Answer
$F_{MT}=4.77\times10^{26}N$
Work Step by Step
$F_G=G\frac{m_E m_M}{r^2}$
$F_{MS}=6.67\times 10^{-11}\frac{N \times m^2}{kg^2}\times \frac{7.35\times10^{22}kg\times 1.99\times10^{30}kg}{(150\times10^6km)^2}=4.33\times10^{26}N$
$F_{ME}=6.67\times 10^{-11}\frac{N \times m^2}{kg^2}\times \frac{7.35\times10^{22}kg\times 5.98\times10^{24}kg}{(384\times10^3km)^2}=1.99\times10^{26}N$
The earth and the sun are at right angles to each other so to find the total force of gravity on the moon, we can use the Pythagorean Theorem.
$F_{MT}=\sqrt{\big(4.33\times10^{26}N\big)^2+\big(1.99\times10^{26}N\big)^2}$$=4.77\times10^{26}N$
