91Ó°ÊÓ

The mass of the Moon is $m_M=7.35×{10}^{22}\mathrm{kg}$.

The mass of the Earth is $m_E=5.98×{10}^{24}\mathrm{kg}$.

The mass of the Sun is $m_S=1.99×{10}^{30}\mathrm{kg}$.

In this problem, it is considered that the distance from the Earth to the Sun is equivalent to the distance from the Moon to the Sun.

The relation of horizontal force on Moon and the Earth is given by,

$F_{ME}=G\frac{m_m{m}_E}{{r_{ME}}^2}$

Here, G is the gravitational constant and $r_{ME}$is the distance from the moon to the Earth.

The relation of vertical force on Moon and the Sun is given by,

$F_{MS}=G\frac{m_m{m}_S}{{r_{MS}}^2}$

Here, $r_{MS}$is the distance from the moon to the Sun.

The relation of net force is given by,

$\begin{array}{l}{F}_n=\sqrt{{F_{ME}}^2+{F_{MS}}^2}\\ F_n=\sqrt{{\left(G\frac{m_m{m}_E}{{r_{ME}}^2}\right)}^2+{\left(G\frac{m_m{m}_S}{{r_{MS}}^2}\right)}^2}\\ F_n=G{m}_m\sqrt{{\left(\frac{m_E}{{r_{ME}}^2}\right)}^2+{\left(\frac{m_S}{{r_{MS}}^2}\right)}^2}\end{array}$

On plugging the values in the above relation.

$\begin{array}{l}{F}_n=\left(6.67×{10}^{-11}\mathrm{N}·{\mathrm{m}}^2/{\mathrm{kg}}^2\right)\left(7.35×{10}^{22}\mathrm{kg}\right)\sqrt{{\left(\frac{5.98×{10}^{24}\mathrm{kg}}{{\left(384×{10}^6\mathrm{m}\right)}^2}\right)}^2+{\left(\frac{1.99×{10}^{30}\mathrm{kg}}{{\left(149.6×{10}^9\mathrm{m}\right)}^2}\right)}^2}\\ F_n=4.79×{10}^{20}\mathrm{N}\end{array}$

Thus, $F_n=4.79×{10}^{20}\mathrm{N}$is the magnitude of net force.

The relation to calculate the direction of the net force is given by,

$\begin{array}{l}\tan \mathrm{θ}=\frac{F_{ME}}{F_{MS}}\\ \tan \mathrm{θ}=\left[\frac{G\frac{m_m{m}_E}{{r_{ME}}^2}}{G\frac{m_m{m}_S}{{r_{MS}}^2}}\right]\\ \tan \mathrm{θ}=\left(\frac{m_E}{{r_{ME}}^2}\right)\left(\frac{{r_{MS}}^2}{m_S}\right)\end{array}$

On plugging the values in the above relation.

$\begin{array}{c}\tan \mathrm{θ}=\left(\frac{5.98×{10}^{24}\mathrm{kg}}{{\left(384×{10}^6\mathrm{m}\right)}^2}\right)\left(\frac{{\left(149.6×{10}^9\mathrm{m}\right)}^2}{1.99×{10}^{30}\mathrm{kg}}\right)\\ \mathrm{θ}=24.5°\end{array}$

Thus, $\mathrm{θ}=24.5°$is the required direction of net force.