Question

The period of the moon's rotation of earth is nearly 29 days. If moon’s mass were 2 times its present value, and all other things remain unchanged, then the period of Moon’s rotation will be nearly:
A. $29\sqrt 2$ days
B. $\dfrac{{29}}{{\sqrt 2 }}$ days
C. $29 \times 2$ days
D. $29$ days

Answer 1

Recall the formula for orbital velocity to derive the formula for the time period of the revolution of the satellite around the planet once. Identify the values each variable represents in the formula, and check whether there is a value for the mass of the satellite.

Complete step by step solution:
We know that the formula of time period of a satellite orbiting around a planet of mass $M$ is,
$T = \dfrac{{Circumference}}{{{V_{Orbital}}}} = \dfrac{{2\pi r}}{{{V_{Orbital}}}} \\\ T = 2\pi r\sqrt {\dfrac{r}{{GM}}} = 2\pi \sqrt {\dfrac{{{r^3}}}{{GM}}} \\\$
Here,
$T$ represents the time period of an orbit
$r$ represents the radius of the orbit
$G$ represents the universal gravitational constant
And $M$ represents the mass of the planet around which the satellite revolves.
Since there is no dependency of the satellite’s mass in the formula for orbital time period, therefore the mass of the satellite does not affect the time period of the orbit.
Thus, the time period of the moon’s rotation around the Earth will remain the same, i.e., $29$ days, even if the mass of the moon became 2 times its present value and all the other things remained unchanged.
Thus, option D is the right choice.

Note:
Orbital velocity is the velocity at which an object revolves around another object. Such objects that travel in a uniform circular motion around the Earth are called to be in Earth’s orbit.