Three masses of 1 kg, 2kg, and 3 kg are placed at the vertices of an equilateral triangle of side 1m. Find the gravitational potential energy of this system
(Take, `G = 6.67 xx 10^(-11) "N-m"^(-2)kg^(-2)`)

Explanation

Gravitational potential energy of the three particles system
`U=-G((m_(3)m_(2))/(r_(32))+(m_(3)m_(1))/(r_(31))+(m_(2)m_(1))/(r_(21)))`

Here, `r_(32)=r_(31)=r_(21)=1m`
`m_(1)=1kg,m_(2)=2kg and m_(3)=3kg`
Substituting in above, we get
`U=-(6.67xx10^(-11))((3xx1)/(1)+(3xx1)/(1)+(2xx1)/(1))`
or `U=-7.337 xx10^(-10)J`