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photoshop1234 [79]
2 years ago
13

During a solar eclipse, the Moon is positioned directly between Earth and the Sun. Find the magnitude of the net gravitational f

orce 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 .
Physics
1 answer:
dezoksy [38]2 years ago
8 0

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.

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