Question

An artificial satellite is revolving around a planet of radius \( R \) in a circular orbit of radius \( a \). If the time period of revolution of the satellite, \( T \propto a^{3/2}g^xR^y \), then the values of \( x \) and \( y \) are respectively:

• A. \( 1, \dfrac{1}{2} \)
• B. \( -\dfrac{1}{2}, 1 \)
• C. \( -\dfrac{1}{2}, \dfrac{1}{2} \)
• D. \( -\dfrac{1}{2}, -1 \)

Hint:

To express time period in terms of \( g \) and \( R \), always use the substitution \( GM = gR^2 \), which helps eliminate the mass term.

Answer 1

The Correct Option is D

Step 1: Start with Kepler’s Third Law \[ T \propto \sqrt{\dfrac{a^3}{GM}} \] But \( g = \dfrac{GM}{R^2} \Rightarrow GM = gR^2 \) 
Step 2: Substitute \( GM \) in the original formula \[ T \propto \sqrt{\dfrac{a^3}{gR^2}} = a^{3/2}g^{-1/2}R^{-1} \] Step 3: Compare with given form Given: \[ T \propto a^{3/2}g^xR^y \Rightarrow x = -\dfrac{1}{2}, \quad y = -1 \]