Gravitational Force of Attraction on a Circular Path – Rankers Physics

Acceleration Due to Gravity and its variation: Practice Problem & Solution

Two masses each equal to \(M\) are moving on a circular path of radius \(R\) about a common centre. The gravitational force of attraction between the masses has magnitude
\(F = \frac{GM^2}{R^2}\)
\(F = \frac{GM^2}{4R^2}\)
\(F = \frac{4GM^2}{R^2}\)
\(F = \frac{GM^2}{2R^2}\)

Short Solution (Best for NEET):

To solve this problem, we apply the core principles of Acceleration Due to Gravity and its variation. Understanding the underlying formula is key to arriving at the correct answer below:

For two identical masses to move on a circular path of radius \(R\) about a common centre, they must always be diametrically opposite. The distance between them is \(2R\). Thus, \(F = \frac{GM^2}{(2R)^2} = \frac{GM^2}{4R^2}\).

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