Question

The gravitational potential energy of a system of three masses $ m $, $ 2m $, and $ 3m $ placed at the three vertices of an equilateral triangle of side $ a $ is Options:

• A. \( \frac{-11 Gm}{a} \)
• B. \( \frac{-11 Gm^2}{a^2} \)
• C. \( \frac{-11 Gm^2}{a} \)
• D. \( \frac{-11 Gm}{a^2} \)

Hint:

To find the total gravitational potential energy of a system of point masses, sum the potential energies between each pair of masses in the system.

Answer 1

The Correct Option is C

The gravitational potential energy between two masses \( m_1 \) and \( m_2 \) separated by a distance \( r \) is given by: \[ U = -\frac{G m_1 m_2}{r} \] For the three masses \( m \), \( 2m \), and \( 3m \) placed at the vertices of an equilateral triangle of side \( a \), the potential energy is the sum of the potential energies between each pair of masses.
Step 1: Calculate potential energy between each pair of masses.
Between \( m \) and \( 2m \): \[ U_{m, 2m} = -\frac{G m \cdot 2m}{a} = -\frac{2 G m^2}{a} \] Between \( 2m \) and \( 3m \): \[ U_{2m, 3m} = -\frac{G \cdot 2m \cdot 3m}{a} = -\frac{6 G m^2}{a} \] Between \( m \) and \( 3m \): \[ U_{m, 3m} = -\frac{G \cdot m \cdot 3m}{a} = -\frac{3 G m^2}{a} \] Step 2: Total gravitational potential energy.
The total potential energy is the sum of the individual potential energies: \[ U_{\text{total}} = U_{m, 2m} + U_{2m, 3m} + U_{m, 3m} \] \[ U_{\text{total}} = -\frac{2 G m^2}{a} - \frac{6 G m^2}{a} - \frac{3 G m^2}{a} = -\frac{11 G m^2}{a} \] Final Answer: \[ \boxed{\frac{-11 G m^2}{a}} \]