Question 108: Gravitation

Question

The time period of a satellite, revolving above earth's surface at a height equal to $$\mathrm{R}$$ will be
(Given $$g=\pi^{2} \mathrm{~m} / \mathrm{s}^{2}, \mathrm{R}=$$ radius of earth)

Accepted Answer

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

Answer

$$\sqrt{32 R}$$

Answer

$$\sqrt{4 \mathrm{R}}$$

Answer

$$\sqrt{8 R}$$

Answer

$$\sqrt{2 R}$$

Gravitation

The time period of a satellite, revolving above earth's surface at a height equal to $\mathrm{R} will be (Given g=\pi^{2} \mathrm{~m} / \mathrm{s}^{2}, \mathrm{R}=$ radius of earth)

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