# Does the sun or the moon have a greater affect on the tides?

## Homework Statement

I know the moon does. I know it is because tidal forces fall off as 1/r3. But why? Mathematically, I mean.

F = GMm/r2

## The Attempt at a Solution

None

phinds
Gold Member
I know the moon does. I know it is because tidal forces fall off as 1/r3. But why? Mathematically, I mean.
In what way does this statement not contain both the question and the answer?

I should have been more specific. How do you get from F = GMm/r2 to an equation for tides that has a 1/r3 in it?

gneill
Mentor
I should have been more specific. How do you get from F = GMm/r2 to an equation for tides that has a 1/r3 in it?
Did you try a search on Tidal Force? Even the Wikipedia article on Tidal Force shows a short derivation (granted it's for the locations lying along the line joining the centers of the two interacting bodies, but it avoids the vector math required for the more general solution for points located anywhere on the surface of the smaller body).

I should have been more specific. How do you get from F = GMm/r2 to an equation for tides that has a 1/r3 in it?
Tides result because of the variation of the force of gravity. Points on the side of the earth near to the moon are more strongly attracted than points on the far side.

In other words, the tides are not caused by the force of gravity which is proportional to 1/r2 but rather by the variation in that force over a distance the size of the earth, that is to say the spatial derivative . And the derivative of 1/r2 is ...