JEE Main 2022 · 28 June · morning · Single correct · +4 / −1
Two planets A and B of equal mass are having their period of revolutions T A and T B such that T A = 2T B . These planets are revolving in the circular orbits of radii r A and r B respectively. Which out of the following would be the correct relationship of their orbits?
- (a) 2r_A² = r_B³
- (b) r_A³ = 2r_B³
- (c) r_A³ = 4r_B³
- (d) T_A² - T_B² = (π ²)/GM( r_B³ - 4r_A³ )
Answer and solution
Answer (c) r_A³ = 4r_B³
Solution
Physics sir · 3 steps
Square the period ratio, cube the radius ratio, and set them equal: that is all Kepler's third law asks.
- Kepler's third law. For orbits around the same star, T² ∝ r³. The planet's own mass does not appear, so the equal masses do not matter here.
T_A²/T_B² = r_A³/r_B³
- Put in the ratio. T_A = 2T_B, so T_A²/T_B² = 2² = 4. That gives r_A³/r_B³ = 4.
- So. r_A³ = 4r_B³, which is option (c).
The idea to keep. In Kepler's third law the period is squared and the radius is cubed. Square the period ratio first, then compare cubes of the radii.
Answer from the NTA final answer key. Solution by PrepWiser.