# Kepler's Laws and altitude of orbit

1. Sep 22, 2004

### Ellimist

A orbiting satellite stays over a certain spot on the equator of (rotating) Mars. What is the altitude of the orbit (called a "synchronous orbit")?

That is all the information given...

I have been trying for awhile now... anyone know?

2. Sep 22, 2004

### tony873004

20,400 km from the center of Mars is where an Areosychronous satellite must orbit, or 17006 altitude.

edit..
misspelling
changed aeorsychronous to areosychronous

Last edited: Sep 22, 2004
3. Sep 22, 2004

### Ellimist

um.... wow. how did you know that/figure it out?

and thank you... very, very much.

Last edited: Sep 22, 2004
4. Sep 22, 2004

### tony873004

$$r = (t / (2pi)) ^ {2 / 3} * (G * M) ^ {1 / 3}$$

where r is the radius of your orbit (distance from the center of Mars), t is the period of your orbit, which is the same as the rotational period of Mars (88643 seconds), G is the gravitational constant $$6.67 * 10^{-11}$$, M is the mass of Mars ($$6.4185 * 10^{23} kg$$).

Subtract from this the radius of Mars (3390 km) to get your altitude.

redo the math. I rounded off big time in my first post.

5. Sep 23, 2004

### BobG

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.

6. Sep 23, 2004

### Janus

Staff Emeritus
Well, not exactly. Kepler's third law deals with the relationship of the orbits of two bodies. Two get Kepler's Third Law from this equation you need to divide it by the same equation using different variables for the distance and period, For instance, t1,t2, and d1,d2.

This will then reduce to Kepler's third law:

$$\frac{t_{1}^2}{t_{2}^2}=\frac{d_{1}^3}{d_{2}^3}$$

Now you can solve the problem just uisng this form, but to do so you need to know the period and distance of a body already oribiting Mars. We have two to choose from: Phobos and Deimos.

Let's use Deimos

If we make d1 the distance of the stationary orbit we are looking for then t1 equals the rotation priod of Mars, (1.03 days), d2 equals the mean orbital radius. of Deimos (23,000 km) and t2 equals the period of Deimos(1.26 days), and we get

$$\frac{(1.03 days)^2}{(1.26 days)^2}= \frac{d_{1}^3}{(23,000km)^3}$$

rearranged to solve for d1:

$$d_{1}= 23,000 km \sqrt[3]{ \left(\frac{1.03 days}{1.26 days}\right)^2}$$

$$d_{1}= 20108 km$$

or 16711 km in altitude.

7. Sep 23, 2004

### tony873004

Trivia...
With Phobos orbiting ~10,000 km interior to a sychronous satellite, and Deimos orbiting ~ 3,000 km exterior from a synchronous satellite, is the sychronous region stable? Or will Phobos and / or Deimos tug the satellite out of sychronisity?

I already know the answer. I just want to see the guesses of others.

8. Sep 24, 2004

### BobG

The orbits of Phobos and Deimos are very nearly in resonance with each other and Mars's rotation period (the same set of relative locations repeats itself roughly every 4 days). If in perfect resonance, the effects of having a moon (or moons) tugging from behind the satellite would be cancelled by the effects of the moon(s) when it's ahead of the satellite.

With the orbit of Deimos being 3.956 times greater than Phobos, a synchronous orbit wouldn't be perfectly stable, but it would be close.

9. Sep 25, 2004

### tony873004

Bob is correct. The orbit is close to stable. It is stable over the lifetime of a communications satellite. Of course, like Earth, minor correction burns would be necessary from time to time to maintain perfection.

I'm not sure about the resonance being responsible, though. With Deimos being much closer to the sychronous spacecraft than Phobos, and Phobos being much more massive than Diemos, and the resonances being not quite perfect, it seems that it's a pretty complicated cancelling game.

The reason the orbit is stable is that Phobos and Deimos are just not massive enough to make a significant difference. Phobos and Deimos are so un-massive, that their Hill Spheres are below their surfaces. (ie. a spacecraft can not orbit either one of them).

If their mass were about 10% the mass of Earth's Moon, a sychronous satellite would immediately be pulled out of sychronisity. Any more massive than ~that would send the satellite's orbit into total chaos.

10. Nov 19, 2004

### CharlesP

What is a Hill Sphere?

11. Nov 21, 2004

### tony873004

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.

12. Nov 23, 2004

### Jenab

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

Jerry Abbott

13. Nov 24, 2004

### tony873004

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....

14. Nov 24, 2004

### Jenab

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.

With initially circular orbits having radii from 90% to 95% of the Hill radius, the sun's gravity causes fluctuations in the semimajor axis length and a rapid procession of its orientation, causing the trace of the test particle's motion over time to resemble one of those designs you might draw with the "spirograph" toy set. Stable, but weird.

Jerry Abbott

15. Nov 25, 2004

### tony873004

The weird spirgraph stuff actually starts as close as the Moon's orbit. This is the Sun's pull rotating the Moon's orbital nodes. That's what gives us the 18 year cycle of eclipses. But it's rather subtle at the Moon's distance. Slightly beyone the moon are where things start getting weird. I wonder if Hill radius means "able to complete one orbit" because you're right, things get ripped away a bit closer than that after completing just a few orbits.

What program are you using?

16. Nov 25, 2004

### pervect

Staff Emeritus
From what I can tell, the original usage "The Hill Sphere" is correct - the term "the Hill radius" is similar but omits some important proportionality factors.

The results I get are based on exact theory rather than simulations (but with some numerical approximations due to a series expansion so that the formula is only good for large mass ratios. The mass ratio of the earth/sun is very large, however).

The formula can be found in the

Wikipedia

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.

Here a is the orbital semimajor axis, m is the mass of the smaller body, and M is the mass of the primary.

The exact theory is only good when there are only three bodies, though - with more than three bodies, objects undoubtedly need to be inside the Hill sphere for the orbit to be stable.

I'll go a bit into the theory behind this result.

There is a constant of motion of the 3 body problem, called the Jacobi integral. You can see the mathematical formula for it

here

It's basically the Hamiltonian of the problem re-written in terms of position and it's derivatives (like the Lagrangian). This form of the Hamiltonian is often known as "the energy function". Because it's a constant of motion, for any orbit this quantity is the same everywhere.

There are some sample plots of the behavior of this function, plotted as an iinequality for some specific mass ratios. The function is plotted as an inequality because it involves both position and it's derivative, and the two dimension plot only shows the position terms. These plots are located

here

As long as the Jacobi integral is above a critical value, there is a bounded region that the third body cannot leave, as it does not have sufficient energy.

The critical value of the Jacobi integral occurs at the Lagrange points L1 and L2. This is where the numerical approximations come in - there is no exact formula for the location of the Lagrange points, but they are approximately located at

a (m/3M)^(1/3)

from the secondary, which gives the radius of the Hill sphere.

Note that for the mass ratios of .01 plotted, the so-called "Hill sphere" is really rather egg-shaped. For the Earth-sun case, the value of the mass ratio should be much larger, and the region should look much more spherical.

17. Nov 25, 2004

### pervect

Staff Emeritus
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.

18. Nov 25, 2004

### Jenab

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.

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 only

Rmax = 0.97 (1/3) (Me/Ms)^(1/3)

If D = 1.471E+8 meters (Earth's perihelion distance),

Rmax = 686300 km

Jerry Abbott

19. Nov 25, 2004

### Jenab

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.

Jerry Abbott

20. Nov 25, 2004

### pervect

Staff Emeritus
Very interesting results, what sort of program did you use to simulate the orbits?

I did some double checking for what happens when the mass ratio was equal to the earth-sun mass ratio of approximately 3e-6. The region of stability did NOT appear to be particularly spherical even for this low value of u. So we are left with the interesting result that the Hill sphere is not particularly spherical!

With a mass ratio of 3e-6, one expects the Lagrange points to be at (u/3)^(1/3), which is .01 of the distance from the primary to the secondary.

The plot of the stable region was significantly larger in the x direction than the y direction. The coordinate system is a rotating one (it rotates with the orbital period of the secondary around the primary), with the primary very near x=0 (it's actually at x=-u), and the secondary very near x=1 (it's actually at x=1-u), with u=3e-6. This puts the center of mass at the origin.

The (just barely) stable region extended approximately from x=1 +/- .01 in the x direction as expected, i.e. it went out to the Lagrange points L1 and L2. In the y direciton, however, the stable region extended only from 0 +/- .0066

This is still larger than your region, which I calculate to be .00466 for u=3e-6 by the forumla 0.97 * (1/3) * (3e-6)^(1/3). Of course the details are still different - this region represents a zero velocity surface in the rotating coordiante system, not a circular orbit in an inertial coordinate system.

The formula for J for this case for anyone interested in "playing with" the formula is:

1.0*y^2+1.0*x^2+1.9999940/((x+.30e-5)^2+y^2)^(1/2)+.60e-5/((x-.9999970)^2+y^2)^(1/2)

and the critical value of J is approximately 3.00089. The "interesting" range of J is very small.

It might be interesting to calculate the J values for circular orbits and double check the simulation results, but I haven't gone that far.

One good test of the simulator would be to see how well the Jacobi integral function was conserved by the numerical integrator, though.

As a sidenote, there are some very interesting things being done with three body dynamics to create some highly efficient orbital transfer for space missions near Jupiter. The mathematical level is extremely nasty, however. On the plus side, they do have some plots of Hill's regions, both in 2d and 3d One such link is at

this location

21. Nov 26, 2004

### Jenab

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

22. Nov 27, 2004

### tony873004

I'd love to see your program. It sounds similar to one I wrote:

http://orbitsimulator.com/gsimBeta/GravitySimulatorBeta.zip

Below are some screenshots testing the Earth Hill Sphere.

This universe contains the Sun, and all 9 planets, but no moons. Earth's moon is left out on purpose.

I started by letting the program generate 50 test particles (purple objects) in circular prograde orbits (0 inclination to the ecliptic). It choose random semi-major axis for the test particles of 1,500,000 km +/- 80%

I then added 50 additional test particles (green objects) in circular retrograde orbits (180 inclination to the ecliptic), also 1,500,000 +/- 80%

The first screen shot shows the system a few months into the simulation. All the particles beyond ~1,500,000 are in the process of getting pulled away by the Sun.

The second screen shot is from about 5.5 years later. After everything settles down, the Earth system is left with prograde particles that never exceed ~750,000 km as an apogee, and retrograde particles that never exceed ~1,500,000 km as an apogee.

http://orbitsimulator.com/gravity/eSOI1.GIF [Broken]
http://orbitsimulator.com/gravity/eSOI2.GIF [Broken]

Last edited by a moderator: May 1, 2017
23. Nov 27, 2004

### pervect

Staff Emeritus
What sort of integrator does "orbitsim" use?

From a theoretical point of view, my reading seems to indicate that a symplectic integrator is the best choice. It doesn't necessarily conserve energy or the Jacobi integral for the actual system being simulated - but it does represent the actual time evolution of some system that's very similar to the system being simulated.

Making some reasonable assumptions, I found that a stability value of about half (actually .49064) of the Hill radius was about right for an approximatly circular orbit. The way I determined this was to give the body a (prograde) circular orbit around the secondary, totally ignoring the primary. I then calculated the value of J at the point in the approximated orbit (the real orbit would not be circular because of the presence of m1) that was most distant from the primary. (I had to choose some point, this was the most convenient).

Some of the numerical sensitivity issues that come up become obvious when doing an actual calculation of J with a mass ratio of 3e-6, which represents the actual mass ratio of the Earth and sun.

The escape value of J is 3.00089 at x=.99, y=0. A more tightly bound situation, where the object is at half the critical radius, is repreented by x=.995, y=0, and J=3.00126. Thus one can see that at least 5 figures of accuracy are needed in the calculation of J. This accuracy must be maintained over the numerical integration process, a rather demanding requirement.

For anyone who wants to calculate J, or to see how well their simulator conserves it, a few more notes:

The normalized form of the Jacobi integral can be found in the references I posted earlier in the thread. The non-normalized form is

$$J = -{{\it xdot}}^{2}-{{\it ydot}}^{2}+{w}^{2} \left( {x}^{2}+{y}^{2} \right) +{\frac {2\,G{\it m1}}{\sqrt { \left( x-{\it x1} \right) ^{2} +{y}^{2}}}}+{\frac {2\,G{\it m2}}{\sqrt { \left( x-{\it x2} \right) ^{ 2}+{y}^{2}}}}$$

w is the angular velocity of the rotation of the system. x and y are the coordinates of the third mass measured in a co-rotating coordiante system. x1 is the x coordinate of m1, x2 is the x coordinate of m2. The origin must be at the center of mass (i.e. m1*x1+m2*x2 = 0). Otherwise one needs to adust the w^2(x^2+y^2) terms to what they would be if the origin was at actually at the center of mass.

The normalized form one can find on the WWW is found by making the substitutions and assumptions that

G=1, m1=u, m2=1-u, x2-x1 = 1

These assumptions imply that w=1 as w^2 = G(m1+m2)/r^3

24. Nov 27, 2004

### pervect

Staff Emeritus
On the topic of calculating J - taking another hard look at the defining equations, I _think_ it can be calculated in a non-rotating coordinate system as

[edit-minus sign]

$$J = -2(E - \vec{w} \cdot \vec{L})/m$$

here m is the small orbiting mass, m << m1 , m<<m2, E is the total energy of m (both gravitational potential energy + kinetic energy) in the center-of-mass frame of the system, L is the angular momentum of mass m around the center of mass of the system, and w represents the angular velocity at which the large primary and secondary masses m1 and m2 revolve around the center of mass.

This also makes the difference between J which is conserved, and E, which is not conserved, clearer.

Last edited: Nov 27, 2004
25. Nov 27, 2004

### tony873004

It uses a first order Euler method, more or less this:
Code (Text):

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

I'm debugging a Runge-Kutta 4 method, but since math is not my strong point, it's not quite ready yet.

Jacobi integral? Is that how much a body wanders from its position in a frame rotating with its average period? For example, in the following 2 screen shots, the test particle at 300,000 km from Earth traces a very small ellipse when the screen is made to rotate with its average period. But the test particle at 500,000 does quite a dance.

http://orbitsimulator.com/gravity/r1.GIF [Broken]
http://orbitsimulator.com/gravity/r2.GIF [Broken]

Last edited by a moderator: May 1, 2017