Question

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?

Answer 1

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
     

F_r is resultant of these two forces. Horizontal component of these forces cancel out.
hence F_r = 2F cos 30 = (3)^1/2F = (3)^1/2 GM² /4
given that F_r = 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