Question

Two satellites of masses \( m \) and \( 1.5m \) are revolving around the Earth with different speeds in two circular orbits of heights \( R_E \) and \( 2R_E \) respectively, where \( R_E \) is the radius of the Earth. The ratio of the minimum and maximum gravitational forces on the Earth due to the two satellites is

• A. \( 2:5 \)
• B. \( 2:3 \)
• C. \( 1:2 \)
• D. \( 1:5 \)

Hint:

The gravitational force decreases with the square of the distance from the Earth's center. Always consider the total radial distance when calculating gravitational force.

Answer 1

The Correct Option is D

Satellite 1: Mass \( m_1 = m \), Height \( R_E \), Orbital radius \( r_1 = R_E + R_E = 2R_E \), Gravitational force \( F_1 = G \frac{m M_E}{(2R_E)^2} = G \frac{m M_E}{4R_E^2} \).
Satellite 2: Mass \( m_2 = 1.5m \), Height \( 2R_E \), Orbital radius \( r_2 = R_E + 2R_E = 3R_E \), Gravitational force \( F_2 = G \frac{1.5m M_E}{(3R_E)^2} = G \frac{1.5m M_E}{9R_E^2} = G \frac{m M_E}{6R_E^2} \).
Ratio of Forces: \( \frac{F_1}{F_2} = \frac{G \frac{m M_E}{4R_E^2}}{G \frac{m M_E}{6R_E^2}} = \frac{6}{4} = \frac{3}{2} \). This means \( F_1 = \frac{3}{2} F_2 \) or \( F_2 = \frac{2}{3} F_1 \).
The question asks for the ratio of the minimum and maximum forces exerted on the earth. The minimum force is \( F_2 \), and the maximum force is \( F_1 \). Therefore, \( F_2:F_1 = 2:3 \).
The correct ratio of \(F_2:F_1\) is \(2:3\).