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(A) For a satellite revolving close to the surface of a planet, the orbital velocity $v$ is given by $v = \sqrt{gR}$, where $g$ is the acceleration du

Vedclass · Accepted Answer

$\sqrt{3\pi/\rho G}$

Vedclass · Answer

(A) For a satellite revolving close to the surface of a planet, the orbital velocity $v$ is given by $v = \sqrt{gR}$, where $g$ is the acceleration due to gravity and $R$ is the radius of the planet.Since the satellite is close to the surface, the time period $T$ is given by $T = \frac{2\pi R}{v} = \frac{2\pi R}{\sqrt{gR}} = 2\pi \sqrt{\frac{R}{g}}$.The acceleration due to gravity $g$ is related to the density $ho$ by the formula $g = \frac{GM}{R^2}$, where $M$ is the mass of the planet.Since $M = 	ext{Volume} 	imes 	ext{Density} = \frac{4}{3}\pi R^3 ho$, we have $g = \frac{G(\frac{4}{3}\pi R^3 ho)}{R^2} = \frac{4}{3}\pi R ho G$.Substituting this value of $g$ into the time period formula:$T = 2\pi \sqrt{\frac{R}{\frac{4}{3}\pi R ho G}} = 2\pi \sqrt{\frac{3}{4\pi ho G}} = 2\pi \frac{\sqrt{3}}{2\sqrt{\pi ho G}} = \sqrt{\frac{4\pi^2 \cdot 3}{4\pi ho G}} = \sqrt{\frac{3\pi}{ho G}}$.

$A$ satellite is revolving around a planet in a circular orbit close to its surface. Let $\rho$ be the mean density and $R$ be the radius of the planet; then the period of the satellite is $(G = \text{Universal constant of gravitation})$

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