Gravitational force at the center of a circular ring – 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

Solution Explained:

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 the symmetrical distribution of mass in a circular ring, the gravitational field at the center is zero. Therefore, the force on any mass placed at the center is zero.

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