I am preparing an elementary derivation of Newton/Laplace's equilibrium theory of tides.
The derivation of the tractive force is understood, as is the derivation of the differential equation for change in ocean height (assuming a spherical water covered Earth).
However, I am stuck with this following reasoning on page 14 of the link below.
https://www.uaf.edu/files/sfos/Kowalik/tide_book.pdf
The authors claim that the differential equation is as follows:
$$ \frac{dh}{d \theta} = - \frac{3r}{2} \frac{m}{M} \left( \frac{r}{R} \right) ^3 \sin(2 \theta) $$
[Their I.40]
where:
- $h$ is the change in height of the water
- $r$ and $M$ are the radius and mass of the Earth
- $R$ is the distance from the Earth to the Moon
- $m$ is the mass of the Moon.
- $\theta$ is the angle from the the north pole (ignoring axial tilt, assuming Moon is equatorial)
The authors solve it to give:
$$ h = r \frac{m}{M} \left( \frac{r}{R} \right) ^3 \left(\frac{3}{2} \cos^2\theta + c \right) $$
Here's the trouble: to determine $c$, the authors quote (Proudman, 1953) which I cannot access. They claim that the condition for conservation of mass of water is:
$$\int_{0}^{\pi} \left(\frac{3}{2} \cos^2\theta + c \right)\sin\theta d\theta$$
I don't understand why there is a sine in this integral. What is its physical significance?
I suppose that $\sin \theta d \theta$ comes from the fact that the volume element in spherical coordinates is $dV=\rho^2 \sin \theta d \rho d\theta d\varphi$ and the symmetry of the problem is such that the variation of the volume , that must be null, depends only from the polar angle $\theta$.