Three masses of `1kg`, `2kg`, and `3kg`, are placed at the vertices of an equilateral triangle of side `1m`. Find the gravitational potential energy of this system.
Take `G= 6.67xx10^(-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`