# 3.3: Geoid

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?

FIGURE g points in r

$$g$$ 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?

FIGURE uniform bigger mass

Close up u will follow the new shape. Far away, however, it will look more spherical.

Zooming in,

FIGURE zoomed in u

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.

FIGURE extra mass less mass Rp

$U=\frac{Gm}{r}$

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

• When mass increases,  r increases to keep Up constant.
• When mass decreases, r decreases to keep Up constant.

$$U$$ surfaces become not spherical

FIGURE 3D images of potential field

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

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 J2>>other Js. This defines the ellipsoidal shape of the earth.

Figure "real" shape of U

The perfect sphere represents the reference ellipsoid, and the uneven line is the "anomalies" due to smaller-scale mass anomalies.

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

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

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 $$f=\frac{c-a}{a}$$ FIGURE uref follows the same shape

and that Uref follows the same shape as the figure. From observation, were can look at fhyd 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.

From theory we can derive that

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

Solve for $$\frac{C}{Ma^2}$$, substitute values for fhyd (J2), $$\omega$$, a, and M, and calculate $$\frac{C}{Ma^2}$$ for the planet.