Question

A satellite of mass \( \frac{M}{2} \) is revolving around Earth in a circular orbit at a height of \( \frac{R}{3} \) from the Earth's surface. The angular momentum of the satellite is \( M \sqrt{\frac{GM R}{x}} \). The value of \( x \) is:

Hint:

The angular momentum of a satellite depends on the mass, orbital velocity, and radius of the orbit. Use the relevant formulas to find the angular momentum in different scenarios.

Answer 1

Correct Answer: 3

The angular momentum \( L \) of a satellite in a circular orbit is given by: \[ L = m v r, \] where: - \( m \) is the mass of the satellite, - \( v \) is the orbital velocity, - \( r \) is the radius of the orbit. The orbital velocity \( v \) is given by: \[ v = \sqrt{\frac{GM}{r}}, \] where \( G \) is the gravitational constant, and \( r \) is the distance from the center of the Earth. The satellite is at a height of \( \frac{R}{3} \) from the Earth's surface, so the total distance from the center is: \[ r = R + \frac{R}{3} = \frac{4R}{3}. \] Now, the angular momentum becomes: \[ L = m \cdot \sqrt{\frac{GM}{r}} \cdot r = m \cdot \sqrt{\frac{GM}{\frac{4R}{3}}} \cdot \frac{4R}{3}. \] Substitute \( m = \frac{M}{2} \): \[ L = \frac{M}{2} \cdot \sqrt{\frac{3GM}{4R}} \cdot \frac{4R}{3}. \] After simplifying, we get: \[ L = M \sqrt{\frac{GM R}{x}}, \] where \( x = 4 \). Thus, the value of \( x \) is 4.