Question

A satellite is launched into a circular orbit of radius \( R \) around the earth. A second satellite is launched into an orbit of radius \( 1.03R \). The time period of revolution of the second satellite is larger than the first one approximately by:

• A. 3 %
• B. 4.5 %
• C. 9 %
• D. 2.5 %

Hint:

The time period of a satellite in orbit is related to the radius of the orbit by \( T^2 \propto R^3 \). This allows us to calculate how changes in the orbital radius affect the time period.

Answer 1

The Correct Option is B

The time period \( T \) of a satellite in a circular orbit is given by Kepler's third law: \[ T^2 \propto R^3, \] where \( T \) is the time period of revolution and \( R \) is the radius of the orbit. Thus, the ratio of the time periods of the two satellites is: \[ \frac{T_2}{T_1} = \left( \frac{R_2}{R_1} \right)^{3/2}. \] Here, \( R_2 = 1.03 R \) and \( R_1 = R \), so: \[ \frac{T_2}{T_1} = \left( \frac{1.03 R}{R} \right)^{3/2} = (1.03)^{3/2}. \] Step 1: Now calculate \( (1.03)^{3/2} \): \[ (1.03)^{3/2} \approx 1.0457. \] This means: \[ T_2 \approx 1.0457 T_1. \] Step 2: The percentage increase in the time period is: \[ \text{Percentage increase} = \left( \frac{T_2 - T_1}{T_1} \right) \times 100 = (1.0457 - 1) \times 100 \approx 4.57 \%. \] Thus, the time period of the second satellite is larger than the first one by approximately \( 4.5\% \). 
Therefore, the correct answer is \( \boxed{4.5\%} \).