Time period of revolution of a satellite around a planet of radius R is T. Period of revolution around another planet. Whose radius is 3R but having same density is

The correct option is T.

Let us derive the time period of a satellite orbiting close to the surface of a planet to understand why it depends only on the density of the planet.

The orbital speed $v$ of a satellite revolving very close to the surface of a planet of radius $R$ and mass $M$ is given by the formula:
$v=\sqrt{\frac{GM}{R}}$
where $G$ is the universal gravitational constant.

The time period of revolution $T$ is the total distance of one orbit (the circumference of the planet) divided by the orbital speed:
$T=\frac{2\pi R}{v}$

Substituting the expression for $v$ into the time period formula:
$T=\frac{2\pi R}{\sqrt{\frac{GM}{R}}}=2\pi \sqrt{\frac{R^3}{GM}}$

Next, we express the mass $M$ of the planet in terms of its uniform density $\rho$ and radius $R$:
$M=\text{Volume}\times \text{Density}=\frac{4}{3}\pi R^3\rho$

Substituting this value of mass $M$ back into the equation for the time period $T$:
$T=2\pi \sqrt{\frac{R^3}{G\left(\frac{4}{3}\pi R^3\rho \right)}}$

Simplifying the expression by canceling $R^3$ from the numerator and denominator:
$T=2\pi \sqrt{\frac{3}{4\pi G\rho }}=\sqrt{\frac{3\pi }{G\rho }}$

From the final formula, we can see that the time period of a close-orbiting satellite depends only on the gravitational constant $G$ and the density $\rho$ of the planet. It is completely independent of the planet's radius $R$.

Since the second planet has the same density $\rho$, the period of revolution of a satellite orbiting close to it remains unchanged. Therefore, the period of revolution is still T.