Solution 2.7
From Mechanics
\displaystyle F=\frac{G{{m}_{1}}{{m}_{2}}}{{{d}^{2}}}
where \displaystyle F is the force on the particle, \displaystyle {{m}_{1}} is the mass of the moon, \displaystyle {{m}_{2}} is the mass of the particle and \displaystyle d is the radius of the moon.
As \displaystyle F={{m}_{2}}a where \displaystyle a is the acceleration of the particle
\displaystyle a=\frac{G{{m}_{1}}}{{{d}^{2}}}
As
\displaystyle G=6\textrm{.}67\times {{10}^{-11}}\text{ k}{{\text{g}}^{\text{-1}}}{{\text{m}}^{\text{3}}}{{\text{s}}^{\text{-2}}}
we get
\displaystyle \begin{align}
& a=\frac{6\textrm{.}67\times {{10}^{-11}}\text{ k}{{\text{g}}^{\text{-1}}}{{\text{m}}^{\text{3}}}{{\text{s}}^{\text{-2}}}\times \text{7}\textrm{.}\text{38}\times \text{1}{{0}^{\text{22}}}\text{kg}}{{{\left( \text{1}\textrm{.}\text{73}\times \text{1}{{0}^{\text{6}}}\text{m} \right)}^{2}}} \\
& =\frac{6\textrm{.}67\times \text{7}\textrm{.}\text{38}}{\text{1}\textrm{.}\text{7}{{\text{3}}^{2}}\times 10}\text{m}{{\text{s}}^{-2}}=1\textrm{.}64\text{m}{{\text{s}}^{-2}} \\
\end{align}