Answer
a) $F_E=1.98\times 10^{20}N$
b) $F_S=4.36\times 10^{20}N$
c) It makes more sense to think of the Moon as orbiting the Sun with a small effect due to the Earth.
Work Step by Step
(a) We can find the force exerted by the Earth on the Moon as
$F_E=G\frac{M_M\space M_E}{R_{EM}^2}$
We plug in the known values to obtain:
$F_E=6.67\times 10^{-11}\times \frac{7.35\times 10^{22}(5.97\times 10^{24})}{(3.84\times 10^8)^2}$
$F_E=1.98\times 10^{20}N$
(b) The force exerted by the Sun on the Moon is given as
$F_S=G\frac{M_M\space M_S}{R_{SM}^2}$
We plug in the known values to obtain:
$F_S=6.67\times 10^{-11}\times \frac{7.35\times 10^{22}(2.00\times 10^{30})}{(1.50\times 10^{11})^2+(3.84\times 10^8)^2}$
$F_S=4.36\times 10^{20}N$
(c) We can see that the force exerted by the Sun is more as compared to that of the Earth on the Moon, thus it makes more sense to think of the Moon as orbiting the Sun with a small effect due to the Earth.
