Answer
a) $F_E=3.56\times 10^{22}N$ toward the Sun.
b) $F_M=2.40\times 10^{20}N$ toward the Sun.
c) $F_S=3.58\times 10^{22}N$ toward the Earth-Moon system.
Work Step by Step
(a) We can find the required force as
$F_E=GM_E(\frac{M_s}{r_{Es}^2}+\frac{M_m}{r_{Em}^2})$
We plug in the known values to obtain:
$F_E=(6.67\times 10^{-11})(5.97\times 10^{24})[\frac{2.00\times 10^{30}}{(1.50\times 10^{11})^2}+\frac{7.35\times 10^{22}}{(3.84\times 10^8)^2}]$
$F_E=3.56\times 10^{22}N$ toward the Sun.
(b) We can find the required force exerted on the Moon as
$F_E=GM_m(\frac{M_s}{r_{sm}^2}+\frac{M_E}{r_{Em}^2})$
We plug in the known values to obtain:
$F_E=(6.67\times 10^{-11})(7.35\times 10^{22})[\frac{2.00\times 10^{30}}{(1.50\times 10^{11}-3.84\times 10^8)^2}+\frac{5.97\times 10^{24}}{(3.84\times 10^8)^2}]$
$F_M=2.40\times 10^{20}N$ toward the Sun.
(c) We can find the required force on the Sun as
$F_S=GM_S(\frac{M_m}{r_{sm}^2}+\frac{M_E}{r_{sE}^2})$
We plug in the known values to obtain:
$F_S=(6.67\times 10^{-11})(2.00\times 10^{30})[\frac{7.35\times 10^{22}}{(1.50\times 10^{11}-3.84\times 10^8)^2}+\frac{5.97\times 10^{24}}{(1.50\times 10^{11})^2}]$
$F_S=3.58\times 10^{22}N$ toward the Earth-Moon system.
