JEE Main 2025 Apr 3 Shift 1, Physics Q21: Satellite, Planetary Motion And Kepler’s Law

JEE Main2025Apr 3, Shift 1Physics

Three identical spheres of mass $m$ , are placed at the vertices of an equilateral triangle of length $a$ . When released, they interact only through gravitational force and collide after a time $T = 4$ seconds. If the sides of the triangle are increased to length $2 a$ and also the masses of the spheres are made $2 m$ , then they will collide after ______ seconds.

Your answer

Solution

Use dimensional analysis. Let $T \propto m^{x} G^{y} a^{z} .$

Dimensions: $[T] = [M]^{x} [M^{- 1} L^{3} T^{- 2}]^{y} [L]^{z} = M^{x - y} L^{3 y + z} T^{- 2 y} .$

Matching powers: $- 2 y = 1 \Rightarrow y = - \frac{1}{2}$ ; $x - y = 0 \Rightarrow x = - \frac{1}{2}$ ; $3 y + z = 0 \Rightarrow z = \frac{3}{2} .$

So $T \propto \frac{a ^{3}}{G m} .$

Ratio: $\frac{T ^{'}}{T} = \frac{( 2 a ) ^{3} / ( 2 m )}{a ^{3} / m} = \frac{8}{2} = 2.$

$T^{'} = 2 \times 4 = 8 s .$

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JEE Main 2025 Apr 3 Shift 1, Physics Q21: Satellite, Planetary Motion And Kepler’s Law

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