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.

Solution

Physics sir · 4 steps

Do not solve the motion at all: find how the collapse time scales with a and m, then compare.

1. What T depends on. The motion is set only by G, m and a. Each sphere's acceleration is proportional to Gm/a², and the distance it must cover is proportional to a.
2. Build the scaling. Distance is roughly acceleration times time squared, so a ∝ Gm/a²T². That gives the time scale below. The triangle has the same shape in both cases, so the constant is the same too.

   T ∝ √(a³/Gm)

3. Take the ratio. With a → 2a and m → 2m: T'/T = √(((2a)³/(2m))/(a³/m)) = √(8/2) = 2.
4. So. T' = 2 × 4 s = 8 s, so the answer is 8.

The common slip: 16. Forgetting the square root and using T ∝ a³/m gives 4 × 8/2 = 16, not 8.

The idea to keep. For any gravitational collapse or orbit, the time goes as √(a³/(GM)), just like Kepler's third law. Use that scaling and compare ratios instead of integrating the motion.

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