Question

A moon completes one revolution around the Earth in $ 27 $ days. If the size (mass) of the moon becomes four times its original size, what will be the new time period?

• A. \( 54 \, \text{days} \)
• B. \( 13.5 \, \text{days} \)
• C. \( 108 \, \text{days} \)
• D. \( 6.75 \, \text{days} \)

Hint:

If mass increases by a factor \( n \), and assuming central force conditions hold, the time period becomes \( \frac{T}{\sqrt{n}} \).

Answer 1

The Correct Option is B

Time period for circular motion under gravity is given by: \[ T = \frac{2\pi r}{v}, \quad \text{and since} \quad v = \sqrt{\frac{GM}{r}}, \quad T = 2\pi \sqrt{\frac{r^3}{GM}} \] If the mass of the orbiting body (moon) increases, centripetal force must match gravitational force: \[ \frac{GMm}{r^2} = \frac{mv^2}{r} \Rightarrow v = \sqrt{\frac{GM}{r}} \Rightarrow T \propto \frac{1}{\sqrt{m}} \] Given: \[ T = 27 \, \text{days}, \quad m' = 4m \Rightarrow T' = \frac{T}{\sqrt{4}} = \frac{27}{2} = 13.5 \, \text{days} \]