Question

The period of rotation of the moon around the earth is nearly $29\text{days}$. If the mass of the moon were $2$ fold as its present value, and all other things remain the same, then the period of rotation of the moon would be nearly given as,
$\begin{align}
  & A.29\sqrt{2}days \\
 & B.\dfrac{29}{\sqrt{2}}days \\
 & C.29\times 2days \\
 & D.29days \\
\end{align}$

Answer 1

Hint: The time period of rotation has been given by taking the ratio of the product of $2\pi $ and the square root of the cube of the radius of the orbit to the square root of the product of the gravitational constant and the mass of the earth. From this we can derive the relation between the time period and radius of the orbit. There is no role for the mass of the earth in this relation. This will help you in solving this question.

Complete step by step answer:
The time period is found by an equation which has been given as,
$T=\dfrac{2\pi {{r}^{\dfrac{3}{2}}}}{\sqrt{GM}}$
Where $r$ be the radius of the orbit, $G$ be the gravitational constant and $M$ be the mass of the earth.
As the gravitational constant, mass of the earth and $2\pi $are found to be constants, the time period will vary in accordance to the radius of the orbit mentioned as,
\[T\propto {{r}^{\dfrac{3}{2}}}\]
That is the square of the time period will be proportional to the cube of the radius of the orbit. As we look into this relation, we can see that this relation is not having any kind of relation with the mass of the earth. Hence the period of the rotation of the moon around the earth will remain unchanged even when we change the mass of the moon.
That is, the period of rotation is given as,
\[T=29days\]
This has been mentioned as the option D.

Note:
Kepler’s first law states that the orbits will be elliptical with the sun placed at one of its focus. The second law says that the distance to a planet from the sun will sweep out the same areas in equivalent times. The third law is the basis for solving this question.