JEE Main 2023 — Gravitation Question with Solution

For a satellite orbiting the Earth at a height equal to Earth's radius, its distance from the center of the Earth will be 2R, where R is the radius of the Earth.

Using the formula for the gravitational force:

$$F=G\frac{Mm}{r^2}$$

where F is the gravitational force, G is the gravitational constant, M is the mass of the Earth, m is the mass of the satellite, and r is the distance from the center of the Earth.

The centripetal force acting on the satellite is given by:

$$F_c=\frac{m{v}^2}{r}$$

Equating the gravitational force and the centripetal force, we get:

$$G\frac{Mm}{(2R{)}^2}=\frac{m{v}^2}{2R}$$

Solving for the orbital speed v, we get:

$$v^2=\frac{GM}{2R}$$

The circumference of the satellite's orbit is given by:

$$C=2\pi (2R)=4\pi R$$

The time period T of the satellite's orbit can be calculated as the ratio of the circumference to the orbital speed:

$$T=\frac{C}{v}=\frac{4\pi R}{\sqrt{\frac{GM}{2R}}}$$

Given that the acceleration due to gravity at the Earth's surface is $$g={\pi }^{2}m/{\mathrm{s}}^2$$, we can express the gravitational constant G in terms of the Earth's radius R and mass M:

$$g=\frac{GM}{R^2}\Rightarrow GM=g{R}^2={\pi }^2{R}^2$$

Substituting the expression for GM into the equation for the time period T, we get:

$$T=\frac{4\pi R}{\sqrt{\frac{{\pi }^2{R}^2}{2R}}}=\frac{4\pi R}{\sqrt{\frac{{\pi }^{2}R}{2}}}$$

$$T=\frac{4\pi R}{\sqrt{{\pi }^2}\sqrt{\frac{R}{2}}}=\frac{4\pi R}{\pi \sqrt{\frac{R}{2}}}=\frac{4R}{\sqrt{\frac{R}{2}}}$$

Multiplying the numerator and denominator by $$\sqrt{2}$$, we get:

$$T=\frac{4R\sqrt{2}}{\sqrt{R}}=4\sqrt{2R}=\sqrt{32R}$$