Question

The time period of a satellite revolving above earth's surface at a height equal to R will be
(given: \(g= \pi^2 \ m/s^2\), R= radius of earth)

• A. \(\sqrt {4R}\)
• B. \(\sqrt {2R}\)
• C. \(\sqrt {8R}\)
• D. \(\sqrt {32R}\)

Answer 1

The Correct Option is D

Step 1: Using the time period formula for a satellite: \[ T^2 = \frac{4\pi^2 r^3}{GM} \] Step 2: Substituting \( r = 2R \): \[ T^2 = \frac{4\pi^2 (2R)^3}{GM} = \frac{4 \times 8 \times \pi^2 R^3}{GM} = \frac{4 \times 8 \times g \times R^3}{GR^2} \] Step 3: Simplifying the expression: At the surface of the Earth, we have \( g = \frac{GM}{R^2} \). 
So, substituting \( GM = gR^2 \) and \( \pi^2 = g \), we get: \[ T^2 = 32R \quad \Rightarrow \quad T = \sqrt{32R} \]