Answer
$4.79\times 10^{20}N$ at $24.4^{\circ}$ toward the Earth off the line from the Moon to the Sun.
Work Step by Step
We can find the required force as follows:
$F_S=G\frac{M_M M_S}{r_{MS}^2}$
$F_S=(6.67\times 10^{-11}N.m^2/Kg^2)\frac{(7.35\times 10^{22}Kg)(2.00\times 10^{30}Kg)}{(1.50\times 10^{11}m)^2}=4.36\times 10^{20}N$
The force exerted on the Earth is given as
$F_E=G\frac{M_M M_E}{r_{ME}^2}$
$F_S=(6.67\times 10^{-11}N.m^2/Kg^2)\frac{(7.35\times 10^{22}Kg)(5.97\times 10^{24}Kg)}{(3.84\times 10^{11}m)^2}=1.98\times 10^{20}N$
Now $F=\sqrt{F_x^2+F_y^2}$
We plug in the known values to obtain:
$F=\sqrt{(4.36\times 10^{20})^2+(1.98\times 10^{20})^2}$
$F=4.79\times 10^{20}N$
and the direction can be determined as
$\theta=tan^{-1}\frac{F_y}{F_x}$
$\theta=tan^{-1}(\frac{1.98\times 10^{20}}{4.36\times 10^{20}})=24.4^{\circ}$ toward the Earth off the line from the Moon to the Sun.
