Consider two satellites S₁ and S₂ with periods of revolution 1 hr. and 8 hr. respectively revolving around a planet in circular orbits. The ratio of angular velocity of satellite S₁ to the angular velocity of satellite S₂ is :

Solution

Physics sir · 3 steps

Angular velocity is just 2π over the period, so the faster satellite has the bigger ω.

1. Link omega and T. One full revolution is an angle of 2π, covered in time T. So the angular velocity is ω = 2π/T.

   ω₁/ω₂ = (2π/T₁)/(2π/T₂) = T₂/T₁

2. Put in numbers. T₁ = 1 hr and T₂ = 8 hr, so ω₁/ω₂ = 8/1. The units cancel, so hours are fine here.
3. So. ω₁ : ω₂ = 8 : 1, which is option (b). S₁ goes round 8 times while S₂ goes round once.

The idea to keep. Since ω = 2π/T, angular velocity is inversely proportional to period. Flip the ratio of periods to get the ratio of angular velocities.

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