A satellite of mass M/2 is revolving around earth in a circular orbit at a height of R/3 from earth surface. The angular momentum of the satellite is M √(GMR/x). The value of x is _________ , where M and R are the mass and radius of earth, respectively. ( G is the gravitational constant)

Solution

Physics sir · 3 steps

Measure the orbit from the centre of the earth, and L becomes m times the square root of GMr.

1. Orbit radius. The satellite is at height R/3 above the surface, so its distance from the earth's centre is r = R + R/3 = 4R/3.
2. Orbital speed. Gravity supplies the centripetal force: GMm/r² = mv²/r, so v = √(GM/r). Then L = mvr = m√GMr.

   L = M/2√(GM × 4R/3) = M/2 × 2√(GMR/3) = M√(GMR/3)

3. So. Matching with M√(GMR/x) gives x = 3.

The common slip: 12. Taking the orbit radius as the height R/3 instead of 4R/3 gives L = M√(GMR/12), so x = 12.

The idea to keep. For any circular orbit, L = m√GMr, with r measured from the centre of the planet. Always add the height to the radius before you substitute.

Answer from the NTA final answer key. Solution by PrepWiser.