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 …

Question

asked Jul 15, 2020 228k views

1 Answer

Deepak Goswami · Answer 1 · 2020-07-19T09:24:17+0000

Answer:

$F_(resultant) = 2.373* 10^(20)\ N$

Solution:

As per the question:

We know that:

Mass of the moon,
$M_(m) = 7.36* 10^(22)\ kg$

Mass of the Earth,
$M_(e) = 5.98* 10^(24)\ kg$

Mass of the sun,
$M_(s) = 1.99* 10^(30)\ kg$

Mean distance between the Earth and the sun, R =
$1.5* 10^(11)\ m$

Mean distance between the Earth and the moon, R' =
$3.84* 10^(8)\ m$

Now,

We know that the gravitational force is given by:

F = (GMm)/(r^(2))

Thus

Gravitational force between the Earth and the Moon:

F_(em) = (6.67* 10^(- 11)* 5.98* 10^(30)* 7.36* 10^(22))/((3.84* 10^(8))^(2))

$F_(em) = 1.9908* 10^(20)\ N$

The above force is towards the earth:

The force of gravitation between the sun and moon:

F_(sm) = (6.67* 10^(- 11)* 1.99* 10^(30)* 7.36* 10^(22))/((1.5* 10^(11) - 3.84* 10^(8))^(2))

$F_(sm) = 4.364* 10^(20)\ N$

Thus net force is given by:

$F_(resultant) = F_(sm) - F_(em) = 4.364* 10^(20) - 1.9908* 10^(20) = 2.373* 10^(20)\ N$

The resultant force is directed towards the sun.