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Four particles each of mass m are placed at the verticles of square of side a. What is the potential energy of the system?
Option: 1
$\frac{-\sqrt{2}Gm^{2}}{a}\left ( 2-\frac{1}{\sqrt{2}} \right )$

Option: 2
$\frac{-2Gm^{2}}{a}\left ( 2+\frac{1}{\sqrt{2}} \right )$

Option: 3
$\frac{-\sqrt{2}Gm^{2}}{a}\left ( \sqrt{2}-\frac{1}{\sqrt{2}} \right )$

Option: 4
$\frac{-\sqrt{2}Gm^{2}}{a}\left ( \sqrt{2}+\frac{1}{\sqrt{2}} \right )$

Answers (1)

best_answer

As we learn

Gravitational Potential energy of discrete distribution of masses -

$U=-G\left [ \frac{m_{1}m_{2}}{r_{12}}+\frac{m_{2}m_{3}}{r_{23}}+\cdot \cdot \cdot \right ]$

$U\rightarrow$ Gravitatinal Potential Energy

$r_{12},r_{23}\rightarrow$ Distance of masses from each other

- wherein

if $r=\infty$

$U$ becomes Zero (maximum)

$P.E. =\frac{-Gm_{1}m_{2}}{r}$

Short trick to calculate no of pair

$=\frac{n (n-1)}{2}$ [n= no of masses]

$=\frac{4*3}{2}=6$

Pair calculation $(1,2), \: \: (1,3),\: \: (1,4),\: \: (2,3),\: \: (2,4),\: \: (3,4)$

Distance $a$ $\sqrt{2}a$ $a$ $a$ $\sqrt{2}a$ $a$

Four pair of distance = $a$

Two pair of distance = $\sqrt{2}a$

Hence P.E. of the system $=\frac{-4 Gm^{2}}{a}+\frac{-2Gm^{2}}{\sqrt{2}a}$

$P.E=\frac{-2Gm^{2}}{a}\left [ 2+\frac{1}{\sqrt{2}} \right ]$

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chirag

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