Three particles each of mass m are placed at the three corners of an equilateral triangle. The centre of the triangle is at a distance x from either corner. If a mass M be placed at the centre, what will be the net gravitational force on it

Correct Answer: The correct option is Zero.

To find the net gravitational force acting on the mass $M$ placed at the center of the equilateral triangle, we can analyze the gravitational forces exerted on it by the three masses at the corners of the triangle using vector addition.

Step 1: Determine the force due to each particle
Let the three corners of the equilateral triangle be denoted as A, B, and C, with a mass $m$ placed at each corner. Let the center of the triangle be O, where the mass $M$ is placed.
According to Newton's law of universal gravitation, the magnitude of the force of attraction between the mass $M$ at the center and a mass $m$ at any corner (which is at a distance $x$) is given by:

$F=\frac{GMm}{x^2}$

Thus, the individual forces acting on the mass $M$ are:
- $F_A$ of magnitude $F$ directed along OA (towards corner A)
- $F_B$ of magnitude $F$ directed along OB (towards corner B)
- $F_C$ of magnitude $F$ directed along OC (towards corner C)

Step 2: Use vector addition to find the resultant force
Due to the symmetric nature of an equilateral triangle, the angle between any two adjacent lines connecting the center to the vertices (OA, OB, and OC) is exactly 120°.
Let us first find the resultant of the two forces $F_B$ and $F_C$. The formula for the resultant $F_{\mathrm{BC}}$ of two vectors of equal magnitude $F$ at an angle of 120° is:

$F_{\mathrm{BC}}=\sqrt{F^2+F^2+2FFcos(120°)}$

Since $cos(120°)=-0.5$, we substitute this value in the equation:

$F_{\mathrm{BC}}=\sqrt{F^2+F^2+2F^2(-0.5)}$

$F_{\mathrm{BC}}=\sqrt{F^2+F^2-F^2}=F$

The resultant force $F_{\mathrm{BC}}$ has a magnitude of $F$ and acts along the bisector of the angle between OB and OC, which points directly opposite to the vector OA (away from A).

Step 3: Combine all forces to get the net force
We now have two forces remaining: the force $F_A$ of magnitude $F$ pointing towards corner A, and the resultant force $F_{\mathrm{BC}}$ of magnitude $F$ pointing in the exact opposite direction. Adding these two collinear and opposite vectors yields:

$F_{\mathrm{net}}=F_A-F_{\mathrm{BC}}=F-F=0$

Therefore, the net gravitational force acting on the mass $M$ placed at the center of the triangle is Zero.