91Ó°ÊÓ

mass of ball $1,$$m_1=10.0kg$

mass of ball $2,m_2=20.0kg$

mass of ball $3,m_3=10.0kg$

distance between ball $1$and $3$$l_3=10cm=0.1m$

Newton's gravitational constant$G=6.67×{10}^{-11}{m}^3/kg{s}^2$

From the given , the distance between mass $1$and $2$will be calculated by Pythagoras theorem, role="math" localid="1649917576011" $l_1=\sqrt{{(0.2)}^2+{(0.5)}^2}=0.54m$

the distance between mass $2$and $3$will be calculated by Pythagoras theorem,

$l_2=\sqrt{{(0.2)}^2+{(0.5)}^2}=0.54m$

the gravitational potential energy for mass to is calculated as,

$U_{1-2}=\frac{G{m}_1{m}_2}{l_1}=\frac{6.67×{10}^{-11}×10×20}{0.54}=2.5×{10}^{-8}J$

The gravitational potential energy for mass $2$to $3$is calculated as,

role="math" localid="1649917792861" $U_{2-3}=\frac{G{m}_2{m}_3}{l_2}=\frac{6.67×{10}^{-11}×20×10}{0.54}=2.5×{10}^{-8}J$

The gravitational potential energy for mass $1$to $3$is calculated as,

$U_{1-3}=\frac{G{m}_1{m}_3}{l_3}=\frac{6.67×{10}^{-11}×10×10}{0.1}=6.8×{10}^{-8}J$

The total gravitational potential energy

$U_{net}=U_{1-3}+U_{1-2}+U_{2-3}=(2.5+2.5+6.8)×{10}^{-8}J=11.8×{10}^{-8}J$