Relative acceleration of two identical particles – Rankers Physics

Newton's Law of Gravitation: Practice Problem & Solution

Two identical particles of combined mass \(M\), placed in space with certain separation, are released. Interaction between the particles is only of gravitational in nature and there is no external force present. Acceleration of one particle with respect to the other when separation between them is \(R\), has a magnitude :
\(\frac{GM}{2R^2}\)
\(\frac{GM}{R^2}\)
\(\frac{2GM}{R^2}\)
not possible to calculate due to lack of information

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:

Each particle has mass \(m = M/2\). The force is \(F = \frac{G m^2}{R^2} = \frac{GM^2}{4R^2}\). Acceleration of each is \(a = \frac{F}{m} = \frac{GM}{2R^2}\). Relative acceleration is \(a_{\text{rel}} = 2a = \frac{GM}{R^2}\).

Leave a Reply

Your email address will not be published. Required fields are marked *