If a satellite orbiting the Earth is 9 times closer to the Earth than the Moon, what is the time period of rotation of the satellite? Given rotational time period of Moon =27 days and gravitational attraction between the satellite and the moon is neglected.

Solution

Physics sir · 3 steps

Same planet at the centre, so Kepler's third law links the two orbits: T² goes as r³.

1. Same central body. The satellite and the Moon both orbit the Earth. For any two bodies orbiting the same mass, the square of the period is proportional to the cube of the orbit radius.

   Tₛ/Tₘ = (rₛ/rₘ)^(3/2)

2. Put in numbers. The satellite is 9 times closer, so rₛ/rₘ = 1/9. Then (1/9)^(3/2) = 1/27, because √9 = 3 and 3³ = 27.
3. So. Tₛ = 27 × 1/27 = 1 day. The answer is (d).

The idea to keep. For orbits around the same body, compare them with T² ∝ r³, so T ∝ r^(3/2). The central mass cancels, and a closer orbit always has a shorter period.

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