Question

What is the ratio of the sun’s gravitational force on the moon to
the earth’s gravitational force on the moon?

Short answer

The ratio of the sun’s gravitational force on the moon to the earth’s gravitational force on the moon is 2.26.

Step-by-step solution

Given information

Mass of moon: $m=7.49\times {10}^{22}\mathrm{kg}$
Gravitational constant $G=6.67\times {10}^{-11}{m}^{3\cdot }k{g}^{-1}\cdot s^{-2}$
Mass of earth: $M_e=5.97\times {10}^{24}\mathrm{kg}$
Distance from earth to moon: $r_{em}=3.84\times {10}^8\mathrm{m}$
Mass of sun: $M_s=1.98\times {10}^{30}\mathrm{kg}$
Distance from sun to moon: $r_{sm}=1.47\times {10}^{11}\mathrm{m}$

Explanation

The gravitational force is calculated by

$F=\frac{GMm}{r^2}$

Use this to calculate gravitational force of sun on moon

$$\begin{gathered} F_{sm}=\frac{G{M}_{s}m}{r_{sm}^2} \\ =\frac{(6.67\times {10}^{-11}{m}^{3\cdot }k{g}^{-1}\cdot s^{-2})\left(1.98\times {10}^{30}\mathrm{kg}\right)\left(7.49\times {10}^{22}\mathrm{kg}\right)}{{\left(1.47\times {10}^{11m}\right)}^2} \\ =4.58\times {10}^{20}\mathrm{N} \end{gathered}$$

And gravitational force on earth on moon is

$$\begin{gathered} F_{em}=\frac{G{M}_{e}m}{r_{em}^2}=\frac{(6.67\times {10}^{-11}{m}^{3\cdot }k{g}^{-1}\cdot s^{-2})\left(5.97\times {10}^{24}\mathrm{kg}\right)\left(7.49\times {10}^{22}\mathrm{kg}\right)}{{\left(3.84\times {10}^8\mathrm{m}\right)}^2} \\ =2.02\times {10}^{20}\mathrm{N} \end{gathered}$$

Now find the ratio

$\frac{F_{sm}}{F_{em}}=\frac{4.58\times {10}^{20}\mathrm{N}}{2.02\times {10}^{20}\mathrm{N}}=2.26$