Question

If $\rho$ is the density of the planet, the time period of near by satellite is given by

• A. $\sqrt{\frac{4\pi }{3G\rho }}$
• B. $\sqrt{\frac{4\pi }{G_P}}$
• C. $\sqrt{\frac{3\pi }{G\rho }}$
• D. $\sqrt{\frac{\pi }{G_P}}$

Answer 1

The correct option is C

$\sqrt{\frac{3\pi }{G\rho }}$

Explanation for the correct option:

Step 1: Finding the density of a sphere

The density of a sphere is given by mass per unit volume, that is

$\begin{array}{c}\rho =\frac{M}{V}\\ =\frac{M}{\frac{4}{3}\pi r^3}\end{array}$

$\Rightarrow M=\rho \cdot \frac{4}{3}\pi r^3$

Step 2: Apply the formula of the time period of the satellite

The formula of the time period of the satellite is given by $T=2\pi \sqrt{\frac{r^3}{GM}}$, where $r$ is the distance between the center of the earth and the satellite $G$ is the gravitational constant and $M$ is the mass of the earth.

As the satellite is revolving near the planet, therefore the radius of the planet is taken as equal to the distance between the center of the planet and the satellite, then the time period of the nearby satellite is given by

$\begin{array}{c}T=2\pi \sqrt{\frac{r^3}{GM}}\\ =2\pi \sqrt{\frac{r^3}{G\times \rho \cdot \frac{4}{3}\pi r^3}}\\ =2\pi \sqrt{\frac{3}{G\cdot \rho \cdot 4\pi }}\\ =\sqrt{\frac{3\pi }{G\cdot \rho }}\end{array}$

Hence, the correct option is (C).