Question

During a solar eclipse, the Moon is positioned directly between Earth and the Sun. Find the magnitude of the net gravitational force acting on the Moon during the solar eclipse due to both Earth and the Sun. The masses of the Sun, Earth, and the Moon are 1.99×10301.99×1030 kg, 5.98×10245.98×1024 kg, and 7.36×10227.36×1022 kg, respectively. The Moon's mean distance from Earth is 3.84×1083.84×108 m, and Earth's mean distance from the Sun is 1.50×10111.50×1011 m. The gravitational constant is ????=6.67×10−11 Nm2/kg2G=6.67×10−11 Nm2/kg2 .

Answer 1

Answer:

$F= 2.3733 x10^{20} N$

Explanation:

Let's define the variables to proceed with the operations,

So,

The masses

$M_ {sun} = 1.99 * 10^{30} Kg$

$M_ {Earth} = 5.98 * 10 ^ {24} Kg$

$M_ {Moon} = 7.36 * 10 ^{22} Kg$

Average distances

$\bar {x} _ {Sun \rightarrow Earth} = 1.5 * 10 ^ {11} m$

$\bar {x} _ {Earth \rightarrow Moon} = 3.84 * 10 ^ 8m$

Gravitational constant

$G = 6.67 * 10^ {-11} \frac {Nm ^ 2} {kg ^ 2}$

The formula of the Gravitational Force between the Moon and the Earth would be,

$F = \frac {GM_ {Earth} M_ {Moon}} {(\bar {x} _ {Earth \rightarrow Moon}) ^ 2}$

$F= \frac{(6.67*10^{-11} \frac{Nm^2}{kg^2})(5.98*10^{24}Kg)(7.36*10^{22}Kg)}{(3.84*10^8m)^2}$

$F = 1.9908 * 10 ^{20} N$

This force is in the direction of the earth.

We perform the same process but now between the Sun and the Moon, like this,

$F_2 = \frac {GM_ {Sun} M_ {Moon}} {(\bar {x} _ {Sun \rightarrow Earth} - \bar {x} _ {Earth \rightarrow Moon}) ^ 2}$

$F_2 = \frac{(6.67*10^{-11} \frac{Nm^2}{kg^2})(1.99*10^{30}Kg)(7.36*10^{22}Kg)}{(1.5*10^{11}m-3.84*10^8m)^2}$

$F_2 = 4.3641*10^-{ 20} N$

This force is in the direction of the Sun

The net force must be

$F_ {net} = F_2-F$

$F_ {net} = 2.3733*10^{20} N$

This in the direction of the Sun.