Q: Limit on a moon's orbital radius due to sun's gravity

[Buzz Bloom]

I recently came across a Wikipedia article about somebody's (?) law regarding limits on a moon's orbital radius because the sun's gravitational influence is greater than the planet's at some distance from the planet. As I recall, the law had two different names associated with it. In addition to the obvious relationship (shown below) for the values when the gravitational forces on the moon are equal for the sun and the planet, there was a useful diagram showing this effect together with consideration of the sun's influence on the Lagrangian points, L1 and L2.
(r_p/r_m)² = M_s/M_p, where
r_p is the distance between sun and planet
r_m is the distance planet and moon
M_s is the mass of the sun
M_p is the mass of the planet

Unfortunately I did not take notes, and now I can not find that article. Can anyone help me? Does anyone know a name of this law?

 

[Janus]

The Hill sphere:
https://en.wikipedia.org/wiki/Hill_sphere

 

[Buzz Bloom]

The Hill sphere:

Hi Janus:

Thank you very much for your prompt help.

Regards,
Buzz

 

[tony873004]

In a prograde direction, the Moon can be almost twice as far and still orbit the Earth. In a retrograde direction, it can go much farther than that. But in either case, its well short of the Hill Sphere.

Try it yourself. This simulation will run in your web browser. Press [>] on the Time Step interface to play. If the screen gets too cluttered, press C for clear.

http://orbitsimulator.com/gravitySimulatorCloud/simulations/1446088723620_outermost_moons.html

120 objects orbit the Earth in with semi-major axes evenly-spaced between 400,000 - 1,600,000 km. The green ones orbit in a prograde direction. The red ones are retrograde. The blue one is for scale. It orbits at the Moon's distance, but it is massless, so it does not influence the simulation.

The frame is constantly rotated to keep the Sun above the top of the screen, as the phases indicate. So the Earth / L1 is above the Earth, and L2 is below Earth. Particles that escape do so through the L1 and L2 regions.

If the Moon were not massless in this simulation, it would destabilize all prograde orbits external to its own. Some retrograde ones would still exist. To try, press E for edit, and change the Moon's mass to 0.0123 Em (Earth masses). Pause the simulation before making the edit.

 

[Buzz Bloom]

This simulation will run in your web browser.

Hi tony:

Thanks for your post.

The simulation page you cited is quite interesting. Although moon-to-moon interactions are not applicable, the simulation is actually more relevant to another thread

https://www.physicsforums.com/threads/why-is-there-no-dm-halo-around-the-solar-system.838583/​

than I expected it to be. I get it that retrograde motion of the moon reduces it's velocity around the sun, while prograde increases it.

In the thread I cited above, the"moon" is a DM particle, the "planet" is the sun, and the "sun" is the mass at and near the center of the Milky Way. There I am seeking help exploring how much DM our sun may have gravitationally captured in its orbit around the Milky Way center.

Regards,
Buzz