Question

# Three identical balls are arranged so there is one ball at each corner of an equilateral...

Three identical balls are arranged so there is one ball at each corner of an equilateral triangle. Each side of the triangle is exactly 2 meters long.

(a) If each ball experiences a net force of 3.40 × 10-9 N because of the other two balls, what is the mass of each ball?

(b) What is the total gravitational potential energy associated with the interactions between these three balls?

Gravitational force between two point masses
F = G(M1)(M2)/R2    where R is distance between two masses.
Force F experienced by one of the masses kept at corners of equilateral triangles is given by
F = GM2 / (2)2
Forces from two masses are are at angle of 60 deg with each other as shown

Fr is resultant of these two forces. Horizontal component of these forces cancel out.
hence Fr = 2F cos 30 = (3)1/2F = (3)1/2 GM2 /4
given that Fr = 3.4x10-9 we get
3.4x10-9 = (3)1/2 G M2 / 4
M = 10.85 kg

b) Gravitational potential energy of two point masses is given by - Gm1 m2 / R
for more than two masses it is sum sum potantial energies of all possible pairs of masses.
With three masses , there are three pairs of masses. For all pairs distance is 2 m
Hence total potential energy is
- 3 G M2 /R = 1.18x10-8 J