Question

An artificial satellite is revolving around a planet in a circular orbit close to its surface. The frequency of revolution (n) of a satellite in terms of density '\(\rho\)' and radius 'R' of the planet is ( G is the universal gravitational constant)

• A. \(\sqrt{\frac{G\rho }{2\pi }}\)
• B. \(\sqrt{\frac{G\rho }{3\pi }}\)
• C. \(\sqrt{\frac{3\pi }{G\rho }}\)
• D. \(\sqrt{\frac{2\pi }{G\rho }}\)

Hint:

Equate gravitational pull to centripetal force for an orbit near the surface and write the planet's mass using its density.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
For an orbit very close to the surface the orbital radius equals \(R\). Gravity provides the centripetal force: \(\frac{GM}{R^{2}}=R\omega^{2}\).

Step 2: Calculation
The mass is \(M=\frac{4}{3}\pi R^{3}\rho\). So \(\omega^{2}=\frac{GM}{R^{3}}=\frac{4\pi G\rho}{3}\). With \(\omega=2\pi n\): \(4\pi^{2}n^{2}=\frac{4\pi G\rho}{3}\).

Step 3: Frequency
\(n^{2}=\frac{G\rho}{3\pi}\), so \(n=\sqrt{\frac{G\rho}{3\pi}}\). It does not depend on \(R\).

Final Answer:
The frequency of revolution is the square root of G rho over three pi. \[ \boxed{\text{(B) }\sqrt{\dfrac{G\rho}{3\pi}}} \]