QJEEM23Ph60c5: JEE Main 2023 Physics - Class 11 JEEM23Ph60c5 Solution: The time period of a satellite, revolving above earth's surface …

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:

$F_c = \frac{mv^2}{r}$

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

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

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 $g = \pi^2 \,\mathrm{m/s^2}$ , we can express the gravitational constant G in terms of the Earth's radius R and mass M:

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

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:

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