Feasibility check for double planet system

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  • Thread starter AotrsCommander
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  • #1

Main Question or Discussion Point

For the background of one of my alien races, I am having them come from a double-planet. (One roughly Earth-sized, one smaller and tide-locked to each other.) My current issue is to to how far apart to set them to get a reasonable balance between stability (because both planets have developed life) and rotational speed.

Data
Large planet (Urrot):
Radius: 6453
Mass: 5.74042E+24 kg (less dense than Earth, approx 0.93g)
Orbital radius (approximate mean, derived from solar flux equivilent to Earth from it's star):18 million km (1.2AU)

Small planet (Haron)
Radius: 4553
Mass:2.21396E+24 kg (approx 0.72g)

Star:
Radius: 765050km (1.1 Sol)
Mass: 2.20729E+30(1.11 Sol)


My start for ten was to set the smaller planet Haron's angular diameter at 35'. This gave an orbital radius of 894000km and a rotation period of about 84 days. As I toyed with this today, I realise that put it at 51% of the primary's Hill Sphere - perhaps a bit too high, so I could probably afford to bring it in just slightly whatever.

Tidal acceleration functions suggest that the tidal forces on the primary would be approximately 2.4 times Luna's on Earth - though obviously, the planets are tide-locked, I assume this will have some effect of the tides (which I assume will still exist, but only on the level of the effect the alignment of the planet's to the sun on the "standing tide" caused by the pull between the planets).

Coming back to this idea after a long break, I had been thinking of a roughly 104-hour rotational period before crunching the astrophysics (and remembering I already had made that first pass on my spread sheet).

As an experiment to see what this value would result in, I changed the angular diameter to 253' (making the smaller planet appear eight-and-a-half times larger than the sun; if nothing else, it'd LOOK spectacular...!), which changed the rotational period to 4.3 days. The orbital radius of the small planet around the primary was now 123500km (7% of the primary's hill sphere radius, perhaps a bit close.)

The tidal acceleration on the planet was now 920 times Luna to Earth, which I would assume would impart not only some significant disortion to the planet along that alignment, but also have some significant effect on the tides (or rather, on the "standing tide" as it were). I am sort of half-assuming the effect of the sun (which would be pretty insignificant to that sort of tidal value) would be fairly small on the actual solar tidal height variance? Or would it rather be that the solar tides would be enormous?

Do either of the values sound reasonable, or would you think I would need something more in-between the two (perhaps to get with the 1/3-1/2 optimal Hill sphere radius). Is the lower radius feasible, or would those sort of tidal forces make planets holding atmosphere and/or habitability impractical?

Some suggestions would be very helpful.
 

Answers and Replies

  • #2
34,053
9,916
Orbital radius (approximate mean, derived from solar flux equivilent to Earth from it's star):18 million km (1.2AU)
I think there is a digit missing.

If your planets are tidally locked, tides from the planets don't exist. The planets won't have an exactly spherical shape but that effect is not relevant. You only get weaker tides from the angle relative to the sun, even weaker than the effect of Sun here on Earth (but a bit weaker as you are at 1.2 AU with 1.11 Msol).

The Hill sphere just gives an upper limit, the lower limit is given by the Roche radius.

Keep in mind that tidal locking leads to days the length of the mutual orbit - you really don't want them to be too long, otherwise you need a really thick atmosphere and/or life on the surface will be really challenging.
 
  • #3
I think there is a digit missing.
Yes, you'e quite right: 180 million km!

If your planets are tidally locked, tides from the planets don't exist. The planets won't have an exactly spherical shape but that effect is not relevant. You only get weaker tides from the angle relative to the sun, even weaker than the effect of Sun here on Earth (but a bit weaker as you are at 1.2 AU with 1.11 Msol).
Right. So the level of the "standing tide" (by which I mean the amount of the bulge of the water at the equator) isn't really of meaningful consequence, the stellar tides are also largely inconequential.

So even with a very high (900 times Luna) tidal acceleration, it won't make an atmosphere infeasible or something?

(I have said the primary has a Pangeas-equvilient super continent on the secondary-facing hemisphere which is just starting to break up, which has a kind of logic to it having a strong pull in that direction...?)


The Hill sphere just gives an upper limit, the lower limit is given by the Roche radius.
Oh crud; a thing I have not done a calculation for on my spread sheet.

*frantic panic calculations on other star systems much time had been spent on*

Phew, okay the roche limit is waaay smaller than any orbital radius I've got anything orbiting at; even just using the solid satelite calculation and assuming roughly double for the fluid (based on wiki's comparions of data calculated from our solar system).

Keep in mind that tidal locking leads to days the length of the mutual orbit - you really don't want them to be too long, otherwise you need a really thick atmosphere and/or life on the surface will be really challenging.
I knew the first part, noted on the second part. Stabilising cloud feeback would do a lot to help the latter problem (and the conceit is these worlds are particularly aqueus), but I would like to get the rotation cycle down to a fairly low level (more towards the 104 hours than the 84 day limit) anyway.

So, am I gathering, then, that we we take the second case, with the 104 hour rotational period (and 120k orbital radius of secondary around primary) is both feasible and for the case of life, perhaps more benificial?
 
  • #4
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(I have said the primary has a Pangeas-equvilient super continent on the secondary-facing hemisphere which is just starting to break up, which has a kind of logic to it having a strong pull in that direction...?)
You don't notice any pull. The surface is an equipotential surface, moving along the surface does not change your potential energy (in the rotating coordinate system). As long as you are outside the Roche limit, everything is fine, and that limit is quite close for bodies of similar density.
So, am I gathering, then, that we we take the second case, with the 104 hour rotational period (and 120k orbital radius of secondary around primary) is both feasible and for the case of life, perhaps more benificial?
Certainly better than longer periods.
 
  • #5
So I can pretty much set any distance I like, then, within the boundaries of the Roche limit and the Hill sphere?

Good to know. Cheers.

(My last experimention along these lines was putting an orbiting moon1 around a tidelocked planet and trying to keep the tidal acceleration reasonably small so that the moon (despite being a late capture) would not have either stopped rotating (yet) or started the planet.)


1Principally just to have SOME basis for working out time would be measured by the native civilisations developed thereon, but alss because it would look cool.)
 
  • #6
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9,916
So I can pretty much set any distance I like, then, within the boundaries of the Roche limit and the Hill sphere?
In terms of orbital dynamics: yes.
 
  • #7
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151
I think there is a digit missing.

If your planets are tidally locked, tides from the planets don't exist. The planets won't have an exactly spherical shape but that effect is not relevant. You only get weaker tides from the angle relative to the sun, even weaker than the effect of Sun here on Earth (but a bit weaker as you are at 1.2 AU with 1.11 Msol).
No - you´ll have tidal strength arbitrary within a wide range, because of the eccentricity of the mutual orbit.
Note different periodicity, though. Tides due to sun have high tide twice a day (at noon and midnight). Tides due to earth have high tide once a day (at perigee in subterranean and antiterranean points).
 
  • #8
34,053
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Ah well, I assumed a circular orbit. An eccentric orbit would lead to tides, yes.
 

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