# What Earth gravity model is used for different types of calculations?

When performing orbital calculations, under what circumstances should I assume:

• You need not only to model the shape of the Earth, you also need an assumption about the distribution of mass. Constant density and sperical shape is the easiest case. – Uwe Jun 10 '17 at 9:04

Since this answer is a bit longish, a TL;DR is in order. Depending on the application, one should use

1. A point mass / spherical mass distribution model.
2. A model that incorporates the effects of the Earth's oblateness. This is subtly different from using the reference ellipsoid.
3. A model that incorporates the effects of Earth's not-quite ellipsoidal shape, based on very detailed analyses of the orbits of existing satellites. For the Earth, there is but one approach, which is to use a spherical harmonics model of the Earth's gravitational potential field.
4. A model that accounts for temporal variations in those spherical harmonics models. For the Earth, the largest temporal variations (by far) are short term variations due to the tides. Longer term variations are observable by specially designed satellites.
5. A very different kind of model for small bodies such as asteroids or comets that look more like lumpy potatoes than distorted beach balls.

When performing orbital calculations, under what circumstances should I assume:

• a perfectly spherical Earth

A point mass model works quite nice when you are doing rather simplistic mission planning, analyzing an object that is very far from the Earth, or doing undergraduate homework. Beyond that, you'll need to account for the fact that the Earth is not quite spherical.

• a reference ellipsoid

Calculating gravitation from an ellipsoid is nontrivial. There is missing information (e.g., a density model), and even with that information, the calculation will lead elliptical integrals. However, the largest deviation from a point mass / spherical mass distribution model is captured quite nicely by the Earth's second dynamic form factor $J_2$. With this factor, the gravitational potential as a function of orbital radius $r$ and geocentric latitude $\phi$ is $$U(r,\theta) = -\frac{\mu_E} r \left( 1-J_2\left(\frac a r\right)^2 \left(\frac{3\sin^2\phi -1}2\right) \right)$$ where

• $a$ is the Earth's equatorial radius,
• $J2$ is the Earth's second dynamic form factor, which is caused by the Earth's equatorial bulge, and
• $\mu_E$ is the Earth's gravitational parameter1.

Take the gradient of the potential, negate, and voila! you'll have the gravitational acceleration.

There's another issue with using the reference ellipsoid: The value of $J_2$ would be exactly described by the Earth's oblateness and its rotation rate if the Earth was in hydrostatic equilibrium. The observed value of $J_2$ and the calculated value based on the reference ellipsoid and the Earth's rotation rate differ slightly. The reason is that the Earth is not quite in hydrostatic equilibrium. It is instead still recovering from the huge masses of ice that covered large tracts of the Northern Hemisphere up until about 12000 years ago.

• a geoid

Just as you don't want to use the reference ellipsoid, you don't to use the geoid, either. Calculating gravitation from a geoid model is ridiculously difficult. What you want instead is a spherical harmonics model. Geoid models are calculated from spherical harmonics models. The spherical harmonics model is what you want. The $J_2$ term discussed above is the leading term in the non-spherical part of the Earth's spherical harmonics model.

As a function of distance to the center of the Earth $r$, geocentric latitude $\phi$, and geocentric longitude $\lambda$2, the spherical harmonics expansion of the Earth's gravitational potential is $$U = -\frac{\mu_E}r \left(1 + \sum_{n=2}^N\sum_{m=0}^n \left(\frac a r\right)^n\,\overline{P_{n,m}}(\cos\phi)\left(\overline{C_{n,m}}\cos(m\lambda) + \overline{S_{n,m}}\sin(m\lambda)\right)\right)$$ where

• $\overline{P_{n,m}}(\cos\phi)$ are the fully normalized associated Legendre functions of the first kind, and
• $\overline{C_{n,m}}$ and $\overline{S_{n,m}}$ are the fully normalized spherical harmonics coefficients for the model.

Take the gradient of the potential, negate, and voila! you'll have the gravitational acceleration.

The fully normalized associated Legendre function and the fully normalized coefficients are used for two reasons. One is that the unnormalized coefficients (which is what physicists tend to use) tend result in all kinds of numerical problems. The other is that the fully normalized coefficient are the de facto standard. These coefficients are available for the Earth and for many bodies besides the Earth.

You can find detailed descriptions in Vallado and many other texts. A free somewhat dated online paper that describes spherical harmonics as used in gravitation is The Evolution of Earth Gravity Models Used in Astrodynamics. A very recent open paper that describes how one such global gravity model was constructed is A GOCE only gravity model GOSG01S and the validation of GOCE related satellite gravity models.

You can find multiple implementations online, in a number of languages, that use spherical harmonics to model gravitation. Regarding the coefficients themselves, the International Center for Global Gravity Field Models maintains a catalog of static global gravity models at http://icgem.gfz-potsdam.de/tom_longtime.

The models listed on the page cited above are static. You'll need to account for temporal variations if you want to be extremely accurate. The largest temporal variations result from how the gravitational forces by the Moon and Sun distort the shape of the Earth. These solid Earth tides subtly affect satellite orbits out to GEO and beyond.

There are also seasonal effects such as the buildup and melting of snow in Siberia and the buildup and drying of soil moisture in tropical rain forests. These are observable by specially designed satellites. Even longer term, the ice sheets over Greenland and Antarctica are melting, and lands in the far north are still rebounding from the glaciation that ended 12000 years ago. How to model these temporal effects is beyond the scope of this answer.

no particular format at all

There's always going to be some format / model. A spherical harmonics model does not work well for a small object whose shape is more like a lumpy potato than distorted ball. (Note well: This does not apply to the Earth.) This is getting into PhD land, quite literally. For example, here's a PhD thesis on how to compute the gravitation in the vicinity of a potato-shaped object.

## Footnotes

1 Regarding the gravitational parameter: Conceptually, this is the product of the gravitational constant $G$ and the Earth's mass $M_E$: $\mu_E = GM_E$. In practice, the Earth's mass is calculated from the gravitational parameter: $M_E = \mu_E/G$. The problem is that $G$ is only known to four or five places of accuracy while $\mu_E$ is known to about nine places of accuracy. This means that almost all of the uncertainty in the estimates of the Earth's mass is due to the uncertainty in the gravitational constant. Never use $GM$ if you know the gravitational parameter to more than five places of accuracy.For the Earth, we know $\mu_E$ to about nine places of accuracy.

2 That you need to know longitude means you need a model of the Earth's rotational state. These range from the very simple (the Earth rotates at a constant rate) to the ridiculously complex (thousands and thousands of terms). The ridiculously complex models target the milliarcsecond level of accuracy needed by radio astronomers. A very simple, constant rate model might be good for a few orbits, but you'll still need a good initial rotational state. The Standards of Fundamental Astronomy provides a library functions (both in Fortran and C) that calculate the Earth's orientation for you. So does JPL's SPICE Toolkit.

• Would a tl;dr escalation path be 1) point/sphere, 2) include J2 effects with a perturbation model, 3) use a standard spherical harmonic expansion? – uhoh Jun 10 '17 at 5:52
• @uhoh: 4) include individual gravimetric map of the planet as generated by satellites. – SF. Jun 10 '17 at 13:46
• @SF. No. Those surface gravimetric maps are quite useless in space. What you want is one of the static global gravity models described and cataloged here: icgem.gfz-potsdam.de/tom_longtime. Since these are static models, the next step, (4), would be to model time variations, the primary one being the effects of the solid Earth tides on the vehicle's orbit. For a vehicle orbiting at a few hundred km altitude, the effects of the Earth tides are generally smaller than are the uncertainties in drag. – David Hammen Jun 10 '17 at 15:46
• @uhoh - I added some content, and since the answer is getting long, I added a tl;dr as well. – David Hammen Jun 10 '17 at 17:25
• @DavidHammen that's great! I never would have thought about that. Now I have, so I couldn't help but ask this follow-up question! – uhoh Jun 10 '17 at 17:38

Called2Voyage is right. Depends on application - and distance.

For influence on probes far at other planets, point mass is sufficient.

For spaceflight around the Moon or vicinity of Lagrange points, sphere is okay.

You'll want a geoid model for LEO satellites in any orbits other than equatorial.

You'll need an even more accurate model for sun-synchronous satellites, taking local gravitational anomalies into account.

• How does using a sphere differ from using a point mass "for spaceflight around the moon or vicinity if Lagrange points"? – JiK Jun 9 '17 at 20:31
• @JiK for orbit propagation they're equivalent, but if you're describing a spherical model you need its radius; not for a point. – pericynthion Jun 9 '17 at 21:08
• You don't want a geoid model for LEO. They are of little use for calculating gravitation. You'll instead want a spherical harmonics model, even for an equatorial orbit. The first term in a planet-centered spherical harmonics model is spherical gravity. The next non-zero term is the second dynamic form factor $J_2$, caused by the Earth's equatorial bulge. This affects all orbits, even equatorial ones, out to the Moon and beyond. If you want to develop a global geoid model, you'll also want a spherical harmonics model of Earth's gravitational field. – David Hammen Jun 10 '17 at 0:40
• SF. you need to change this answer. According to the Shell theorem first proposed by Isaac Newton, "A spherically symmetric body affects external objects gravitationally as though all of its mass were concentrated at a point at its centre." As long as your are not underground, there is no gravitational difference between a spherically symmetric Earth and a point Earth. This is why a point Earth is not mentioned in the question. – uhoh Jun 10 '17 at 1:41
• @pericynthion stackexchange answers are not written to show how much one knows. The answer draws a clear distinction where there is none and that could be confusing or misleading to future readers. Unless the orbit passes through a tunnel below the surface of the Earth, you would always use a point - the term "spherical model" does not even make mathematical sense here. We should take responsibility to correct wrong stuff we write when it is pointed out to us. – uhoh Jun 10 '17 at 10:51