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Could Earth Capture a SECOND Satellite the size of the Moon?

by Taymith
Tags: capture, earth, moon, satellite, size
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tony873004
#19
Jan21-14, 08:09 PM
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Quote Quote by Borek View Post
Large satellites orbiting planets have some interesting properties when it comes to the dynamics of the system. I believe I have read somewhere (possibly even on PF) that lack of other satellites of the Earth is due to the fact Moon interferes with their orbits, effectively ejecting them into space.

IOW, it is not just a matter of putting something on the orbit, it is also a matter of keeping it in that orbit and not destroying the already existing system.
I think we have discussed this here before. I like simulating this stuff. There are no prograde orbits around Earth exterior to the Moon that are stable. Any object put in any prograde orbit exterior to the Moon gets ejected from the system in just a few orbits. Retrograde orbits exterior to the Moon are fine. But they're not too common in the solar system. Prograde orbits interior to the Moons are stable. But the Moon was once much closer to Earth. It would have swept that area (volume?) clear as it migrated to where it is now.
|Glitch|
#20
Jan26-14, 10:08 AM
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Quote Quote by Keln View Post
This is what I understand to be a major "problem" in having a second large satellite in the Earth system. As moons go, our Moon is quite dense and massive, and simply inserting a large satellite into the system isn't going to fly. Small objects for mining and the like seem feasible, since they could possibly be constantly manipulated to keep them in orbit, but putting, say, Ceres into orbit, not even considering the energy required to do that, seems destined for failure over the long term. The interaction between the Earth, the moon, and the smaller, yet large satellite would, I think, end up with the smaller object getting booted out.
I agree. Just out of curiosity, I created a simulation using the Sun, Earth, Moon, and Ceres. The only way I could keep Ceres in a stable orbit was by placing it in Earth-Moon Lagrange Points 3, 4, and 5. Everywhere else (including Earth-Moon Lagrange Points 1 and 2), Ceres either crashes into the Moon, or gets flung out of the Earth-Moon orbit.

In absolutely every case (even with Ceres in the Earth-Moon Lagrange Points 3, 4, and 5), the extra mass was sufficient to slightly slow down the Earth's velocity around the Sun (from 29.8 km/s to 29.3 km/s), causing Earth's orbit to increase slightly (1.02 AU SMA) and increasing the length of a year (adding about 8.4 hours per year).

The amount of energy required to move an object the size of Ceres would be huge. Well beyond all the energy of our combined nuclear weapons. However, once in either the Earth-Moon Lagrange Points 3, 4, or 5, no additional energy would be required to keep it in a stable orbit.
D H
#21
Jan26-14, 11:00 AM
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Of course an object at the first and second Lagrange points is going to be booted out of the system. Those points aren't stable. So is the third. Only the two triangular Lagrange points are stable.

That you couldn't find any stable orbits suggests that you might just be seeing an artifact of your numerical integrator rather than something real. If you are using the classical fourth order Runge-Kutta integrator, you can almost certainly blame the booting on the integrator rather than reality. The same applies for low order symplectic integrators. The large truncation errors associated with these techniques means the results are pure fiction after a few dozen orbits. A good integrator for the four body problem (the Sun is going to be a big perturber) is hard to come by.


There has been a lot of interest of late in selenocentric distant retrograde orbits (google that term). These orbits appear to be stable for hundreds of revolutions given the right evection angle (between -90 and +90 degrees, more or less). Some are very distant indeed. At a distance of 70,000 km from the Moon, these orbits are anything but elliptical when viewed from the perspective of a Moon-centered frame, and they are rather exotic when viewed from the perspective of an Earth-centered frame.
mfb
#22
Jan26-14, 12:44 PM
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Quote Quote by TumblingDice View Post
That doesn't make sense. If there were more *unknown* asteroids, how could you know?
Most known planets are in our galaxy, but we know our galaxy is not unusual in that respect - other galaxies have stars with planets as well, we just don't have the technology to detect them (with a few candidates as exception).
In the same way, you can know that the Kuiper belt has many objects we did not discover so far, as our telescopes are not good enough and/or did not scan the whole sky with the same precision.
tinypositrons
#23
Jan26-14, 05:40 PM
P: 28
It would screw up our lives. That's all I can say.
|Glitch|
#24
Jan26-14, 09:31 PM
P: 54
Quote Quote by D H View Post
Of course an object at the first and second Lagrange points is going to be booted out of the system. Those points aren't stable. So is the third. Only the two triangular Lagrange points are stable.

That you couldn't find any stable orbits suggests that you might just be seeing an artifact of your numerical integrator rather than something real. If you are using the classical fourth order Runge-Kutta integrator, you can almost certainly blame the booting on the integrator rather than reality. The same applies for low order symplectic integrators. The large truncation errors associated with these techniques means the results are pure fiction after a few dozen orbits. A good integrator for the four body problem (the Sun is going to be a big perturber) is hard to come by.


There has been a lot of interest of late in selenocentric distant retrograde orbits (google that term). These orbits appear to be stable for hundreds of revolutions given the right evection angle (between -90 and +90 degrees, more or less). Some are very distant indeed. At a distance of 70,000 km from the Moon, these orbits are anything but elliptical when viewed from the perspective of a Moon-centered frame, and they are rather exotic when viewed from the perspective of an Earth-centered frame.
I did not consider putting Ceres in a lunar orbit. When I place Ceres 70,000 km in a lunar orbit (30.1 km/s with an eccentricity of 0.014, or 19.1 days per orbit around the Moon), using RK4, in about 4.5 (~90 days) orbits the eccentricity of Ceres' orbit increases to 0.56 and is flung out of the Earth-Moon orbit into an orbit around the Sun. Furthermore, the eccentricity of the Moon increased, bringing the Moon almost 100,000 km closer to Earth at its semi-minor axis.
D H
#25
Jan26-14, 10:15 PM
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You can't use RK4 and expect anything remotely resembling reality after even one orbit. It is not a stable integrator.
tony873004
#26
Jan27-14, 06:23 PM
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The RK4 integrator I use in Gravity Simulator can hold a Ceres-mass object in the Moon's L4 or L5 indefinitely. If I input solar system data for today, and run it backwards, I can accurately postdict solar eclipses from hundreds of years ago. My predictions for future eclipses are in excellent agreement with the predictions given in Nasa's eclipse website. This is after not just one orbit, but thousands of lunar orbits, hundreds of Earth orbits, and hundreds of orbits of the other planets that perturb the system.

My original Euler-Cromer integrator can do these things too provided the time step is low enough.

Beyond hundreds of years, the postdictions start to fail, not because of integrator error, but because of forces not modeled (i.e. Earth is not a perfect sphere like I pretend it to be). Likewise, I can integrate forward and find the circumstances of Venus' next transit of the Sun, and be in excellent agreement with published values. But I can't predict the next time Io will occult Europa, as Jupiter's oblateness is significant but not modeled in the code.
|Glitch|
#27
Jan27-14, 10:01 PM
P: 54
Quote Quote by tony873004 View Post
The RK4 integrator I use in Gravity Simulator can hold a Ceres-mass object in the Moon's L4 or L5 indefinitely. If I input solar system data for today, and run it backwards, I can accurately postdict solar eclipses from hundreds of years ago. My predictions for future eclipses are in excellent agreement with the predictions given in Nasa's eclipse website. This is after not just one orbit, but thousands of lunar orbits, hundreds of Earth orbits, and hundreds of orbits of the other planets that perturb the system.

My original Euler-Cromer integrator can do these things too provided the time step is low enough.

Beyond hundreds of years, the postdictions start to fail, not because of integrator error, but because of forces not modeled (i.e. Earth is not a perfect sphere like I pretend it to be). Likewise, I can integrate forward and find the circumstances of Venus' next transit of the Sun, and be in excellent agreement with published values. But I can't predict the next time Io will occult Europa, as Jupiter's oblateness is significant but not modeled in the code.
I had the same experience holding a Ceres-mass object in the Moon's L4 and L5 points indefinitely. However, D. H. referred me to selenocentric distant retrograde orbits. When I place Ceres 70,000 km in a lunar orbit (30.1 km/s with an eccentricity of 0.014, or 19.1 days per orbit around the Moon), using RK4, in about 4.5 (~90 days) orbits the eccentricity of Ceres' orbit increases to 0.56 and is flung out of the Earth-Moon orbit into an orbit around the Sun. According to selenocentric distant retrograde orbits, that should not be happening. Ceres should be in a stable orbit around the moon.

I also agree that the margin for error is directly proportional to the step increment.
tony873004
#28
Jan27-14, 10:29 PM
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Quote Quote by |Glitch| View Post
...When I place Ceres 70,000 km in a lunar orbit (30.1 km/s ...
I'm probably not understand you, but 30.1 km/s is typical of a solar orbit. Lunar orbits are much slower than that, even when they're near the moon's surface, let alone 70,000 km.
|Glitch|
#29
Jan27-14, 11:28 PM
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Quote Quote by tony873004 View Post
I'm probably not understand you, but 30.1 km/s is typical of a solar orbit. Lunar orbits are much slower than that, even when they're near the moon's surface, let alone 70,000 km.
I worked it out as a four-body problem, using the Sun, Earth, Moon, and Ceres. In order for the Earth to maintain a solar orbit at 1 AU with an eccentricity of 0.0167, it needs to travel at 29.8 km/s. The Moon orbits the Earth slightly faster than that, and in order for Ceres to orbit the Moon it needs to travel slightly faster than the Moon's velocity.
tony873004
#30
Jan27-14, 11:44 PM
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Quote Quote by |Glitch| View Post
...In order for the Earth to maintain a solar orbit at 1 AU with an eccentricity of 0.0167, it needs to travel at 29.8 km/s. The Moon orbits the Earth slightly faster than that...
The Moon orbits the Earth WAY slower than that. It's more like 1 km/s. Unless you're talking about the Moon's velocity with respect to the Sun. In that case, it spends about half its time orbiting the Sun up to about 1 km/s faster than the Earth, and the other half of the time orbiting the Sun up to about 1 km/s slower than than the Earth. I think you know what you're talking about, but we're just having some trouble communicating.
tony873004
#31
Jan28-14, 12:42 AM
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Quote Quote by D H View Post
You can't use RK4 and expect anything remotely resembling reality after even one orbit. It is not a stable integrator.
I know in my previous post I'm essentially disagreeing with you when you say "You can't use RK4 and expect anything remotely resembling reality after even one orbit. It is not a stable integrator. "
But with all due respect, you helped me get my degree in Physics and Astronomy. Google "tony87004 thanks DH". I used to be an electrical contractor. Now I'm a high school physics teacher :)
|Glitch|
#32
Jan28-14, 05:58 AM
P: 54
Quote Quote by tony873004 View Post
The Moon orbits the Earth WAY slower than that. It's more like 1 km/s. Unless you're talking about the Moon's velocity with respect to the Sun. In that case, it spends about half its time orbiting the Sun up to about 1 km/s faster than the Earth, and the other half of the time orbiting the Sun up to about 1 km/s slower than than the Earth. I think you know what you're talking about, but we're just having some trouble communicating.
I think we are talking about the same thing. Only I am referring to the total orbital velocity, rather than the relative orbital velocity. I completely agree that the Moon only orbits the Earth at some thing just under 1 km/s. Furthermore, Ceres would orbit the Moon at less than 1 km/s as well. However, their total velocity includes Earth's velocity around the Sun since it was a four-body problem. If I worked out the simulation as just a three-body problem (Earth, Moon, and Ceres) then the Moon would have an orbital velocity around Earth of 996 m/s and Ceres would have an orbital velocity around the Moon of 860 m/s (at a distance of 70,000 km).
D H
#33
Jan28-14, 06:47 AM
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Quote Quote by |Glitch| View Post
I worked it out as a four-body problem, using the Sun, Earth, Moon, and Ceres. In order for the Earth to maintain a solar orbit at 1 AU with an eccentricity of 0.0167, it needs to travel at 29.8 km/s. The Moon orbits the Earth slightly faster than that, and in order for Ceres to orbit the Moon it needs to travel slightly faster than the Moon's velocity.
There's your problem.

If you are using native floating point (specifically, double precision; if you are using floats you are completely lost) and if want any hope of capturing the dynamics you need to set up a hierarchy of reference frames. If you want to do all of your computations in a common solar system barycenter frame, you *must* use some kind of extended precision arithmetic to avoid the huge truncation error problems in your integrators that would otherwise result with using double precision numbers.

For example, it's best to represent an object orbiting the Earth using an "Earth-centered inertial" (aka ECI) frame. I put that in quotes because (a) it's not an inertial frame and (b) it is a very commonly used term. There's even a wikipedia page on Earth-centered inertial: http://en.wikipedia.org/wiki/Earth-centered_inertial.

This ECI frame is not inertial. It is an accelerating frame. The astronomical / aerospace engineering term for the fictitious forces that results from using this accelerating frame is "third body force". (Physicists would call these apparent forces "tidal forces", but that term is used to denote something else in modeling solar system dynamics.) You need to model these apparent forces or you need to use extended precision arithmetic. If you use the latter, your integration will proceed extremely slowly.


Quote Quote by tony873004 View Post
I know in my previous post I'm essentially disagreeing with you when you say "You can't use RK4 and expect anything remotely resembling reality after even one orbit. It is not a stable integrator. "
These selenocentric distant retrograde orbits (SDROs) are in the class of "exotic orbits", particularly when the orbital distance is greater than the distance to the Earth-Moon L1 point (for example, a 70,000 km orbit about the Moon). These very distant SDROs don't look anything close to elliptical when viewed from the perspective of a Moon-centered inertial frame. They look more like a square with rounded corners. They do look roughly elliptical when viewed from the perspective of an Earth-centered inertial frame, but the Earth is not at one of the foci of the ellipse.

These very distant SDROs are of increasing interest to NASA, other space agencies, and also to the (not anywhere close to ready for prime time) asteroid mining community. So we've studied them. You need a very good integrator to model the behaviors. RK4 just doesn't cut it. Things get wonky after just an orbit or so when one uses RK4 against these exotic orbits.

But with all due respect, you helped me get my degree in Physics and Astronomy. Google "tony87004 thanks DH". I used to be an electrical contractor. Now I'm a high school physics teacher :)
Thanks! I'm blushing.


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