Answer:
The magnitude of the acceleration ae of the earth due to the gravitational pull of the moon is [tex] \mathbf{3.3187\times10^{-5}}\frac{m}{s^{2}} [/tex]
Explanation:
By Newton's gravitational law, the magnitude of the gravitational force between two objects is:
[tex] F=G\frac{Mm}{r^{2}} [/tex](1)
With G the gravitational constant, M the mass of earth, m the mass of the moon and r the distance between the moon and the earth, a quick search on physics books or internet websites give us the values:
[tex] M=5.972\times10^{24}\,kg[/tex]
[tex]m=7.34767309\times10^{22}\,kg [/tex]
[tex]r=384400\,km [/tex]
[tex] G=6.674\times10^{-11}\,\frac{N\,m^{3}}{kg^{2}}[/tex]
Using those values on (1)
[tex]F=(6.674\times10^{-11})*\frac{(5.972\times10^{24})(7.34767309\times10^{22})}{(384400\times10^{3})^{2}} [/tex]
[tex] F\approx1.98193\times10^{20}N[/tex]
Now, by Newton's second Law we can find the acceleration of earth ae due moon's pull:
[tex]F=M*ae\Longrightarrow ae=\frac{F}{M}=\frac{1.98193\times10^{20}}{5.972\times10^{24}}\approx\mathbf{3.3187\times10^{-5}} [/tex]