JEE Main 2021 — Gravitation Question with Solution

Kepler's Third Law states that the square of the period of a satellite's orbit is proportional to the cube of the semi-major axis of its orbit. This relationship can be written as:

$$T^2\propto r^3$$

where:

• $T$ is the orbital period
• $r$ is the radius of the circular orbit (which serves as the semi-major axis in this case)

Considering two satellites, one with period $T$ and radius $R$, and another with unknown period $T^{\prime }$ and radius $9R$, we can form an equation:

$$\frac{{T^{\prime }}^2}{T^2}=\frac{(9R{)}^3}{R^3}$$

This simplifies to:

$$\frac{{T^{\prime }}^2}{T^2}=729$$

Taking the square root of both sides, we get:

$$T^{\prime }=T\times \sqrt{729}=T\times 27$$

So, the period of another satellite in a circular orbit of radius $9R$ is $27T$.