The orbital angular momentum of a satellite is L, when it is revolving in a circular orbit at height h from earth surface. If the distance of satellite from the earth centre is increased by eight times to its initial value, then the new angular momentum will be -

Solution

Physics sir · 3 steps

Orbital speed falls as you go higher, so L grows only as the square root of r, not as r itself.

1. Orbital speed. In a circular orbit, gravity supplies the centripetal force: GMm/r² = mv²/r. So v = √(GM/r), and a higher satellite moves more slowly.

   L = mvr = m√(GM/r) r = m√GMr ⇒ L ∝ √r

2. New distance. The distance is increased by eight times its initial value. The increase is 8r, so the new distance is r + 8r = 9r.
3. So. L'/L = √(9r/r) = 3, so the new angular momentum is 3L, option (d).

The idea to keep. For circular orbits, write v from the gravity balance first. Then L = m√GMr, so L goes as √r. Also read the wording: "increased by n times" means the new value is (n+1) times the old.

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