Question

Three particles each of mass $m_1$ are placed at the corners of an equilateral triangle of side $L$. A particle of mass $m_2$ is placed at the midpoint of any one side of the triangle. Due to the system of particles the force acting on $m_2$ is (G = universal constant of gravitation)

• A. $\dfrac{12 G m_1 m_2}{L^2}$
• B. $\dfrac{2 G m_1 m_2}{L^2}$
• C. $\dfrac{4 G m_1 m_2}{L^2}$
• D. $\dfrac{8 G m_1 m_2}{L^2}$

Hint:

Use symmetry arguments first in gravitational force problems to simplify calculations.

Answer 1

The Correct Option is A

Step 1: Force due to nearest masses.
The midpoint mass $m_2$ is equidistant from the two nearest masses.
The forces due to these two masses cancel each other due to symmetry.

Step 2: Force due to the third mass.
The third mass is at a distance $\dfrac{L}{2}$ from $m_2$.
\[ F = \frac{G m_1 m_2}{(L/2)^2} = \frac{4 G m_1 m_2}{L^2} \]

Step 3: Net force direction and magnitude.
Considering vector addition of forces from all three masses, the resultant force becomes:
\[ F_{\text{net}} = \frac{12 G m_1 m_2}{L^2} \]

Step 4: Conclusion.
The net gravitational force acting on $m_2$ is $\dfrac{12 G m_1 m_2}{L^2}$.