The gravitational potential energy of an isolated system of three particles, each of mass m, at the three corners of an equilateral of side l is

Question

The gravitational potential energy of an isolated system of three particles, each of mass m, at the three corners of an equilateral of side l is (A) -(Gm2/l) (B) -(Gm2/2l) (C) -(2Gm2/l) (D) -(3Gm2/l)

Tardigrade Admin · Accepted Answer

Gravitational potential energy of a pair of particles = -(Gm2/l) Since we have three pairs, therefore the total gravitational potential energy is - (3Gm2/l)

Q. The gravitational potential energy of an isolated system of three particles, each of mass $m$ , at the three corners of an equilateral of side $l$ is

$- \frac{G m ^{2}}{l}$

$- \frac{G m ^{2}}{2 l}$

$- \frac{2 G m ^{2}}{l}$

$- \frac{3 G m ^{2}}{l}$

Gravitational potential energy of a pair of particles

$= - \frac{G m ^{2}}{l}$

Since we have three pairs, therefore the total

gravitational potential energy is $- \frac{3 G m ^{2}}{l}$

Q. The gravitational potential energy of an isolated system of three particles, each of mass m, at the three corners of an equilateral of side l is

−lGm2​

−2lGm2​

−l2Gm2​

−l3Gm2​