant
${{M}_{e}}$ is the mass of the earth
Let this be equation 2.
Coming to our question, we are required to find the time period of revolution of the moon when the mass of the earth is considered to be twice its original mass. Also, the distance of the moon from the earth or the orbital radius of the moon is considered to be the same, as provided. Clearly, if ${{M}_{e}}'$ represents twice the original mass of the earth, then, ${{M}_{e}}'$ is given by
${{M}_{e}}'=2{{M}_{e}}$
where
${{M}_{e}}$ is the original mass of the earth
Let this be equation 3.
Now, if $T{}_{m}'$ represents the time period of revolution of the moon around the earth, whose mass is considered to be ${{M}_{e}}'$, then, using equation 1, we have
${{T}_{m}}'=2\pi \sqrt{\dfrac{{{r}_{m}}^{3}}{G{{M}_{e}}'}}$
where
${{T}_{m}}'$ is the time period of revolution of the moon around the earth, whose mass is considered to be ${{M}_{e}}'$
${{r}_{m}}$ is the orbital radius of the moon
$G$ is the gravitational constant
Let this be equation 4.
Substituting equation 3 and equation 2 in equation 4, we have
${{T}_{m}}'=2\pi \sqrt{\dfrac{{{r}_{m}}^{3}}{G{{M}_{e}}'}}=2\pi \sqrt{\dfrac{{{r}_{m}}^{3}}{G(2{{M}_{e}})}}=\dfrac{{{T}_{m}}}{\sqrt{2}}$
Let this be equation 5.
Now, we know that the time period of revolution of the moon around the earth is equal to $28$ days, after doing the necessary substitutions and calculations. Therefore, equation 2 can be rewritten as
${{T}_{m}}=2\pi \sqrt{\dfrac{{{r}_{m}}^{3}}{G{{M}_{e}}}}=28days$
Using this value of ${{T}_{m}}$ in equation 5, we have
${{T}_{m}}'=\dfrac{{{T}_{m}}}{\sqrt{2}}=\dfrac{28}{\sqrt{2}}=14\sqrt{2}days$
Therefore, the time period of revolution of the moon around the earth, whose mass is considered as twice its original mass, is equal to $14\sqrt{2}days$.

Hence, the correct answer is option $C$.

Note:
Students need to be aware of the time period of revolution of the moon around the earth$(=28days)$, in order to solve this question easily. Similar questions on Kepler’s third law of planetary motion can also be asked regarding the time period of revolution of the earth around the sun too$(=365days)$. Therefore, it is important to be aware of these basic values.
Also, in this question, it is mentioned that the mass of the moon is considered to be half its original mass, in the second case. Since mass of the moon is not required to calculate its time period, students need not give importance to this statement.