Question

A satellite of mass ' m ' is orbiting the earth of radius ' R ' at height ' h ' from the surface of earth. The total energy of the satellite is ( g = acceleration due to gravity at the earth's surface)

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

Hint:

- Total energy of orbiting satellite is always negative - $E = -\frac{GMm}{2r}$

Answer 1

The Correct Option is A

Concept:
Total energy of a satellite in circular orbit: \[ E = -\frac{GMm}{2r} \]

Step 1: Express $GM$ in terms of $g$.
At earth’s surface: \[ g = \frac{GM}{R^2} \Rightarrow GM = gR^2 \]

Step 2: Substitute in energy formula.
\[ E = -\frac{gR^2 \cdot m}{2r} \]

Step 3: Use orbital radius.
\[ r = R + h \]

Step 4: Final expression.
\[ E = -\frac{mgR^2}{2(R+h)} \]