Gravitational Potential Energy of Three-Mass System – Rankers Physics

Gravitational Potential Energy: Practice Problem & Solution

Two masses each equal to \(M\) are moving on a circular path of radius \(R\) about another fixed mass \(M\) (at the centre of the circular path). The gravitational potential energy of the system is:
\(-\frac{GM^2}{2R}\)
\(-\frac{GM^2}{R}\)
\(-\frac{2GM^2}{R}\)
\(-\frac{5GM^2}{2R}\)

Solution Explained:

To solve this problem, we apply the core principles of Gravitational Potential Energy. Understanding the underlying formula is key to arriving at the correct answer below:

The total GPE of the three-mass system is \(U = -\frac{GMM}{R} - \frac{GMM}{R} - \frac{GMM}{2R} = -\frac{5GM^2}{2R}\) since the outer masses are at a distance of \(2R\) from each other and \(R\) from the center.

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