Question

Two satellites A and B of mass 'M' are orbiting the earth in circular orbits at altitudes 2R and 5R respectively, where R is the radius of the earth. The ratio of kinetic energies of satellite A and B will be

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

Hint:

For a satellite in a circular orbit, remember these energy relationships:
Kinetic Energy \(KE = \frac{GMm}{2r}\) (Always positive)
Potential Energy \(PE = -\frac{GMm}{r}\) (Always negative)
Total Energy \(E = KE + PE = -\frac{GMm}{2r}\)
Notice that \(KE = -E\) and \(KE = -\frac{1}{2}PE\). This can save time in problems involving ratios of different types of energy.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The kinetic energy of a satellite in a circular orbit depends on its mass, the mass of the central body (Earth), and its orbital radius. The orbital radius is the sum of the Earth's radius and the satellite's altitude.
Step 2: Key Formula or Approach:
For a satellite of mass \(m\) in a circular orbit of radius \(r\) around a central body of mass \(M_E\), the gravitational force provides the necessary centripetal force:
\[ \frac{G M_E m}{r^2} = \frac{m v^2}{r} \] From this, we can find the kinetic energy, \( KE = \frac{1}{2} m v^2 \):
\[ m v^2 = \frac{G M_E m}{r} \] \[ KE = \frac{1}{2} m v^2 = \frac{G M_E m}{2r} \] This shows that the kinetic energy is inversely proportional to the orbital radius \(r\).
Step 3: Detailed Explanation:
1. Determine the orbital radii:
The orbital radius \(r\) is measured from the center of the Earth. \( r = R_{earth} + \text{altitude} \).
For satellite A: altitude \(h_A = 2R\). So, orbital radius \(r_A = R + 2R = 3R\).
For satellite B: altitude \(h_B = 5R\). So, orbital radius \(r_B = R + 5R = 6R\).
2. Calculate the ratio of kinetic energies:
The mass of both satellites is given as 'M'.
\[ KE_A = \frac{G M_E M}{2 r_A} = \frac{G M_E M}{2(3R)} \] \[ KE_B = \frac{G M_E M}{2 r_B} = \frac{G M_E M}{2(6R)} \] The ratio is:
\[ \frac{KE_A}{KE_B} = \frac{\frac{G M_E M}{2(3R)}}{\frac{G M_E M}{2(6R)}} = \frac{1/(3R)}{1/(6R)} = \frac{6R}{3R} = \frac{2}{1} \] Step 4: Final Answer:
The ratio of the kinetic energies of satellite A to satellite B is 2:1.