Question

three point masses of mass m are placed at the vertices of an equilateral triangle of side a. The gravitational potential due to these masses at the centroid of the triangle is

Answer 1

Answer:

Refer to the attached image for diagram.

According to the question, three masses 'm' are placed at three corners of an equilateral triangle.

Now we know that formula for calculating Gravitational Potential is:

$\boxed{\text{Gravitational Potential}=\dfrac{-GM}{R}}$

Now we need to calculate the distance from each vertex to the centroid O of the triangle ABC. According to geometry, equilateral triangle has each angle to measure 60°. Also in the case of an equilateral triangle all the medians act as angle bisectors as well.

Therefore if you consider Δ OBD,

∠ OBD = 30°

Therefore using trigonometry we can find OB.

Assuming each side to be 'a' units we get, BD to be a/2

⇒ Cos 30 = BD / OD

⇒ √3 / 2 = a / 2 OD

⇒ OD = a × 2 / √3 × 2

⇒ OD = a/√3

Therefore distance between the mass and the centroid is a/√3.

Therefore One mass at A the Potential would be:

$\boxed{ \text{ Potential at A} = \dfrac{ -\sqrt{3}G m}{a}}$

Therefore by applying the rule of symmetry we can say that it is equal for the other two points as well. Therefore Net Potential would be:

⇒ Potential at A + Potential at B + Potential at C

⇒ 3 × Potential at A

⇒ -3√3 ( Gm / a )