Finding tidal height difference using two wells

[Esran]

Homework Statement

Newton's model of the tidal height, using two wells dug to the center of Earth (one from the North pole, one from the equator on the side of Earth facing away from the Moon), used the fact that the pressure at the bottom of the two wells should be the same. Assume water is incompressible and find the tidal height difference h due to the Moon using this model.

Homework Equations

Pressure is force per unit area. Assume the wells have perpendicular cross-sectional area of exactly one unit area. Then pressure numerically equals (neglecting units) force. Our two wells connect at the center of Earth, so we must have \int^{x_{max}}_{0}\rho g_{x}dx=\int^{y_{max}}_{0}\rho g_{y}dy, where \rho is the density of water and g_{x} and g_{y} are the gravitational fields active at x (distance from Earth's center in the direction of the North pole well mouth) and y (distance from the Earth's center in the direction of the equator well mouth), respectively.

The Attempt at a Solution

I'm having trouble getting started, beyond what I've outlined above. I have equations for the x and y components of the tidal force. I'm assuming g_{x} is the sum of the gravitational field induced by the Earth and the gravitational field induced by the Moon. Likewise for g_{y}. Now, the gravitational field of the Earth should be about the same in either case, and thus cancel out, leaving the tidal forces to contend with.

Is this the right approach?

 

[Quinzio]

Now, the gravitational field of the Earth should be about the same in either case, and thus cancel out, leaving the tidal forces to contend with.
Is this the right approach?

I'm not really sure what you mean here.

I don't think they cancel out because you are dealing with two integrals, so you really can't do that.

I would start imagining the mass of the Earth and the Moon concentrated in their respective centers (I think we're allowed to do that), then evaluate the attraction forces of the two masses (with the distance center-water point), and solve the integrals.