# Check: explanation of tides

1. Sep 26, 2004

### brian0918

One of my hw problems for GR says:

"Explain why a uniform external gravitiational field cannot raise tides on the Earth."

My explanation is:

Because of the large size of the Earth compared to the distance the Earth is from the moon, the nonuniformity of the moon's gravitational field implies that there is a noticeable difference between the strength of the force on the side of the Earth facing the moon versus either the strength of the force on the center of the Earth or the strength of the force on the side opposite the moon.

So, the water closest to the moon is pulled away from the Earth, and the Earth is pulled away from the far water.

This can't happen in a uniform g field, as the forces at each point would be equal and everything would be pulled toward the moon with the same force.

2. Sep 26, 2004

### mathman

What's the question?

One statement you made is in error - the distance from earth to moon is much greater than the diameter of the earth. However tides still occur. The sun's gravity also has an effect, although smaller than the moon's.

3. Sep 26, 2004

### brian0918

I just wanted you to check if my explanation answers the question. I didn't say the Earth's diameter is larger than its distance to the moon, just that its size is large enough to have a noticeable effect when it comes to tidal forces.

4. Sep 27, 2004

### nrqed

This is good, except that you are using a language that is different from what people use in physics and that may cause some confusion (like in the reply you got). I know what you mean by "large size compared to.." but most people would take this as meaning that the Earth is larger than the distance to the Moon! I know that's not what you meant but that's how people woul interpret it. The correct jargon would be to say that "the size of the Earth is not negligible compared to the distance to the Moon". This means that the Earth is smaller but not small to the point that it can be neglected. Which is what you meant, I think.

That's good. Although maybe you should not mention the Moon in the last sentence. It's clear that we are no longer talking about the Moon if we are discussing a uniform field. I woud probably just finish by "....and everything would be pulled with the same force". But I am nitpicking here.

Overall, your explanation shows that you understand the concept well. I would give full marks if I were your teacher!

Pat

5. Sep 27, 2004

### pervect

Staff Emeritus

The only issue I have with your answer is that you spend a lot of time talking about how tides are raised on the Earth, and not very much time talking about why a uniform gravitational field wouldn't raise tides.

Did they cover the geodesic deviation equation in your course? You might be able to formulate a shorter answer by considering the geodesics in a uniform field on the "near side" and "far side" of the earth.

6. Sep 27, 2004

### Garth

Consider a spherical ball of dust free falling towards a gravitating mass M. The sides of the ball are not going to fall parallel but converge on M. Furthermore the nearest point of the ball to M is under a stronger gravitational attraction/greater curvature than its centre whereas the furthest point is under a weaker attraction than the centre.

Hence the ball is going to be elongated towards M.

Now assume the frame of reference at the centre of the ball. You have a more or less spherical shape with two tidal bulges, one towards the M and one directly away from M.

In a uniform gravitational field all the points of the dust ball will travel parallel to each other and there will be no tidal forces. [There will be no gravitational attraction either for that matter - the presence of a gravitational field always involves tidal forces, although the ball may be so small w.r.t. the distance to M that they are too small to be observed.]

Garth