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The idea is to modify Newtonian gravity so that it fits measurements of orbits around the sun. For example the precession of Mercury's orbit unlike Newtonian $n$-body simulations.

I'm currently not using this in any serious simulation, but I'm just wondering because it'd simplify approximations a lot.

The modified formula for the force I found about half a year ago after googling was similar to this one: $$F =\frac{Gm_1m_2}{r^2} + \frac{Gm_1m_2B^2}{r^4} = \frac{Gm_1m_2}{r^2} (1+\frac{B^2}{r^2})$$

Similar means that I lost the link which I was unable to relocate and this is the only thing I wrote down elsewhere. $B$ stands for the dot product of velocity and unit vector pointing at the other object if I remember right.

This all sounds rather vague and arbitrary. Thus I'm asking whether anyone knows of something like this and the explanation behind it.

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    $\begingroup$ Are you trying to add general relativistic effects to Newtonian theory, or are you trying to empircally model the motions of the planets? Are you only considering the effects of the sun, or are you also considering planet-planet effects? $\endgroup$
    – Zo the Relativist
    Commented Oct 10, 2014 at 19:33
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    $\begingroup$ Looks sorta like MOND. $\endgroup$
    – Kyle Kanos
    Commented Oct 10, 2014 at 19:33
  • $\begingroup$ @JerrySchirmer planet-planet effects are included(that's what I meant by n-body simulation). You could call it relativistic effects, although it's not based on GR. Modeling the motion of the planets to fit observations fits better. $\endgroup$
    – DeinFreund
    Commented Oct 10, 2014 at 19:43
  • $\begingroup$ @KyleKanos I should test that, as it should make an impact on the small forces between planets. $\endgroup$
    – DeinFreund
    Commented Oct 10, 2014 at 19:57
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    $\begingroup$ @DeinFreund: If you're going to model the precession of Mercury, you can look into the Post-Newtonian formalism. It's basically a trick to approximate GR in a weak-field limit. For solar system stuff, you really only need the first couple of terms. $\endgroup$
    – Zo the Relativist
    Commented Oct 10, 2014 at 20:08

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Here are couple of references that describe professional uses of a post-Newtonian formalism to model the planets and the Earth's Moon:

Standish, et al. "Orbital ephemerides of the Sun, Moon, and planets," Explanatory Supplement to the Astronomical Almanac (1992): 279-323.
The relevant equation is 8-1 on page 3.

Petit and Luzum (eds.), "IERS Technical Note No. 36, IERS Conventions (2010)," International Earth Rotation and Reference Systems Service, Frankfurt, Germany (2010). You'll want to look at chapter 10. This describes the Moon, but not the planets. The IERS's focus is the Earth.

Note that the IERS document does describes relativistic time scales. Mixing and matching the non-relativistic time scales we use to keep time on the surface of the Earth with a post-Newtonian formalism doesn't buy much. JPL (the first reference) uses its own relativistic time scale, Teph. This is now very close to one of the officially defined time scales, Barycentric Dynamical Time (TDB). The IERS document uses a different relativistic time scale that doesn't tick with Earth-based clocks.

(Note: The acronym doesn't match the name. That's intentional. The official acronym is short for the French name "Temps Dynamique Barycentrique", regardless of what language you use for the full name.)

Edit: Regarding the simplified version specified in the question

My old, old copy of Marion, Classical Dynamics has something very close to the formula in the question. It extends the Newtonian relation

$$ \frac{d^2}{d\theta^2}\left(\frac 1 r\right) + \frac 1 r = -\frac m {l^2}r^2 F(r) = GM \frac {m^2}{l^2}$$

to

$$ \frac{d^2}{d\theta^2}\left(\frac 1 r\right) + \frac 1 r = GM \frac {m^2}{l^2} + \frac {3GM} {r^2 c^2} $$

from which one the force modified $F(r)$ can be expressed as

$$F(r) = -\frac {GMm}{r^2}\left( 1 + \frac{3 \, l^2}{m^2c^2 r^2} \right)$$

Using $\vec l \equiv \vec r \times (m \vec v)$, this can be rewritten as

$$F(r) = -\frac {GMm}{r^2} \left( 1 + \frac{3 \, ||\vec r \times \vec v||^2}{c^2 r^2} \right) $$

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  • $\begingroup$ The edit looks promising (and a lot simpler than your other two links). I'll wait if there are any other replies, but try this for sure. $\endgroup$
    – DeinFreund
    Commented Oct 11, 2014 at 14:05

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