Which orbit around Earth would have the least amount of perturbation from the moon, Sun etc.? Does an orbit exist where in either direction of travel that the perturbation would be the same?

The Orbital Mass Accelerator Engine Theory


1 Answer 1


tl;dr: GEO! It turns out that moving out from LEO, the falling effect of Earth's $J_2$ meets the rising effect of the Moon at about the geosychronous distance, which turns out to be quite nice bit of luck!

I only have this plot and the accompanying text. Hopefully someone will be able to add an another answer with the appropriate equations that the author mentions!

This is a good question, and we're in luck, there's an answer!

Note that the plot is log-log, and the y axis (on the left) has a tick every 105 (100,000)!

In the figure $GM$ is Earth's main gravity. It is not a perturbation. You can ignore that.

The next line down is $J_2$ which is the main perturbation from the Earth. It's from the Earth's equatorial oblateness and drives all kinds of satellite orbital changes. Since it dominates for most Earth orbit you just follow that line to the right until it intersects something, and that's happens (by mostly a coincidence) right around GEO, where the Moon's perturbation rises to meet it.

You can see that over about a factor of 10 in radius $J_2$ falls by 104 and higher order $J$ terms fall faster because they are higher order multipole moments, while solar radiation pressure stays flat (your always the same 1AU from the Sun) and the effect of Sun and Moon increase very slowly. The reason for that is the larger your orbit, the larger the variation or oscillation in distance from the Sun and the Moon during one orbit.

Below is borrowed from my answer to the question The sorting of perturbational effects by the power

I found the following plot in the book Satellite Orbits; Models, Methods, Applications by Oliver Montenbruck and Eberhard Gill, Springer, 2000. The figure and description can also be found in google books. It's a low quality snapshot but it's hard to capture a dozen different dependencies over 20 orders of magnitude without showing the whole thing.

Satellite Orbits; Models, Methods, Applications, Montenbruck & Eberhard Gill

Here is the bit of text that discusses the figure in more detail:

The effect of various perturbations a s a function of geocentric satellite distance is illustrated in Figure 3.1. For the calculation of the influence of atmospheric drag on circular low-Earth satellite orbits, exospheric temperatures between 500K and 2000K (cf. Sect. 3.5) have been assumed. The area-to-mass ratio used in the computation of non-gravitational forces is 0.01 m2/kg. For specially designed geodetic satellites like LAGEOS, the corresponding value may be smaller by one or two orders of magnitude. The perturbations due to various Geopotential coefficients Jn,m and the lunisolar attraction have been calculated from rule-of-thumb formulas by Milani et al. (1987). For the purposes of comparison it is mentioned that a constant radial acceleration of 10-11 km/s2 changes the semi-major axis of a geostationary satellite by approximately 1 m.

Aside from the aforementioned forces, various minor perturbations are considered in Fig. 3.1 which produce accelerations in the order of 10-15 to 10-12 km/s2. The most are due to the radiation pressure, resulting from the sunlight reflected by the Earth (albedo), as well as relativistic effects and the solid Earth tides.

  • $\begingroup$ what is it called when you travel in the opposite direction of GEO? $\endgroup$
    – Muze
    Dec 9, 2018 at 15:47
  • $\begingroup$ @Muze dunno, ask a question! $\endgroup$
    – uhoh
    Dec 9, 2018 at 15:49
  • $\begingroup$ in astronomy.se $\endgroup$
    – Muze
    Dec 9, 2018 at 15:52
  • $\begingroup$ Excellent answer! Is it possible to explain the differences between J2,0 J2,2 J6,6 and J18,18? $\endgroup$
    – Uwe
    Dec 9, 2018 at 18:13
  • $\begingroup$ @Uwe that's just a very sparse sample of all of the $(n, m)$ spherical harmonics. Most require both a $C$ and and $S$ term (sin, cos) for phase, but that wouldn't make sense in this plot of magnitudes. If I explained further I'd probably say something wrong. Have a look at Ceres gravity from spherical harmonics from Dawn, how to get the coefficients, definitions and potential? and the answer there, and ask a new question if that's not enough. $\endgroup$
    – uhoh
    Dec 9, 2018 at 23:46

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