QJEEM21Phb2b0: JEE Main 2021 Physics - Class 11 JEEM21Phb2b0 Solution: The time period of a satellite in a circular orbit of radius R…

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 $T'$ and radius $9R$ , we can form an equation:

$\frac{{T'}^2}{T^2} = \frac{(9R)^3}{R^3}$

This simplifies to:

$\frac{{T'}^2}{T^2} = 729$

Taking the square root of both sides, we get:

$T' = T \times \sqrt{729} = T \times 27$

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