Question

The total energy of a satellite in a circular orbit at a distance \( (R + h) \) from the center of the Earth varies as

• A. \( \frac{1}{(R + h)^2} \)
• B. \( \frac{1}{(R + h)} \)
• C. \( \frac{1}{(R + h)^3} \)
• D. \( \frac{1}{(R + h)^2} \)

Hint:

For a satellite in a circular orbit, remember that the total energy is negative and inversely proportional to the square of the distance from the center of the Earth.

Answer 1

The Correct Option is D

The total energy \( E \) of a satellite in a circular orbit is given by the sum of its kinetic and potential energy. The kinetic energy \( K \) is: \[ K = \frac{1}{2} m v^2 \] where \( v \) is the orbital velocity of the satellite. The potential energy \( U \) is given by: \[ U = - \frac{G m M}{R + h} \] where \( G \) is the gravitational constant, \( m \) is the mass of the satellite, \( M \) is the mass of the Earth, and \( R + h \) is the distance from the center of the Earth to the satellite. Using the fact that the orbital velocity \( v \) of a satellite is: \[ v = \sqrt{\frac{GM}{R + h}} \] we can substitute this into the kinetic energy equation: \[ K = \frac{1}{2} m \left( \frac{GM}{R + h} \right) \] Thus, the total energy \( E \) is: \[ E = K + U = \frac{1}{2} m \left( \frac{GM}{R + h} \right) - \frac{G m M}{R + h} = - \frac{G m M}{2 (R + h)} \] This shows that the total energy \( E \) is inversely proportional to \( (R + h)^2 \). Therefore, the correct answer is: \[ \frac{1}{(R + h)^2} \]