5
$\begingroup$

I was reading Satellite Orbits - Models, Methods, and Applications by Montenbruck & Gill and was trying to use the equations in it to calculate $J_2$ as used in the GEM and JGM gravity models. While working on this, I found the following statement: "Even though the definition of the geopotential coefficients $C_{nm}$ and $S_{nm}$ is rather complicated at first sight, one may nevertheless derive some simple results if only low-degree and order coefficients are considered, or if one uses an approximate model for the terrestrial density variation."

The book goes on to calculate $C_{00} = 1$ and $C_{10} = 0$ (the second result is valid only if the center of the coordinate system is chosen as Earth's center of mass). The general equation for $C_{nm}$ is $$C_{nm} = \frac{2 - \delta_{0m}}{M_\oplus} \frac{(n-m)!}{(n+m)!} \int \frac{s^n}{R_\oplus^n} P_{nm} (\sin \phi') \cos(m\lambda')\rho(\mathbf{s})d^3\mathbf{s}$$

While calculating $C_{20}$, I got to $$C_{20} = \frac{1}{M_\oplus R_\oplus^2} \int s^2 \left(\frac{3}{2} \sin^2 \phi' - \frac{1}{2}\right)\rho(\mathbf{s})d^3 \mathbf{s}$$ $$C_{20} = \frac{1}{M_\oplus R_\oplus^2} \left( \frac{3}{2} \int z'^2 \rho(\mathbf{s}) d^3\mathbf{s} - \frac{1}{2}\int s^2 \rho(\mathbf{s}) d^3\mathbf{s} \right)$$

I'm not sure how to proceed, or if it's worth trying to continue. If these coefficients are empirically determined for orders 2 and higher, it seems that any analytical solution for these values would require knowledge of some quantity that we can't measure directly (requiring an approximate density model or other simplifications for an analytical solution as stated earlier in the book).

I've read through several papers searching for an answer but haven't been able to find anything conclusive:

Some of the early papers seem to be calculating values from satellite orbits, but it's not explicitly stated that these coefficients can't be found analytically.

To summarize, are $J_2$, $J_3$, etc. empirical or not?

Improve this question
$\endgroup$
0

1 Answer 1

Reset to default
5
$\begingroup$

It will be a challenge because $\rho(\mathbf{s})$ is the mass density of Earth at each point $\mathbf{s}$ and that's only available theoretically, by watching certain satellite orbits and deducing the geopotential coefficients and from those modeling $\rho(\mathbf{s})$, the backwards procedure of what you are talking about.

So unless you have $\rho(\mathbf{s})$ handy I don't think it's possible to roll your own.

To summarize, are $J_2$, $J_3$, etc. empirical or not?

Yes, GM, $J_2$, $J_3$, etc. are all empirically derived from satellite orbits.

Even careful monitoring of the precession (like this) of the very earliest artificial satellites like Sputnik and Vanguard were used as the first measurements of Earth's $J_2$. Those were really exciting times for geophysicists!

Before artificial satellites, $J_2$ could be estimated from Earth's measured oblateness from large scale geographic surveys and from models based on the equilibrium shape of a ball of fluid like this.

International Geophysical Year 1957 Source

Some of the early papers seem to be calculating values from satellite orbits

I would guess that all of them are doing that, and sixty years later this is still how people are doing it, to Earth, and to other solar system bodies.

Further reading:

Improve this answer
$\endgroup$
3
  • 1
    $\begingroup$ This makes sense; I thought the assumption that $\rho(\mathbf{s})$ was dependent only on radial distance from the planet center might be good enough, since that's what the book does for $C_{10}$, but that does seem too simple now that I think about it. That being the case, why have equations for these coefficients at all if we can't use them? $\endgroup$
    – zaen
    Commented Mar 26, 2019 at 3:50
  • 2
    $\begingroup$ @zaen I really like this book, I've quoted it here a few times before. I can't answer that for sure, but I assume that many scientists would not be happy using coefficients without seeing how they are properly defined. By listing those equations explicitly, there can be no question what $C_{nm}$ and $S_{nm}$ mean and how it should or shouldn't be used. Think of it as "mathematical tracibility" perhaps. $\endgroup$
    – uhoh
    Commented Mar 26, 2019 at 4:14
  • 1
    $\begingroup$ The fundamental physical constant G is much harder to measure on its own than the product GM is to estimate from orbits. G itself is only known to 22 parts per million, but GM_Earth is known to 2 parts per billion from satellite data, and GM_Sun is known to 60 parts per trillion from planetary orbits. $\endgroup$
    – Ryan C
    Commented Sep 3, 2020 at 20:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.