Question

If the mass of earth were $4$ times the present mass, the mass of the moon were half the present mass and the moon was revolving round the earth at twice the present distance, the time period of revolving of the moon would be (in days)
(Take the normal time period of moon be $28$ days)

• A. $56\sqrt{2}$
• B. $28\sqrt{2}$
• C. $14\sqrt{2}$
• D. $7\sqrt{2}$

Answer 1

The correct option is B $28\sqrt{2}$

The time period of a satellite is given by

$T=\frac{2\pi r^{3/2}}{\sqrt{GM}}$

where, $r\to$ radius of orbit
$M\to$ mass of earth

Since, $\pi$ and $G$ are constant

$\therefore T\propto \frac{r^{3/2}}{\sqrt{M}}$

$\Rightarrow \frac{T^{\prime }}{T}={\left(\frac{r^{\prime }}{r}\right)}^{3/2}\left(\sqrt{\frac{M}{M^{\prime }}}\right)$

From question, $r^{\prime }=2r;M^{\prime }=4M$

$\Rightarrow \frac{T^{\prime }}{T}={\left(\frac{2r}{r}\right)}^{3/2}\left(\sqrt{\frac{M}{4M}}\right)=\sqrt{2}$

Present time period of moon in normal condition is $28\text{days}$.

$\Rightarrow T^{\prime }=\sqrt{2}T=28\sqrt{2}\text{days}$

Why this question: Note - The time period of the satellite is independent of its mass.