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According to Wikipedia:

“ ... for $n$ mutually interacting bodies, Newtonian forces on body $i$ from the other bodies $j$ are” given by the summation equation$$\ddot{\mathbf{r}}_{i}=\sum_{j=1,j\neq i}^{n}\frac{Gm_{j}\left(\mathbf{r}_{j}-\mathbf{r}_{i}\right)}{r_{ij}^{3}},$$

“with all vectors being referred to the barycenter of the system.”

The summed over accelerations denoted by the right side of this equation do not appear to take into account gravitational forces between the $j$ other bodies? In other words, all the accelerations are functions of $\mathbf{r}_{i}$, the position vector of the body acted on by the $j$ other bodies. Is that interpretation correct?

EDIT

So if, for example, I wanted to calculate the Earth's acceleration due to the effect of the Sun and Jupiter, I would use the equation three times. Once with $i$ representing Earth to find the Earth's acceleration; once with $i$ representing the Sun to find the Sun's acceleration; and once with $i$ representing Jupiter to find Jupiter's acceleration. If I knew the initial position and velocity of the three bodies I could then numerically integrate these three equations to find positions and velocities after a certain time. Is that right?

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  • $\begingroup$ Great question! For a rather inefficient example of how that might be done see deriv_Newton_Only(X, t) in the script in this answer to How to calculate the planets and moons beyond Newtons's gravitational force?. $\endgroup$
    – uhoh
    Aug 6, 2022 at 22:20
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    $\begingroup$ @uhoh - thanks for that. Using an Excel spreadsheet and Cowell's method I've managed to calculate the position of Phobos to within 308km of where JPL says it will be after 48 hours (about six orbits of Mars). I included the gravitational effects of all the planets and used a one minute time interval. I started off using a six hour time interval, and then wondered why my Phobos was hurtling off into the cosmos. The same spreadsheet predicts the position of Mars to within 5,911km after 687 days (using a six hour time interval). $\endgroup$
    – Peter
    Aug 9, 2022 at 8:03
  • $\begingroup$ Excellent, sounds fun! It's quite a challenge integrating equations of motion and n-body interactions in Excel, did you use a spreadsheet or Virtual Basic for the calculation? $\endgroup$
    – uhoh
    Aug 9, 2022 at 11:01
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    $\begingroup$ @uhoh - I used a single worksheet (no Visual Basic) and the leapfrog method of numerical integration. Nine columns for each of the planets and the Sun (three each for distance, velocity and acceleration), except for Mars where I had an additional column to give the Mars-Sun distance (using Pythagorean theorem). It was good fun but would have been much, much easier if I were able to code. $\endgroup$
    – Peter
    Aug 9, 2022 at 15:24
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    $\begingroup$ @uhoh - Thanks for that. I actually ordered a beginner's guide to Python yesteday. I'm looking forward to replicating (if I can) my spreadsheet results using Python. $\endgroup$
    – Peter
    Aug 10, 2022 at 19:09

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Yes, that interpretation is correct. The given formula only denotes the (integral) acceleration seen by a single mass element at a certain time $t$ (as implicitly $\mathbf{r_i} = \mathbf{r_i(t)}$ where $i$ could be in your example of the system of the three bodies Earth, Sun and Jupiter either of those three).

In order to solve the overall system you have to calculate the acceleration on each mass element in an identical manner. You can get the evolution of the overall system by choosing sufficiently small time steps and forward integrating it. There are other, possibly more suitable methods, like implicit methods, which do not suffer so much from adding up small errors in the integration which will happen naturally due to discretization. A number of methods to solve N-body problems are listed on the corresponding wikipedia page with useful references or on these lecture notes along with its references. You will find a plethora of information on how to approach these types of problems when you search for "N-body simulations".

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  • $\begingroup$ I think I've got that. Could you have a look at my edit to confirm I'm on the right track? Thanks. $\endgroup$
    – Peter
    Aug 3, 2022 at 15:21
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    $\begingroup$ yes, that's right. The accuracy depends on how small you choose your timesteps (and what actual integration method you choose). The numerics of solving these type of problems efficiently and with damped errors cover whole lecture series. $\endgroup$
    – planetmaker
    Aug 3, 2022 at 15:32
  • $\begingroup$ I think the OP has just discovered why it’s so complicated (some say impossible) to calculate the long-term evolution of the Solar System… $\endgroup$
    – Pierre Paquette
    Aug 3, 2022 at 21:20

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