Question

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

• A. \(\sqrt{\frac{4π}{ρG}}\)
• B. \(\sqrt{\frac{2π}{ρG}}\)
• C. \(\sqrt{\frac{π}{ρG}}\)
• D. \(\sqrt{\frac{3π}{ρG}}\)

Hint:

For a satellite close to the surface, the orbital radius equals R and gravity supplies the centripetal force.

Answer 1

The Correct Option is D

Step 1: Use g at the surface
Near the surface, the orbital speed satisfies $\dfrac{v^2}{R} = g$, so $T = 2\pi\sqrt{\dfrac{R}{g}}$.

Step 2: Express g
$g = \dfrac{GM}{R^2} = \dfrac{G}{R^2}\cdot\dfrac{4}{3}\pi R^3\rho = \dfrac{4}{3}\pi G\rho R$.

Step 3: Substitute
$T = 2\pi\sqrt{\dfrac{R}{\frac{4}{3}\pi G\rho R}} = 2\pi\sqrt{\dfrac{3}{4\pi G\rho}}$.

Step 4: Simplify
$T = \sqrt{\dfrac{4\pi^2\cdot3}{4\pi G\rho}} = \sqrt{\dfrac{3\pi}{G\rho}}$.

Final Answer:
The period is sqrt(3 pi/(rho G)). This is option (D). \[ \boxed{\text{(D) }\sqrt{\frac{3\pi}{\rho G}}} \]