Question

For a given density of a planet, the orbital speed of satellite near the surface of the planet of radius \(R\) is proportional to

• A. \(R^{1/2}\)
• B. \(R^{3/2}\)
• C. \(R^{-1/2}\)
• D. \(R^0\)

Hint:

For given planetary density, the orbital period $T = \sqrt{3\pi/(G\rho)}$ is independent of $R$; hence $v_0 = 2\pi R/T \propto R$.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
Orbital speed near the surface: \(v_0 = \sqrt{GM/R}\). Express \(M\) in terms of density \(\rho\) and \(R\).

Step 2: Detailed Explanation:
\(M = \dfrac{4}{3}\pi R^3 \rho\), so: \[ v_0 = \sqrt{\frac{G \cdot \frac{4}{3}\pi R^3 \rho}{R}} = \sqrt{\frac{4\pi G\rho}{3}}\cdot R \] Applying Newton’s second law to a circular orbit, we have \[ \frac{mv^2}{r} = \frac{4\pi^2 r m}{T^2} = \frac{G M m}{r^2} \] where \( m \) is the mass of satellite, and \( v \) is the orbital speed. \( T \) is the time period \[ T = \frac{2\pi r^{3/2}}{\sqrt{G M}} \] For \[ r \approx R \] and \[ M = \frac{4}{3} \pi R^3 \rho \quad (\rho = \text{density of planet}) \] \[ T = \sqrt{\frac{3 \pi}{\rho G}} \] i.e. is independent of \( R \).
Step 3: Final Answer:
orbital speed \(\propto R^0\) (independent of radius for given density).