Gravitational force at the center of a bent rod – Rankers Physics

Newton's Law of Gravitation: Practice Problem & Solution

A thin rod of length \(L\) is bent to form a circle. Its mass is \(M\). What force will act on the mass \(m\) placed at the centre of the circle ?
\(\frac{4 \pi^2 GMm}{L^2}\)
\(\frac{GMm}{4 \pi^2 L^2}\)
\(\frac{2 \pi GMm}{L^2}\)
zero

Short Solution (Best for NEET):

To solve this problem, we apply the core principles of Newton's Law of Gravitation. Understanding the underlying formula is key to arriving at the correct answer below:

Due to symmetry, the gravitational forces exerted by all symmetric parts of the ring at the center cancel each other out, resulting in a net force of zero.

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