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zhar2
2009-Feb-23, 11:23 PM
Im trying to find the stable orbit for a hypothetical moon around a hypothetical planet for a system im doing, The system itself was generated in stargen but the generator fails to give the distance of a moon to its parent planet, it did tell me though that the planet moon had a day of 18.3 hours and i figured that either the moon is far enough to have its own independent rotation from its orbit period or its close enough to be tidaly locked and the 18.3 hours would be the same as its orbit period, using the second assumption i managed to find the distance between the two bodies near 42,000 km but i dont know if it can be right as one body has 2.071 Earth masses and a radius of 8071.9 km, while its moon has a mass of 0.418 Earth masses and a radius of 4794.7 km, i worked out that at that distance the moon was beyond the roche limit so it would remain intact, but i dont know if this orbit could be stable or possible.

Any help of pointer will be sincerelly appreciated.

: )

EDG
2009-Feb-24, 12:41 AM
How far is the planet from its star (and what kind of star is it?).

The orbit will be stable if the solar tides aren't big enough to affect it seriously, and if the moon is the only body in orbit around the planet.

As for whether or not it's tidelocked, that woudl depend on the age of the system - but with the planet and moon both being pretty massive I would think that the moon would become tidelocked very rapidly, regardless of how far it was from the planet.

Hornblower
2009-Feb-24, 12:42 AM
Im trying to find the stable orbit for a hypothetical moon around a hypothetical planet for a system im doing, The system itself was generated in stargen but the generator fails to give the distance of a moon to its parent planet, it did tell me though that the planet moon had a day of 18.3 hours and i figured that either the moon is far enough to have its own independent rotation from its orbit period or its close enough to be tidaly locked and the 18.3 hours would be the same as its orbit period, using the second assumption i managed to find the distance between the two bodies near 42,000 km but i dont know if it can be right as one body has 2.071 Earth masses and a radius of 8071.9 km, while its moon has a mass of 0.418 Earth masses and a radius of 4794.7 km, i worked out that at that distance the moon was beyond the roche limit so it would remain intact, but i dont know if this orbit could be stable or possible.

Any help of pointer will be sincerelly appreciated.

: )That looks like a copy of Pluto/Charon, but bigger. I think both bodies would be tidally locked, and the 18.3 hour period looks about right for the separation and the total mass. I see no reason to think that the orbit would be unstable.

zhar2
2009-Feb-24, 08:31 AM
Well the planet is at around 1.2 AU from its parent star which has a mass of 1.02 solar masses.

The star is very much like our sun (thank you so far for the responses)

EDG
2009-Feb-24, 09:27 AM
Well I can at least tell you that the planet won't be tidelocked to the star at that distance from it. Dunno if it would be locked to the moon (though I'm pretty sure the moon would be locked to the planet like our own moon is).

antoniseb
2009-Feb-24, 09:51 AM
One factor in the tidal locking probability is the age of the planet-moon system. The Earth and Moon will eventually be tide-locked (if they both survive the Sun turning into a white dwarf...).

Bob B.
2009-Feb-24, 01:31 PM
To determine period and distance, we don't need to worry about the radii of the bodies, only their masses. The equation needed to solve the problem is:

P^2 = 4 * pi^2 * a^3 / ( G * (M + m) )

where:
P = period, in seconds
a = geocentric semi-major axis, in meters
G = constant of gravitation = 6.67259E-11 N-m^2/kg^2
M = mass of primary body, in kg
m = mass of secondary body, in kg

In your case,
P = 18.3 * 3600 = 65,880 s (assuming moon is tidally locked)
M = 2.071 * 5.9737E+24 = 1.237E+25 kg
m = 0.418 * 5.9737E+24 = 2.50E+24 kg

(mass of Earth = 5.9737E+24 kg)

Therefore,

65880^2 = 4 * pi^2 * a^3 / ( 6.67259E-11 * (1.237E+25 + 2.50E+24) )
a = 47,780,000 m

If we assume your distance of 42,000 km is correct, then the period is,

P = SQRT[ 4 * pi^2 * 42000000^3 / ( 6.67259E-11 * (1.237E+25 + 2.50E+24) ) ]
P = 54,294 s = 15.08 hours

I think the first solution is more likely, i.e. that the moon is tidally locked.

zhar2
2009-Feb-24, 02:31 PM
Well thank you all, now i see its possible.

Though it seems this planet might have some dramatic & destructive tides due to its massive moon but thats just speculation.

This planet is a for a proyect on speculative evolution, now ill try to calculate the tidal force and find a ratio to earths, and from there find an average height or so, not sure if it will work.

chornedsnorkack
2009-Feb-24, 03:35 PM
This planet is a for a proyect on speculative evolution, now ill try to calculate the tidal force and find a ratio to earths, and from there find an average height or so, not sure if it will work.
I am not certain how you could calculate it.

If you have a tidally locked body, you could plausibly have no tides due to the tidally locked body at all. If the twin planet has exactly circular orbit, then the tidal forces are of course big, but the tidal bulges are frozen in place so there is no dramatic or destructive effect. The only tide would be that due to the sun, which is about 1,7 times weaker than the solar tides on Earth (1,2 au).

If the twin has some orbital eccentricity then the tidal bulges will shrink and expand, and also move about, so that there are tidal effects. But the strength of those tides is completely up to what the eccentricity is.

Is the eccentricity of a satellite a free parametre? Or do the disturbances from the Sun force each satellite to have a specific eccentricity of orbit?

zhar2
2009-Feb-24, 03:45 PM
Well ive been looking and found a tide equayion (differential force), which is :

Ftide= -2G * (M*m*x)/r^3

X being the distance between the two orbiting particles around the planet ( i used x as 1).

and after i found the ration between the force of earth and this planet, i found that the tide would be 36000 times higher, if we take the earths normal change of 0.6 meters the tide would be 21.6 km high !!!, higher than the planets highest mountain, it cant be right can it?

zhar2
2009-Feb-24, 03:52 PM
I am not certain how you could calculate it.

If you have a tidally locked body, you could plausibly have no tides due to the tidally locked body at all. If the twin planet has exactly circular orbit, then the tidal forces are of course big, but the tidal bulges are frozen in place so there is no dramatic or destructive effect. The only tide would be that due to the sun, which is about 1,7 times weaker than the solar tides on Earth (1,2 au).

If the twin has some orbital eccentricity then the tidal bulges will shrink and expand, and also move about, so that there are tidal effects. But the strength of those tides is completely up to what the eccentricity is.

Is the eccentricity of a satellite a free parametre? Or do the disturbances from the Sun force each satellite to have a specific eccentricity of orbit?

Well i suppose it will be noticable as the planet rotates faster than its moon orbital perid (though i doubt it, it might be a mistake from a limitation of the system generator i used), but it seems this planet is turning out to have a hellish weather and conditions so far.

Hornblower
2009-Feb-24, 03:53 PM
Well ive been looking and found a tide equayion (differential force), which is :

Ftide= -2G * (M*m*x)/r^3

X being the distance between the two orbiting particles around the planet ( i used x as 1).

and after i found the ration between the force of earth and this planet, i found that the tide would be 36000 times higher, if we take the earths normal change of 0.6 meters the tide would be 21.6 km high !!!, higher than the planets highest mountain, it cant be right can it?I think you are right. Mind-blowing, but a good estimate. If this planet is not yet tidally locked, the tides will be catastrophic. The worst tsunami on our planet would be a little ripple by comparison.

Murphy
2009-Feb-24, 06:35 PM
If you want to know how long it takes for bodies to become Tide Locked, you have to use a rather complicated formula, like the ones we used in this thread http://www.bautforum.com/space-astronomy-questions-answers/84111-tidal-locking-need-understand.html.

EDG told me about it, maybe he can help you with this.

zhar2
2009-Feb-25, 12:05 AM
Well thank you all, Ive decide the moon was too close and created unreasonable tides for the planets purpose, so im "tweaking" the results from the generator to give the moon a period of 62 hours at a distance of 107,781 km, where tides are just 11.5 times higher. :)

timb
2009-Feb-25, 12:59 AM
Well thank you all, Ive decide the moon was too close and created unreasonable tides for the planets purpose, so im "tweaking" the results from the generator to give the moon a period of 62 hours at a distance of 107,781 km, where tides are just 11.5 times higher. :)

If the planets (what you are describing is a binary planet rather than a planet-moon system) are tidally locked then there are no tides. There is a fixed tidal bulge.

EDG
2009-Feb-25, 02:22 AM
If the planets (what you are describing is a binary planet rather than a planet-moon system) are tidally locked then there are no tides. There is a fixed tidal bulge.

If the planet is close to the star, then the tides from the star may be strong enough to affect the orbits and to sap momentum from the planet-moon system. The stellar influence might even force the eccentricity of the satellite's orbit a bit.

timb
2009-Feb-25, 02:44 AM
If the planet is close to the star, then the tides from the star may be strong enough to affect the orbits and to sap momentum from the planet-moon system. The stellar influence might even force the eccentricity of the satellite's orbit a bit.

True there would be some libration, but that would be a small effect. The OP stated that the planet(s) was 1.2 AU from the sun-like star. Eventually the two planets will collide, but that would take long enough to happen for a even a long sci fi story to be set.

neilzero
2009-Feb-25, 04:26 AM
My guess is your numbers are in the believable range. With double Earth mass, the core will be compressed to a bit higher density, so it needs to be smaller or contain lighter elements than Earth's core. Both are possible, and demonstrated in smaller planets, and large moons in our solar system. Tides will be much larger on both the planet and the moon, but likely life can adapt. 0.418 earth mass for the moon is higher than any moon in our solar system and is higher than the mass of Mercury, or Mars, but about the same as the mass of Venus. Neil

Murphy
2009-Feb-25, 05:27 AM
neilzero:
0.418 earth mass for the moon is higher than any moon in our solar system and is higher than the mass of Mercury, or Mars, but about the same as the mass of Venus. Neil

Huh? Venus has a mass of 0.815 of Earth.

zhar2
2009-Feb-25, 09:30 AM
True there would be some libration, but that would be a small effect. The OP stated that the planet(s) was 1.2 AU from the sun-like star. Eventually the two planets will collide, but that would take long enough to happen for a even a long sci fi story to be set.

I wonder how long it would take for them to collide (if they even collide).

qraal
2009-Feb-25, 09:45 AM
I wonder how long it would take for them to collide (if they even collide).

Conservation of momentum means the two will slowly draw apart due to the tidal forces from the star breaking them out of tidal lock to each other.

One problem with close-in high-mass objects is that their whole masses are distorted and energy is very rapidly dissipated in the semi-fluid bodies of their molten interiors. Such a close orbit just won't last. For example the Earth-Moon system rapidly pulled apart after the initial collision so quickly that by 3.2 billion years ago the Moon's period was about 22 days and the Earth's day was 16 hours.

Bob B.
2009-Feb-25, 01:16 PM
Conservation of momentum means the two will slowly draw apart due to the tidal forces from the star breaking them out of tidal lock to each other.

Furthermore, if the bodies are not yet tidally locked, conservation of angular momentum will cause the bodies to move apart as their rotations slow down. In our own system, the Moon's tidal forces are causing the Earth's rotation to slow down; consequently, the Moon moves away to conserve the momentum of the system.

I think, zhar2, most everyone here agrees that your satellite is, most likely, tidally locked to the primary (unless your system is very young). It is less certain whether the primary is tidally locked to the satellite. If not, it is very possible your primary's rotation is slowing and the satellite is moving away.

zhar2
2009-Feb-25, 01:51 PM
Yes i agree aswell, and will follow your second point about the rotation slowing down and the satellite moving away

EDG
2009-Feb-25, 05:56 PM
Conservation of momentum means the two will slowly draw apart due to the tidal forces from the star breaking them out of tidal lock to each other.

I don't think so actually. If the star wasn't there, the tidal forces between the planet and moon will (a) slow the moon's rotation so that it is the same as its orbital period around the planet, (b) circularise the moon's orbit, (c) slow the planet's rotation down, and (d) push the moon away from the planet.

(a) will happen first (and quickly). (b) will happen next, then (c) and (d) will happen last. Eventually the planet's rotation period will be the same as the moon's orbital period, and they'll have the same face toward eachother all the time, and there'd be no further evolution.

If you throw the star into the mix, however, that adds a factor that is trying to slow the planet's rotation period to equal its orbital period around the star. This is the part I'm fuzzy on - shouldn't this act to push the synchronous orbit outward from the planet, so that the moon now finds itself within it and evolves appropriately? Moons that are within the synchronous orbit should be (a) moving inwards and (b) speeding up the planet's rotation until they collide with the planet?

chornedsnorkack
2009-Feb-25, 08:06 PM
Whether a double tidal lock is stable depends on the mass ratio of the bodies. If the satellite is small, double tidal lock is unstable. If the satellite is modest sized, double tidal lock is stable for long orbital periods, unstable for short orbital periods.

Look at it like this: if a satellite is small like Phobos and Deimos, and happens to be near the areostationary orbit, like Deimos whose orbit takes slightly longer than Mars´ rotation, then Deimos is propelled by tides from Mars. As Deimos moves outwards, its period increases significantly, while Mars slows down to an insignificant degree. Thus Mars will not lock to Deimos, and Deimos moves out till Mars drives it over Hill limit. Phobos orbits faster than areostationary - as Mars slows it down, Phobos sinks and orbits ever faster till it crosses Roche limit and forms rings of Mars.

If the satellite is massive enough, however, it is the effect on the primary rotational period that dominates. So, if the system is disturbed and tidal lock were broken, the tidal lock would be restored at a slightly changed period.

This kind of close-in, massive satellite should lead to a very rapid tidal evolution leading to mutual tidal lock shortly after the system formed. There would be a slight solar tidal influence afterwards; the result would be slow approach of the system and decrease of the period. But I expect that the tidal lock would be maintained, just as the Moon´s rotation gradually slows down even as the Moon is receding from Earth.

zhar2
2009-Feb-25, 08:28 PM
Well i searched around for a gravity simulator and found one, the modelled the system and found vthat the system is indeed stable for a prolonged time (for quite a few billion years) though the moon tended to move away over time.

The simulator did couldnt illustrate tidal locking as planets and moons are just filled in circles of color.

but im convinced.

Bob B.
2009-Feb-25, 09:03 PM
Well i searched around for a gravity simulator and found one

Is this an online application? If so, what's the link? I'd like to take a look at it.

zhar2
2009-Feb-25, 11:42 PM
Yep:

http://www.orbitsimulator.com/gravity/articles/what.html

(if you wana do this planet, i used the 62 hours perid for the moon)

EDG
2009-Feb-26, 04:47 PM
Well i searched around for a gravity simulator and found one, the modelled the system and found vthat the system is indeed stable for a prolonged time (for quite a few billion years) though the moon tended to move away over time.

I'm not sure you did it right then. If you're modelling it with GravSim then running it "for a few billion years" means that your timestep must have been huge and you must have been running your model for weeks or months. I think the maximum timestep should be around 1-5% of the shortest orbital period in the system for you to get reasonably results - if it's bigger then you're not calculating the objects continuously in their orbits (e.g. if the timestep is 1.5 days and the orbital period is 6 days, then you're only plotting the object four times per orbit, which means its influence isn't really being calculated properly).

There's a forum on the Gravity Simulator board where you can ask questions too.

zhar2
2009-Feb-26, 05:24 PM
I used the same time step as for their solar system (till some planets starting flying off, timmed it after i had done the same with mine and i assumed that although it did not take as long for my moon to fluctuate and fly off i assumed that it would have been a rather long time in relation to the effect in the solar system model.

EDG
2009-Feb-26, 07:05 PM
If you only had the star and the planet and moon in the system, then it would take a lot quicker to do the simulation - I had forgotten about that. I'm used to doing sims with a hundred asteroids and lots of planets :), which obviously slows things down a bit! On my old Athlon 3800+ dualcore it took about 48 hours to run 200,000 years of simulated time for that that asteroids/planet sim. I've yet to try it on my shiny new Q9550 quadcore, though I don't think it'd be THAT much faster because gravsim isn't multi-core capable - it just runs on a single core.

zhar2
2009-Feb-26, 08:23 PM
Well in my case its alot quicker as im only using 17 planets, 1 moon. :D

Murphy
2009-Feb-27, 02:28 AM
Hmm, I just downloaded Gravity Simulator and looked up their tutorial, but it's still rather difficult to understand how to get it to work.

For one thing, any system I try to create flies apart fairly soon if you speed it up. Even their own pre-set simulations of the Solar system go crazy once you play them. The Earth was flung out of the Solar system in just 200 years! What's the point of this program? Or am I missing something?

EDG
2009-Feb-27, 03:18 AM
You're probably missing something :)

First pause the sim (Time menu, Pause)

Then go to the View menu, go to Dash Board Elements and set them to "Show All" (vertically). Then you'll be able to see the timestep. Hit the "-" button till it's down to somewhere around 1024 - that does one calculation for every increment of 1024 seconds of simulated time.

Then hit the "||" button on the timestep dashboard element to restart it, and it should go run the sim at a more normal rate. And if you want to speed it up, click the "+" button on the timestep element.

As for "what's the point", it's to simulate gravitational interactions in any astronomical system. You can even make your own too (I use it to test my worldbuilding).

If you go to the Discussion Board there you'll see a lot of examples of how it's used.

Murphy
2009-Feb-27, 05:23 AM
Oh, well maybe that's what I was doing wrong, I just kept hitting the "+" button to make it go faster, I didn't realise that influences the calculations, maybe that's why my planet's were whizzing out of the system at increasable speed. :lol:


If you go to the Discussion Board there you'll see a lot of examples of how it's used.

Yeah, I'll probably have to do that if I want to work it right. Thanks.

EDG
2009-Feb-27, 05:36 AM
Oh, well maybe that's what I was doing wrong, I just kept hitting the "+" button to make it go faster, I didn't realise that influences the calculations, maybe that's why my planet's were whizzing out of the system at increasable speed. :lol:

Oh lawks, yes :). If you bump up the timestep too much then you will get things flying all over the place when they shouldn't be, because the sampling interval is too high and it won't be remotely accurate.

Usually setting the timestep to a maximum of about 5% of the slowest period in the system works. Going faster than that increases the risk of greater inaccuracies.

(I'm on the gravsim boards as "Mal").

Murphy
2009-Feb-27, 05:42 AM
Right, that must be it then, now I have an explanation for why Mimas was Shooting pass Mercury the minute I turned it on. :doh: :lol:

They should probably try explaining that in the tutorial.

zhar2
2009-Feb-27, 11:14 AM
Well in that case ill repeart the simulation and let it run a couple of days to see results more accuratlt, shame the program dosent give you a report of orbit parameter changes every few million yeras or so.

EDG
2009-Feb-27, 04:39 PM
Well in that case ill repeart the simulation and let it run a couple of days to see results more accuratlt, shame the program dosent give you a report of orbit parameter changes every few million yeras or so.

It can. You can get it to output various aspects of the orbital data (File > Output Data) at whatever interval you like, and then you can plot them in excel or in the gravsuim viewer that one of the GS forum members made.