So I've only recent started to try to wrap my brain around geopotential models. I've done work with some special perturbations using $J_2$ before, however I've never dealt with a full spherical harmonic model before.

After reading "Analytical Mechanics of Space Systems", I know that the geopotential field in spherical coordinates can be written as:

$$V(r,\phi,\theta) = - \frac{GM}{r} - \sum_{k=1}^{\infty}\frac{1}{r^{k+1}}\left(A_kP_k(\sin\phi) + \sum_{j=1}^kP_k^j(\sin\phi)(B_k^j\cos j\theta + C_k^j\sin j\theta)\right)$$

It later goes on to discuss how the conventional notation uses:

$$J_k = -\frac{A_k}{r_{eq}^k}$$

So far, thtis makes sense to me. The geopotential wikipedia page appears to have the resultant of this in the more "standard form" that I've seen many other places:

$$V(r,\phi,\theta)=-{\frac{GM}{r}} + \sum_{n=2}^{\infty}{\frac{J_{n}P_{n}^{0}(\sin \phi )}{r^{n+1}}}+\sum _{n=2}^{\infty}\sum _{m=1}^{n}{\frac {P_{n}^{m}(\sin \phi )(C_{n}^{m}\cos m\theta +S_{n}^{m}\sin m\theta )}{r^{n+1}}}$$

But now comes my confusion... My understanding is that the zonal terms ($J_n$) describe only latitude dependence, while the tesseral terms ($C_n^m, S_n^m$) describe both latitude and longitude dependence. Which is why when you assume an axially symmetric field, the above can be simplified down to:

$$V(r,\phi) = -\frac{GM}{r}\left(1 - \sum_{n=2}^\infty \left(\frac{r_{eq}}{r}\right)^n \tilde{J}_n P_n(\sin \phi)\right) = -{\frac{GM}{r}} + \sum_{n=2}^{\infty}{\frac{J_{n}P_{n}^{0}(\sin \phi )}{r^{n+1}}}$$

My assumption therefore would be that if you did not restrict yourself to having an axially symmetric gravity field, then you must need all of the J, C and S coefficients. Indeed, the wikipedia page gives these constants.

However, when I look at models such as EGM-96 and JGM-3, I find that only the tesseral coefficients are given.

As I stated earlier, I had read from this site that:

"The Tesseral terms represent the geopotential that is both latitude and longitude dependent."

Does this mean that the $J$ (zonal) terms are not needed, and only useful if the axially symmetric assumption is used? If so, how would one go about evaluating the potential equation given above, as it clearly requires $J$ terms. Do you need to calculate them from $C$ and $S$ terms?

If you have any guidance on how I should be thinking about this, please let me know! And if there are any resources you might recommend, that would be greatly appreciated!


1 Answer 1


For these type of files, use your same first equation, but in this form:

$$V(r,\phi,\theta)=-{\frac{GM}{r}} +\sum _{n=2}^{\infty}\sum _{m=0}^{n}{\frac {P_{n}^{m}(\sin \phi )(C_{n}^{m}\cos m\theta +S_{n}^{m}\sin m\theta )}{r^{n+1}}}$$

Note that the second summatory now begins at $m=0$. Zonal harmonics are $C^0_{n}=J_n$ and $S^0_{n}=0$, $n=2...\infty$.

Clarification: Note that the answer equation is the same the OP has written but putting the zonal harmonics term within the series expansion. If the OP wants to use its equation he just has to take $J_n=C^0_n$ from the file he is referring

  • 1
    $\begingroup$ Wow, this makes so much sense and I can't believe I never noticed it, since yeah, $C_2^0$ in the jgm-3 coefficients file I linked is just the standard $J_2$. I can't believe I didn't put two and two together. Anyways, thank you so much for clearing that up for me! $\endgroup$
    – Chris Gnam
    Nov 21, 2019 at 18:26
  • $\begingroup$ I'm far more confused about gravitational potentials expressed in spherical harmonics than the OP, thus my interest in your answer. Any thoughts on How can I verify my reconstructed gravity field of Ceres from spherical harmonics? $\endgroup$
    – uhoh
    Nov 22, 2019 at 0:09

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.