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)/R^{2} 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 = GM^{2} / (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)^{1/2}F =
(3)^{1/2} GM^{2} /4

given that F_{r} = 3.4x10^{-9} we get

3.4x10^{-9} = (3)^{1/2} G M^{2} / 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 M^{2} /R = 1.18x10^{-8} J

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