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julsineya [31]
3 years ago
8

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.
Geography
1 answer:
disa [49]3 years ago
4 0

Answer:

F_{resultant} = 2.373\times 10^{20}\ N

Solution:

As per the question:

We know that:

Mass of the moon, M_{m} = 7.36\times 10^{22}\ kg

Mass of the Earth, M_{e} = 5.98\times 10^{24}\ kg

Mass of the sun, M_{s} = 1.99\times 10^{30}\ kg

Mean distance between the Earth and the sun, R = 1.5\times 10^{11}\ m

Mean distance between the Earth and the moon, R' = 3.84\times 10^{8}\ m

Now,

We know that the gravitational force is given by:

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

Thus

Gravitational force between the Earth and the Moon:

F_{em} = \frac{6.67\times 10^{- 11}\times 5.98\times 10^{30}\times 7.36\times 10^{22}}{(3.84\times 10^{8})^{2}}

F_{em} = 1.9908\times 10^{20}\ N

The above force is towards the earth:

The force of gravitation between the sun and moon:

F_{sm} = \frac{6.67\times 10^{- 11}\times 1.99\times 10^{30}\times 7.36\times 10^{22}}{(1.5\times 10^{11} - 3.84\times 10^{8})^{2}}

F_{sm} = 4.364\times 10^{20}\ N

Thus net force is given by:

F_{resultant} = F_{sm} - F_{em} = 4.364\times 10^{20} - 1.9908\times 10^{20} = 2.373\times 10^{20}\ N

The resultant force is directed towards the sun.

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