Answer:
[tex]F= 2.3733 x10^{20} N[/tex]
Explanation:
Let's define the variables to proceed with the operations,
So,
The masses
[tex]M_ {sun} = 1.99 * 10^{30} Kg[/tex]
[tex]M_ {Earth} = 5.98 * 10 ^ {24} Kg[/tex]
[tex]M_ {Moon} = 7.36 * 10 ^{22} Kg[/tex]
Average distances
[tex]\bar {x} _ {Sun \rightarrow Earth} = 1.5 * 10 ^ {11} m[/tex]
[tex]\bar {x} _ {Earth \rightarrow Moon} = 3.84 * 10 ^ 8m[/tex]
Gravitational constant
[tex]G = 6.67 * 10^ {-11} \frac {Nm ^ 2} {kg ^ 2}[/tex]
The formula of the Gravitational Force between the Moon and the Earth would be,
[tex]F = \frac {GM_ {Earth} M_ {Moon}} {(\bar {x} _ {Earth \rightarrow Moon}) ^ 2}[/tex]
[tex]F= \frac{(6.67*10^{-11} \frac{Nm^2}{kg^2})(5.98*10^{24}Kg)(7.36*10^{22}Kg)}{(3.84*10^8m)^2}[/tex]
[tex]F = 1.9908 * 10 ^{20} N[/tex]
This force is in the direction of the earth.
We perform the same process but now between the Sun and the Moon, like this,
[tex]F_2 = \frac {GM_ {Sun} M_ {Moon}} {(\bar {x} _ {Sun \rightarrow Earth} - \bar {x} _ {Earth \rightarrow Moon}) ^ 2}[/tex]
[tex]F_2 = \frac{(6.67*10^{-11} \frac{Nm^2}{kg^2})(1.99*10^{30}Kg)(7.36*10^{22}Kg)}{(1.5*10^{11}m-3.84*10^8m)^2}[/tex]
[tex]F_2 = 4.3641*10^-{ 20} N[/tex]
This force is in the direction of the Sun
The net force must be
[tex]F_ {net} = F_2-F[/tex]
[tex]F_ {net} = 2.3733*10^{20} N[/tex]
This in the direction of the Sun.
The magnitude of the net gravitational force acting on the Moon is equal to [tex]2.37 \times 10^{20} \;Newton[/tex].
Given the following data:
Mass of Sun = [tex]1.99 \times 10^{30}[/tex] kg.
Mass of Earth = [tex]5.98 \times 10^{24}[/tex] kg.
Mass of Moon = [tex]7.36 \times 10^{22}[/tex] kg.
Distance of M-E = [tex]3.84 \times 10^8[/tex] m.
Distance of E-S = [tex]1.50 \times 10^{11}[/tex] m.
Gravitational constant = [tex]6.67 \times 10^{-11}\;Nm^2/kg^2[/tex]
Mathematically, gravitational force between two planetary objects is given by this formula:
[tex]F=\frac{GM_1M_2}{r^2}[/tex]
Where:
For the Moon and Earth, we have:
[tex]F_{ME}=\frac{6.67 \times 10^{-11}\times 7.36 \times 10^{22} \times 5.98 \times 10^{24}}{(3.84 \times 10^8)^2}\\\\F_{ME}=1.99 \times 10^{20}\;Newton[/tex]
For the Moon and Sun, we have:
[tex]F_{MS}=\frac{6.67 \times 10^{-11}\times 7.36 \times 10^{22} \times 1.99 \times 10^{30}}{(1.5 \times 10^{11} - 3.84 \times 10^8)^2}\\\\F_{MS}=4.36 \times 10^{20}\;Newton[/tex]
Now, we can determine the magnitude of the net gravitational force acting on the Moon:
[tex]F_{net}=F_{MS}-F_{ME}\\\\F_{net}= 4.36 \times 10^{-20} - 1.99 \times 10^{20}\\\\F_{net}= 2.37 \times 10^{20} \;Newton[/tex]
Read more on gravitational force here: https://brainly.com/question/19050897
