Homework Help

Compare the gravitational forces acting on the Earth from the Sun and then from the...

user profile pic

parvesh | Student, Grade 11 | eNoter

Posted November 4, 2011 at 12:36 AM via web

dislike 1 like

Compare the gravitational forces acting on the Earth from the Sun and then from the Moon.  Explain why, if the FG of the Sun is larger than that of the Moon, the Moon has a greater effect on the Earth's tides.

1 Answer | Add Yours

Top Answer

user profile pic

mwmovr40 | College Teacher | (Level 1) Associate Educator

Posted November 4, 2011 at 4:33 AM (Answer #1)

dislike 1 like

According to the Universal Law of Gravity, FG = G M1M2/R^2

For the Sun-Earth system:  Fs = G MeMs/Res^2  where G is universal gravitational constant, Me is mass of Earth, Ms is mass of Sun, and Res is the average orbital radius of the Earth about the Sun.

For the Moon-Earth system:  Fm = GMeMm/Rem^2

Taking the ratio of Fs/Fm we get

Fs/Fm = Ms/Mm * (Rem/Res)^2  plugging appropriate values into this equation we get Fs/Fm = 178

So, the gravitational force from the Sun acting on the Earth is approximately 178 times stronger than that of the Moon.

If the Sun has so much more gravitational influence than the Moon, why is it that the Moon has a bigger influence on tides?

Tides are controlled, not by the absolute strength of the force, but by a quantity known as the gradient.  A gradient is a measure of how much something changes from one point to another.  In general, the gradient caused by the force of gravity is dependent on the cube of the distance of separation and not the square of the distance.  When we calculate the gradient of the Moon's gravitational force compared to that of the Sun's gradient we get that the Moon's influence is about twice as much as that of the Sun.  Consequently, the Moon influenced our tides more than the Sun does due to its closer proximity to the Earth.

Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes