As Calvin noted in his answer, the quantity given by ChatGPT is dimensionless and cannot represent a height. In order to derive the shape of the tide it helps to go to earth's non-inertial frame:
- Consider the difference between the acceleration caused by moon on earth's center and a place on earth's surface:
$$\frac{\mathbf{a}_{tidal}(\mathbf r)}{GM_{moon}} = \frac{\mathbf{R}-\mathbf{r}}{|\mathbf R-\mathbf r|^3} - \frac{\mathbf R}{|\mathbf R|^3}\, ,$$
where $\mathbf R$ is a vector pointing from earth's center to moon and $\mathbf r$ is a vector from earth's center to a location on its surface.
- Expanding the above formula in the limit of $r \ll R$ and keeping the leading terms gives us:
$$\frac{\mathbf{a}_{tidal}(\mathbf r)}{GM_{moon}} \approx \frac 1 {R^3}(3\hat {\mathbf R}(\hat {\mathbf R}.\mathbf r)-\mathbf r)\, .$$
(hint: $|\mathbf R - \mathbf r | = \sqrt{(\mathbf R - \mathbf r)\cdot (\mathbf R - \mathbf r)} = \sqrt{R^2 + r^2 - 2 \mathbf R \cdot \mathbf r} \approx R(1-\frac{\mathbf R \cdot \mathbf r}{R^2})$)
- Therefore the tidal force a mass $m$ senses would be:
$$\mathbf F_{tidal}(x,y) \approx \frac{G M_{moon}mr^2}{2R^3}(2x,-y)\, .$$
- By integration we can find the equivalent potential, which gives us:
$$V_{tidal}(r,\theta) \approx \frac{GM_{moon}mr^2}{2R^3}(1-3\cos^2\theta)\, .$$
The shape of the tide can be considered as an equipotential surface; set the above formula to be a constant and derive $r(\theta)$ from it. I've left enough details with enough tips along the path so that you can reconstruct the derivation entirely.
A few points:
- There are two tidal bulges, one facing moon and one away from it. However, conceptually speaking the one away from moon isn't caused by moon pushing it away, but by rather pulling it less than it is pulling the rest of the earth!
- We made several estimations here:
- Implicitly we have assumed earth to be a perfect smooth sphere with oceans covering its entire surface. This yields the ellipsoid we derived. If we were to consider earth's deformities, the equipotential surface would be different; a shape the geologists like to call a Geoid.
- We have ignored earth's rotation. Because earth is rotating around itself faster than moon's orbital period, the tidal bulge is ahead of moon by about $3$ degrees or for the observer on earth it would be about 12 minutes after the moon passing overhead.
- Sun will also have a tide, which we ignored. Interesting enough, the sun's gravity is larger on earth than moon's gravity, however, for tides what matters is the variation in the gravitational field and not the field strength itself. Because moon is closer to us than sun the variation in its gravitational field is also larger.
