The time period of a satellite, revolving above earth's surface at a height equal to R will be
(Given g=π² m / s², R= radius of earth)

Solution

Physics sir · 4 steps

Measure the orbit radius from Earth's centre, so it is 2R, then swap GM for gR².

1. Orbit radius. The satellite is at height R above the surface, so its distance from Earth's centre is r = R + R = 2R.
2. Remove GM. The question gives g, not G or M. At the surface g = GM/R², so GM = gR². Put this into T = 2π√(r³/GM).

   T = 2π√((2R)³/gR²) = 2π√(8R/g)

3. Use g = π². With g = π², √g = π, so T = 2π/π√8R = 2√8R. Taking the 2 inside the root gives √(4 × 8R) = √32R.
4. So. The time period is √32R, which is option (a).

The idea to keep. In orbit questions, r is always measured from the centre of the Earth. If g is given instead of G and M, replace GM with gR² straight away.

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