91Ó°ÊÓ

mass of ball $1,$$m_1=5.0kg$

mass of ball $2,m_2=20.0kg$

mass of ball 3, $m_3=10.0kg$

distance between ball $1$and $2$$l_1=20cm=0.2m$

distance between ball $2$and $3$$l_2=10cm=0.1m$

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

From the given $FIGUREP13.35$, the distance between mass$1$and$3$will be calculated by Pythagoras theorem,$l_3=\sqrt{{l_1}^2+{l^2}_2}=\sqrt{{(0.2)}^2+{(0.1)}^2}=0.22m$

the gravitational potential energy for mass $1$to $2$is calculated as, $U_{1-2}=\frac{G{m}_1{m}_2}{l_1}=\frac{6.67×{10}^{-11}×5×20}{0.2}=3.3×{10}^{-8}J$

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

$U_{2-3}=\frac{G{m}_2{m}_3}{l_2}=\frac{6.67×{10}^{-11}×20×10}{0.1}=13.34×{10}^{-8}J$

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

$U_{1-3}=\frac{G{m}_1{m}_3}{l_3}=\frac{6.67×{10}^{-11}×5×10}{0.22}=1.5×{10}^{-8}J$

The total gravitational potential energy

$U_{net}=U_{1-3}+U_{1-2}+U_{2-3}=(3.3+13.34+1.5)×{10}^{-8}J=18.14×{10}^{-8}J$