Wikipedia's Geopotential model give's the form of the potential component linked to $J_2$ as

$$u = J_2\frac{P^0_2(\sin \theta)}{r^3} = J_2\frac{1}{r^3}\frac{1}{2} \left(3 \sin^2\theta-1 \right) = J_2\frac{1}{r^5}\frac{1}{2} \left(3z^2-r^2 \right)$$

and the acceleration components (gradient of this potential) are:

$$a_x = J_2 \frac{x}{|r|^7} (6z^2 - 1.5(x^2+y^2)) $$

$$a_y = J_2 \frac{y}{|r|^7} (6z^2 - 1.5(x^2+y^2)) $$

$$a_z = J_2 \frac{z}{|r|^7} (3z^2 - 4.5(x^2+y^2)). $$

From Geopotential model and this answer.

I've plotted $u$ and $|\mathbf{a}|$ below for the $xz$ plane, where $z$ is Earth's rotational axis. The distribution in three dimensions is simply this rotated about the $z$ axis.

Question: $J_2$ is the dominant part of Earth's quadrupole moment, but it's only one component.

  1. Does $J_{22}$ represent the rest of Earth's quadrupole moment?
  2. What does $J_{22}$ look like? What is its shape and symmetry?

potential and acceleration due to Earth's J2

import numpy as np
import matplotlib.pyplot as plt

Re = 6378137. # meters
J2 = 1.7555E+25 # m^5/s^2
hw = 400
x = np.linspace(-2.5*Re, 2.5*Re, 2*hw+1)
x, z = np.meshgrid(x, x)
y = np.zeros_like(x)
x2, y2, z2 = [w**2 for w in (x, y, z)]
r2 = x2 + y2 + z2
r = np.sqrt(r2)
r7 = r**7

u = J2 * r**-5 * 0.5 * (3*z2 - r**2)
ax = J2 * (x/r7) * (6*z2 - 1.5*(x2+y2)) 
ay = J2 * (y/r7) * (6*z2 - 1.5*(x2+y2)) 
az = J2 * (z/r7) * (3*z2 - 4.5*(x2+y2)) 

for thing in (u, ax, ay, az):
    thing[r<Re] = np.nan

axz = np.sqrt(ax**2 + az**2)

if True:
    plt.subplot(1, 2, 1)
    umax = np.nanmax(np.abs(u))
    plt.imshow(u, cmap='seismic', vmin=-umax, vmax=umax)
    plt.title('specific potential energy (J/kg)', fontsize=16)
    plt.plot([hw, hw], [0.3*hw, 1.7*hw], '-k')
    plt.xlim(0, 2*hw+1)
    plt.ylim(0, 2*hw+1)
    plt.subplot(1, 2, 2)
    plt.imshow(axz, cmap='inferno')
    plt.title('acceleration magnitude (m/s^2)', fontsize=16)
    plt.plot([hw, hw], [0.3*hw, 1.7*hw], '-k')
    plt.xlim(0, 2*hw+1)
    plt.ylim(0, 2*hw+1)
    plt.suptitle("Earth's geopotential's J2 component", fontsize=16)
  • 1
    $\begingroup$ @gerrit I am duly admonished! I got this the first time i.stack.imgur.com/1XdmH.png from Python 3's matplotlib's default so I quickly used my old Python 2 which defaults to jet. I remember reading a tweet with an old haiku about Jet but I can't find it now. I did include "Don't use Jet!" in this and this plot and apologized for it in one of them as well. I'll look for a better cmap and update ;-) $\endgroup$ – uhoh Jan 27 '20 at 8:36
  • $\begingroup$ @gerrit somehow I think it looks more distracting now, but I've updated the colormaps. $\endgroup$ – uhoh Jan 27 '20 at 9:02
  • 1
    $\begingroup$ I recommend colorbrewer2 :) $\endgroup$ – gerrit Jan 27 '20 at 9:26
  • $\begingroup$ @gerrit I've just asked Using brewer2mpl diverging colormap with matplotlib, gamma gives poor results with values different than 1 $\endgroup$ – uhoh Jan 27 '20 at 12:36

Does J22 represent the rest of Earth's quadrupole moment?

With a more usual notation ldegree, morder, there are the zonal l2m0, tesseral l2m1 and sectoral l2m2 spherical harmonics, but you probably already know.

What does J22 look like? What is its shape and symmetry?

Here's what I get for l2m2:

enter image description here

For comparison, here's what I get for your "acceleration magnitude" graph (l2m0) (it's not clear what 800 represents):

enter image description here

  • 1
    $\begingroup$ Thank you for your answer! To double check, the answer to "Does $J_{22}$ represent the rest of Earth's quadrupole moment?" is No? There is a $J_{21}$ as well? Does $I$ and $J$ represent the same thing? $\endgroup$ – uhoh Jan 28 '20 at 1:43
  • 1
    $\begingroup$ Informally speaking, in geodesy we consider a quadrupole something acting on a plane (gravitational, magnetic, electric quadrupole), hence to represent the Earth's quadrupole moment you need an infinite number of spherical harmonics; so my answer is "no". $\endgroup$ – Cristiano Jan 28 '20 at 10:25

Wikipedia actually goes into some detail regarding the components of Earth's gravitational force field.

In the Wikipedia reference we do not have $J_{m,n}$ for $m>n>0$. Rather this term is broken up into a cosine component $C_n^m$ and a sine component $S_n^m$. This table gives results from one on a series of models called the JGM models, developed by the Goddard Space Center:

Zonal coefficients

$$\begin{array}{|c|c|c|}\ n\\ \hline 2&-0.1082635854D-02\\ 3&0.2532435346D-05\\ 4&0.1619331205D-05\\ 5&0.2277161016D-06\\ 6&-0.5396484906D-06\\ 7&0.3513684422D-06\\ 8&0.2025187152D-06\\ \end{array}$$

Tesseral coefficients [sic; cases with $n=m$ are sectoral]

$$\begin{array}{|c|c|c|} n&m&C&S\\ \hline 2&1&-0.3504890360D-09&0.1635406077D-08\\ 2&2&0.1574536043D-05&-0.9038680729D-06\\ 3&1&0.2192798802D-05&0.2680118938D-06\\ 3&2&0.3090160446D-06&-0.2114023978D-06\\ 3&3&0.1005588574D-06&0.1972013239D-06\\ 4&1&-0.5087253036D-06&-0.4494599352D-06\\ 4&2&0.7841223074D-07&0.1481554569D-06\\ 4&3&0.5921574319D-07&-0.1201129183D-07\\ 4&4&-0.3982395740D-08&0.6525605810D-08\\ \end{array}$$

Here $D$ means multiplying by a power of ten whose exponent follows, thus $D-05$ means multiply by $10^{-5}$. The absolute values of the numbers above represent the relative strength of the gravitational potential arising from the spherical harmonic component, at a distance equal to the average radius of Earth and with the isotropic, "spherical mass" component (which would be $J_0$ in this nomenclature) scaled to $1.0$.

Now suppose that you are putting a satellite into low Earth orbit, at a radial distance essentially equal to the reference used above. Among the quadrupole components clearly $J_2$ is largest, with $C_2^1$ and $S_2^1$ many orders of magnitude smaller, and $C_2^2$ and $S_2^2$ falling in-between. You may therefore defensibly describe the quadrupole with only the $(2,0)$ ($J_2$) and $(2,2)$ ($C_2^2$, $S_2^2$) components, ignoring the very tiny $(2,1)$ pieces.

But that is not the whole story. While the remaining quadrupole component is tiny near Earth's surface, some higher order polar components ($n\ge 3$) are not. If you are including the $(2,2)$ components of the quadrupole to model your low Earth orbit motion, you should also include at least the similarly sized $(3,0)$, $(3,1)$ and $(4,0)$ terms in order to really imprive your accuracy over the rotationally symmetric quadrupole. The deviations ftom a ritationally symmetric quadrupole are more complex than just an additional quadrupole component.

The reason for this is that the deviations from a rotationally symmetric quadrupole are governed by the distribution of large-scale geographic features such as mountain ranges, basins and tectonic subduction boundaries. Such features on Earth are too complex to be captured by one or two spherical harmonic components. If you need to capture the geological complexity of Earth to model your satellite, you will need plenty of spherical harmonic terms to do it.

You can simplify things if you are instead launching your satellite to a high altitude, say a geosynchronous orbit. The components with larger $n$ values fade more rapidly with distance than those with smaller ones, so depending on the accuracy you need you may drop out those $(3,m)$ and $(4,0)$ components. The geographical warts we see up close get smoothed out with a more distant view. That is where you are most likely to succeed with just $J_2$, $C_2^2$, and $S_2^2$.

Earth is not the only body with such gravitational complexities. Any ostensibly hydrostatically equilibrated, but solid, body will likely have its own geographic rough spots. Mars, in particular, has the large altitude difference between north and south, the Hellas Basin, and Olympus Mons. If we want to track a Mars orbiter with high accuracy, we had better brush up on our spherical harmonics.

  • 1
    $\begingroup$ Okay so to double check that I understand, 1) $J_{22}$ is just a magnitude and written as a single number lacks enough information to show the variation with longitude. 2) ditto for $J_{21}$ but it is very small and seldom spoken of by itself. $\endgroup$ – uhoh Jan 28 '20 at 1:54
  • 1
    $\begingroup$ Uh-huh, that's it. $\endgroup$ – Oscar Lanzi Jan 28 '20 at 2:01
  • $\begingroup$ @OscarLanzi you put n = m terms in your "Tesseral coefficients" table, but when n = m that coefficient is called "sectoral". $\endgroup$ – Cristiano Jan 28 '20 at 10:27
  • 1
    $\begingroup$ The table is quoted from the source article. See the annotation I added. $\endgroup$ – Oscar Lanzi Jan 28 '20 at 10:31

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