Question

A figure shows four particles, each of mass 20.5 g, that form a square with an edge length of d = 0.480 m. If d is reduced to 0.160 m, what is the change in the gravitational potential energy of the four-particle system?

Answer 1

A figure shows four particles, that form a square which given below as -

From an above figure, we have

U = - (1/2) G m² (4) [(1/d) + (1/d) + (1/d 2)]

U = - 2 G m² [(2 / d) + (1 / d 2)]

U = - 2 G m² [2 + (1 / 2)] / d

Change in the gravitational potential energy of four-particle system which will be given as :

U = U₁ - U₂ = - 2 G m² [2 + (1 / 2)] [1 / d₁ - 1 / d₂]

where, G = gravitational constant = 6.67 x 10^-11 Nm²/kg²

m = mass of each particles = 0.0205 kg

then, we get

U = - 2 (6.67 x 10^-11 Nm²/kg²) (0.0205 kg)² (2 + 0.7072) [1 / (0.48 m) - 1 / (0.16 m)]

U = - (0.01517693 x 10^-11 Nm²) (-4.16 m^-1)

U = 6.31 x 10^-13 J