# 3.3: Geoid

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- 3521

We've been talking a lot about calculating the gravitational potential field. Now let's dive into what exactly that is. \(U\) is the gravitational potential and has the relationship

\[U=\frac{Gm}{r}.\]

\(g\) is the gravitational gradient for a point mass (or a uniform solid), and will point in the radial direction. \(g\) and \(U\) have the relationship \(g=\triangledown U\).

The relationship between the gravitational potential and the field lines is pretty simple,

\[g=\triangledown U=\frac{dU}{dx}\frac{dU}{dy}\frac{dU}{dz}\]

With gravitational field lines, the gravity field is a point source, thus you can add up the point source to represent any mass.

What exactly does a point source mean?

\(g\) points in \(r\). \(g\) is always perpendicular to \(u\), and \(|g|\) is larger where the gradient in \(u\) is steeper i.e., \(|g|=|\frac{dU}{dr}|\)

But what if the object is not a sphere?

Close up u will follow the new shape. Far away, however, it will look more spherical as the mass anomalies influence u less.

Zooming in,

we can see that locally g is perpendicular to u, and g is in the r direction

The real \(U (g)\) for earth has **lots** of bumps. Some are the equatorial bulge, topography (but not all topography), and internal mass anomalies.

\[U=\frac{Gm}{r}\]

\(U(R_p)=U_p\) constant "equipotential surface"

- When mass increases, r increases to keep U
_{p}constant. - When mass decreases, r decreases to keep U
_{p}constant.

\(U\) surfaces become not spherical

**FIGURE 3D images of potential field Magali**

This figure shows 3D images of potential fields. So how do we represent this mathematically? ⇒ spherical harmonies

The above figure shows visual representations of the first few real spherical harmonics. Positive regions of the function are blue and yellow regions are negative. Spherical harmonics are like sines and cosines, but they depend on (\(\theta,\phi\)) lat/lon. We can sum them together to represent any surface (ups/downs) on a sphere.

\[U=\frac{Gm}{r}\left(1-\sum_{\ell=1}^{\infty}J_2\ell \left[\dfrac{a}{r}\right]^{2\ell}P_{2\ell}(\cos\theta)\right)\]

where \(\frac{Gm}{r}\) is the point mass. At r=a (the earth's average a), the value is 1. For earth, because it is a rapidly rotating "fluid-like" planet, \(\ell\)=1 and thus J_{2}>>other J_{s}. This defines the ellipsoidal shape of the earth.

The perfect sphere represents the reference ellipsoid, and the uneven line is the "real" shape of U.

\[g_{ref}=\left|\frac{dU_{ref}}{dr}\right|\]

\[U\approx-\frac{Gm}{r}+\frac{Gma^2}{2r^3}J_2[3\sin^2\phi-1]-\frac{1}{2}\omega^2r^2\cos^2\phi\]

where the \(\frac{Gm}{r}\) represents the spherical shape, the \(\frac{Gma^2}{2r^3}J_2[3\sin^2\phi-1]\) is the equatorial bulge (ellipsoidal shape), and the \(-\frac{1}{2}\omega^2r^2\cos^2\phi\) is the actual spin of the planet.

Stepping back to moments of inertia for a minute...

We know that \(f=\frac{c-a}{a}\)

and that U_{ref} follows the same shape as the figure. Each surface will have a constant value for U. We want to chose one that is close to the actual surface of the planet, which will be our reference ellipsoid. For earth, we chose sea level.

From observation, we can look at f_{hyd} for gravity.

\[f_{hyd}=\frac{3}{2}J_2+\frac{1}{2}\frac{a^3\omega^2}{GM}\]

where \(\frac{3}{2}J_2\) is the shape and \(\frac{1}{2}\frac{a^3\omega^2}{GM}\) is the spin. J_{2} can be found using satellite orbits.

From theory we can derive that

\[f_{hyd}=\frac{\frac{5}{2}\frac{\omega^2a^3}{GM}}{1+\left(\frac{25}{4}\right)\left(1-\frac{3}{2}\frac{C}{Ma^2}\right)^{\frac{1}{2}}}\]

We then solve for \(\frac{C}{Ma^2}\), substitute values for f_{hyd }(J_{2}), \(\omega\), a, and M, and calculate \(\frac{C}{Ma^2}\) for the planet.