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.

Answer 1

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.