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: Understand the concept
A satellite orbiting close to the surface has orbital radius \(R\). Gravity provides the centripetal force: \(\dfrac{GMm}{R^2} = m\dfrac{4\pi^2}{T^2}R\).

Step 2: Solve for T
\[ T^2 = \frac{4\pi^2R^3}{GM} \]

Step 3: Use the density
The mass is \(M = \rho\cdot\frac{4}{3}\pi R^3\). Substituting:
\[ T^2 = \frac{4\pi^2R^3}{G\rho\cdot\frac{4}{3}\pi R^3} = \frac{3\pi}{G\rho} \]

Step 4: Result
\(T = \sqrt{\dfrac{3\pi}{\rho G}}\), option (D). The radius cancels, so the period depends only on density. Options (A), (B) and (C) have the wrong numerical factors of \(4\pi\), \(2\pi\) and \(\pi\).

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