Question

Two satellites are revolving at a distance of 2.5R and 7.5R from the center of the Earth. Find the ratio of time period of the satellites.

• A. \( \frac{1}{3} \)
• B. \( 1 \)
• C. \( \frac{1}{9} \)
• D. \( 1:2 \)

Hint:

The time period of a satellite is proportional to the square root of the cube of the distance from the center of the Earth.

Answer 1

The Correct Option is C

The time period of a satellite is given by Kepler's Third Law: \[ T \propto \sqrt{r^3} \] where \( T \) is the time period and \( r \) is the distance from the center of the Earth. For two satellites, we can write the ratio of their time periods as: \[ \frac{T_1}{T_2} = \left( \frac{r_1}{r_2} \right)^{3/2} \] Let \( r_1 = 2.5R \) and \( r_2 = 7.5R \). Substituting these values into the equation: \[ \frac{T_1}{T_2} = \left( \frac{2.5R}{7.5R} \right)^{3/2} = \left( \frac{1}{3} \right)^{3/2} = \frac{1}{9} \] 
Thus, the ratio of the time periods is \( \frac{1}{9} \).