BobG said:If you look closely at Tony's equation, you'll notice that it is basically Kepler's third law, but rearranged to solve for the radius instead of the period.
I should have called it a Hill radius. It is the radius around a planet where something can orbit that planet. If a planet were the only thing in the universe, its Hill radius would be infinity. But once you throw a more massive body into the universe, there will be a distance at which the more massive body will strip an orbiting object away from the less massive body. That is the Hill radius.CharlesP said:What is a Hill Sphere?
That's what I used to think the Hill radius was. But I did some simulations on the three-body problem, Earth-Sun-test particle, and it looks like the outer stability radius istony873004 said:I should have called it a Hill radius. It is the radius around a planet where something can orbit that planet. If a planet were the only thing in the universe, its Hill radius would be infinity. But once you throw a more massive body into the universe, there will be a distance at which the more massive body will strip an orbiting object away from the less massive body. That is the Hill radius.
For the Earth, the Hill radius is about 1,500,000 km. Things can orbit the Earth if they are closer than that, but if they are further, the Sun will strip them away. The Earth would have a larger Hill sphere if it were further from the Sun. Jupiter's Hill sphere is huge since it is massive and far from the Sun.
I didn't try retrograde orbits. But I did notice that for prograde orbits, with the test particle's orbit in the ecliptic plane, an initially circular orbit of the test particle around Earth having a radius between ~97% and 100% of the Hill radius led to some exceptionally chaotic (and usually very temporary) movements of the test particle, until at some point there's a weak slingshot event and the sun nabs the test particle anyway.tony873004 said:I don't doubt your formula. I get similar results with Gravity Simulator. I was just using the traditional formula of (off the top of my head) d * (3M/m)^3 or something like that. I'll pull out the textbook tommorow and see how badly my memory fails me :)
But a retrograde orbit does go 1,500,000 and beyond...
Jenab said:That's what I used to think the Hill radius was. But I did some simulations on the three-body problem, Earth-Sun-test particle, and it looks like the outer stability radius is
Rmax = (D/3) { (Me+m) / Ms }^(1/3)
Where
Me is the Earth's mass.
Ms is the sun's mass.
m is the mass of the test particle.
D is the distance between Earth and Sun.
If D=1.4959787E+8 km, Me=5.976E+24 kg, Ms=1.989E+30 kg, and m=0, then
Rmax = 719553 km
Which is about half of your estimate.
Jerry Abbott
I believe that I was confused about what was implied by the term Hill radius. I thought that this was the maximum distance for the longterm stability of a circular orbit against a perturbing mass.pervect said:The relevant formula is
r = a (m/3M)^(1/3)
as per the Wikipedia article. This is different by a factor of roughly 2:1 from your result.
Apparently, what I determined with my numerical trials was the maximum stable radius for prograde satellite orbits. For satellites of negligible mass, my empirically determined maximum radius is onlyWikipedia said:The Hill sphere is but an approximation, and other forces (such as radiation pressure) can make an object deviate from within the sphere. The third object must also be of small enough mass that it introduces no additional complications through its own gravity. Orbits at or just within the Hill sphere are not stable in the long term; from numerical methods it appears that stable satellite orbits are inside 1/2 to 1/3 of the Hill radius (with retrograde orbits being more stable than prograde orbits).
The simulations began with circular orbits and remained so while R << Rmax. As R became larger, there began a tendency for the orbits to ellipticize. As R approached Rmax, the eccentricity of the satellite orbit became highly variable with a rapid precession of the major axis probably resulting from a torque by the perturbing body. In an inertial reference frame the satellite began to trace a "spirograph" kind of pattern. When R exceeded 0.97 Rmax (or closely thereabout), the motion became temporarily chaotic, until the satellite was no longer bound to the planet.pervect said:I should add that the "Hill sphere" is the zero velocity surface of the conserved "energy function". An object with a nearly circualr orbit will have to be well inside the sphere for it to be in the stable region. As I think about it, I suspect that this might be the main factor which explains the difference in the simulation results from the theory - if the simulations were of circular orbits, the stability region would be significantly smaller - by a factor of sqrt(2)/2 in radius.
Jenab said:I was using a straightforward three-body program that I wrote myself, using a time step of about 1 second and constant acceleration during each time interval. I put in the Sun, Earth, and a test body. I omitted the moon for most of my tests because what I was trying to determine was a general value for the orbital stability radius for an earth-sized planet relative to a sun-sized star.
None of the objects was "fixed..." they were all free to move. However, I kept the "field of view" centered on the "Earth" for all runs, the better to show the relative motion of the test body. The coordinate system did not rotate, though of course it translated through space as the "Earth" went around the "sun" in its orbit.
I also used this program, with different initial conditions, to simulate multiple star systems. I got some nice slingshot events sometimes.
Jerry Abbott
for k=1 to objects
for j=k+1 to objects
a = GM1 / d^2
a2 = GM2 / d^2
// update velocities
next j
next k
for k=1 to objects
// update positions
next k
