Solution.
The formula for calculating the acceleration due to gravity is :
$$ g=G \times \frac{M}{R^{2}} $$
Here, Gravitational constant :
$G=6.7 \times 10^{-11} \mathrm{Nm}^{2} / \mathrm{kg}^{2}$
Mass of the moon, $M=7.4 \times 10^{22} \mathrm{~kg}$
Radius of the moon, $R=1740 \mathrm{~km}$
$$
\begin{array}{l}
=1740 \times 1000 \mathrm{~m} \\
=1.74 \times 10^{6} \mathrm{~m}
\end{array} $$
Now, putting these values of $G, M$ and $R$ in the above formula, we get:
$$ g=\frac{6.7 \times 10^{-11} \times 7.4 \times 10^{22}}{\left(1.74 \times 10^{6}\right)^{2}} $$
$$
\text { or } \quad g=1.63 \mathrm{~m} / \mathrm{s}^{2}
$$
Thus, the acceleration due to gravity, $g$, on the surface of the moon is $1.63 \mathrm{~m} / \mathrm{s}$
