# NavList:

## A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding

**Re: The Three-body Problem**

**From:**Frank Reed

**Date:**2021 Oct 28, 09:53 -0700

Of the JPL numerical integrations of the Solar System, Dave Walden wrote:

"It includes the interactions of over 350 bodies."

That's *nothing.* :) There are nearly 5**·**10^{57} massive objects in the Solar System (see note **‡** ). Most are quarks, some are electrons. I'm ignoring particles travelling at, or nearly at the speed of light since their gravitational fields are almost non-existent. Each and every one of those particles has a gravitational field, and each pulls on any test particle with a force depending on its mass and distance from that test particle (proportional to m/r^{2} where the mass of each individual subatomic particle and r is the distance to that specific particle from the test particle's location). If we want the gravitational force created by the Sun, we have to add up all of the gravitational fields of every particle within the Sun. A supercomputer as large as the Galaxy itself could not numerically integrate this vast system! And so, JPL, * j'accuse!* You could not possibly predict the positions of the planets. Yet another NASA lie, no doubt.

Ha.

Of course the most important solution in all of Newtonian gravitational physics is the two-body problem, also known as the "Kepler problem", but a close second is the proof that a big ball of mass (if it's truly spherically symmetric, and if we can ignore details smaller than some scale) acts gravitationally as if it's a point of mass at the physical center of the big ball. So despite the fact that the Solar System has an insanely huge number of particles, each with its own gravitational pull, we can reduce that to a very small number of nearly spherical masses (and, in special cases, a few consisting of multiple particles counted in the hundreds, for example a multipolar expansion of the Earth's gravitational field --the same thing responsible for the "deflection of the vertical" in celestial navigation). The analysis of gravitational systems would be radically more difficult if did not have this magical theorem that let's us replace the gravitational field of trillions of trillions of trillions of particles arranged in a spherical shape with a single "effective" point particle at the center of the original sphere. This is a solution of the N-body problem where N is crazy large (and where we assert the existence of some other non-gravitational forces to guarantee that the N bodies are arranged in some relatively fixed, nearly spherical shape).

While I'm at it, there's another special case solution of the three-body problem that was in the news just last week. You can find clues to that special case solution here: Launch of Lucy.

Frank Reed

**‡** That number, 5**·**10^{57}, which I will write as 5e57 in this note, comes from the mass of the Sun which is a little less than 2e33 grams. Add in everything else in the Solar System, and it's still about 2e33 grams. Next, one gram of anything contains about 6e23 nucleons. Nucleons are neutrons and protons, but mostly protons for the case at hand since the vast majority of the mass in the Solar System is Hydrogen plasma in the Sun. And each Hydrogen atom (same for H plasma) contains three quarks and one electron, so the total number of massive fundamental particles is approximately *total grams* multiplied by *nucleons per gram* multiplied by *four*.