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Date: Monday, 10.10.2016, 04:48 | Message # 826

Galaxy Architect

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I don't know offhand any reason why life could not form on either world. smile

Although, like Mars, Theia might be too small to remain habitable for as long as Earth. It might not retain a thick atmosphere for very long, especially if large impacts blow it off.

How their orbits evolve depends on the initial spin/orbit rate and direction. Suppose that, like in the present Earth-moon system, Theia orbits in the same direction that Earth spins, but the orbital period is longer. This means it lags behind the tidal bulge that it raises on the Earth (since Earth's rotation drags that bulge ahead), so the bulge pulls the moon forward and accelerates it, giving it more orbital energy and expanding its orbit outward. This expansion continues until the bodies become synchronously locked.

But what if Theia were to orbit retrograde -- opposite Earth's spin? Then the bulge raised on the Earth will always be pulling Theia backward, sapping its orbital energy. It will migrate inward.

Date: Monday, 10.10.2016, 16:51 | Message # 827

World Builder

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How to convert solar day to sidereal day (rotation period) ?
It is : T_solar*(Y_solar/Y_sidereal) (with H_solar : Time solar (3600 s = 1 h) in hours, Y_{solar : Year solar in days, Y[sub]sidereal} : Years sideral in days, one day = time solar) ?

Because I tried that to update the planet Thalassa but Space Engine, it's not the same solar day.
Here my code and calculs on comment :

Code

//Thalassa
Planet "Thalassa 1"
{
ParentBody "Thalassa system"
Class "Terra"
Mass  1.285428027
Radius 7015.9507 //hypothetic
RotationPeriod 29.56955461 //h = (29.4325*3600)*(313.966913/312.5116793) = 106450.3966 s (Solar day : 29.4325 h)
Gravity 1.06 //g = 10.3986 m/s² = 9.81*(1+(6/100))
Obliquity 5.71429 //hypothetic

Life
{
Class "Organic"
Type "Multicellular"
Biome "Marine/Terrestrial"
Panspermia    false
}

Surface
{
SurfStyle 0.520806
OceanStyle 0.149282
Randomize (1.079, -0.889, 0.190)
colorDistMagn 0.0415044
colorDistFreq 697.054
detailScale 18045.4
colorConversion true
drivenDarkening -1
seaLevel 0.803431
snowLevel 1
tropicLatitude 0.10789
icecapLatitude 1
icecapHeight 1
climatePole 0.9375
climateTropic 0.353893
climateEquator 0.6875
heightTempGrad 0.583607
tropicWidth 0.0817879
mainFreq 0.584501
venusFreq 1
venusMagn 0
mareFreq 0.372759
mareDensity 0.0714286
terraceProb 0.319919
erosion 0
montesMagn 0.205713
montesFreq 259.412
montesSpiky 0.949549
montesFraction 0.485357
dunesMagn 0.0601864
dunesFreq 42.9586
dunesFraction 0.0593326
hillsMagn 0.0965508
hillsFreq 583.213
hillsFraction 0.891151
hills2Fraction 0.581915
riversMagn 66.8105
riversFreq 3.97417
riversSin 7.73035
riversOctaves 0
canyonsMagn 0.0339486
canyonsFreq 100
canyonFraction 0.520806
cracksMagn 0.130161
cracksFreq 0.824795
cracksOctaves 1
craterMagn 0.032052
craterFreq 59.6143
craterDensity 0
craterOctaves 0
craterRayedFactor 0
volcanoMagn 0.618842
volcanoFreq 0.781987
volcanoDensity 0.103175
volcanoOctaves 1
volcanoActivity 1.33333
volcanoFlows 0.00793651
volcanoRadius 0.544894
volcanoTemp 1761.9
lavaCoverTidal 0
lavaCoverSun 0
lavaCoverYoung 0
stripeZones 1.74933
stripeFluct 0
stripeTwist 0
cycloneMagn 2.64312
cycloneFreq 0.972265
cycloneDensity 0.322386
cycloneOctaves 4
colorSea (0.040, 0.100, 0.200, 1.000)
colorShelf (0.150, 0.480, 0.460, 1.000)
colorBeach (0.400, 0.330, 0.280, 0.000)
colorDesert (0.260, 0.240, 0.220, 0.000)
colorLowland (0.500, 0.420, 0.350, 0.000)
colorUpland (0.240, 0.210, 0.210, 0.000)
colorRock (0.220, 0.210, 0.210, 0.000)
colorSnow (1.000, 1.000, 1.000, 1.308)
colorLowPlants (0.100, 0.160, 0.070, 0.000)
colorUpPlants (0.090, 0.110, 0.040, 0.000)
BumpHeight 76.9841
BumpOffset 61.8515

DiffMapAlpha "Water"
SpecBrightWater 0.65
SpecBrightIce 0.85
SpecularPower 55
Hapke 0
SpotBright 4
SpotWidth 0.05
DayAmbient 0.07
}

Ocean
{
Height 61.8515

}

Atmosphere
{
Greenhouse 4.51    // degrees K
Pressure 1.21    // atm
Density 1.21  // kg/m^3
Model "Earth"

Composition // values in percent
{
N2 77.7729
O2 20.8625
Ar 0.9303
H2O 0.4000
CO2 0.0398
}
}

NoRings true

Orbit
{
RefPlane "Equator"
Period 1.052165173 //years (Sidereal year : (1.052165173 years = ((29.4325*313.1561)/24)/365 => Solar year : (((365*(1+(5/100)))*24)/29.4325 = 312.5116793 days)
Inclination 0.74
}
}

Unless I am mistaken to convert the sidereal year to solar year.
Information for Thalassa :

Solar day : 29.432 h
Sidereal year : 313.1561 solar day
Gravity : 1.06 g = 10.3986 m/s²

Thanks in advance !

Edited by quarior14 - Monday, 10.10.2016, 16:52

Date: Monday, 10.10.2016, 18:24 | Message # 828

Galaxy Architect

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1/P_sidereal = 1/P_solar + 1/P_orbital

Change to a minus sign if it is a retrograde rotating planet.

To derive, notice that in one solar day the planet moves along its orbit by an angle equal to its orbital angular frequency times the length of solar day. During that same period of time, it rotates about its axis by one full rotation relative to the stars, plus an additional bit of angle which is equal to the angle it moved along its orbit (so that the same point on the planet faces the sun again). That full angle of rotation is equal to its angular rotation frequency (relative to the stars) times the length of solar day. Relate these and convert angular frequencies to periods to find the sidereal period.

Date: Monday, 10.10.2016, 18:59 | Message # 829

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Quote Watsisname (

)

1/P_sidereal = 1/P_solar + 1/P_orbital

Ah, already found a solution and a problem :
For me, I thought P_sidereal = P_orbital, a big mistake I see, and I do'nt know P_solar and P_orbital, I just P_sidereal.

Quote Watsisname (

)

Change to a minus sign if it is a retrograde rotating planet.

Ah, to take information but in principle, it rotates in the direction prograde since it has no information in the book about it.

Quote Watsisname (

)

To derive, notice that in one solar day the planet moves along its orbit by an angle equal to its orbital angular frequency times the length of solar day. During that same period of time, it rotates about its axis by one full rotation relative to the stars, plus an additional bit of angle which is equal to the angle it moved along its orbit (so that the same point on the planet faces the sun again). That full angle of rotation is equal to its angular rotation frequency (relative to the stars) times the length of solar day. Relate these and convert angular frequencies to periods to find the sidereal period.

How do you that I find this ? We can not find other means without know the angle ?

Date: Monday, 10.10.2016, 20:02 | Message # 830

Galaxy Architect

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Quote quarior14 (

)

How do you that I find this ? We can not find other means without know the angle ?

Oh, you don't need to. I was describing how the formula itself was found -- that process is all symbolic, no numbers. To use the formula, you only need to know two of the three periods.

You already know the sideral period, and you can find the orbital period by Kepler's Third Law: the square of orbital period equals 4pi²/GM times the cube of the semi-major axis:

where "T" is orbital period and "a" is semimajor axis. "M" is the mass of the star it orbits and "G" is the gravitational constant.

Then once you have the orbital and sidereal periods you can find the solar period with the formula.

Quote quarior14 (

)

Ah, to take information but in principle, it rotates in the direction prograde since it has no information in the book about it.

Yeah, most planets rotate prograde. smile

spacer

Date: Monday, 10.10.2016, 20:27 | Message # 831

Star Engineer

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try to think about that and maybe solve it Watsisname, biggrin

"we began as wanderers, and we are wanderers still"
-carl sagan

-space engine photographer

Date: Monday, 10.10.2016, 20:47 | Message # 832

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]Watsisname, thanks, but I don't have semimajor axis (it is AU ?), I juste sidereal periods, so I have P_sidereal = 1.052165173 years = 313.1561 solar days = 9216.96691325 hours
=> 1/P_sidereal = 1/P_solar + 1/P_orbital <=> 1/P_solar = 1/9216.96691325 - 1/((4π²a³)/6.6738480*10^-11*0.8*1.9891*10³⁰))

Edit : I found this for P_orbital :

(Source : Wikipedia)

Edit 2 : I'm stupid...
1/P_orbital = 1/P_sidereal - 1/P_solar <=> 1/P_orbital = 1/1.052165173 - 1/1.0500000001138413242009132420091 = -0,0019598343188704316143210359517 years
P_orbital = 1/(-0,0019598343188704316143210359517) = -510,24721343606184076731659026361 years
I mistake or it is retrograde ?

Edited by quarior14 - Monday, 10.10.2016, 23:11

Date: Monday, 10.10.2016, 20:47 | Message # 833

Galaxy Architect

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spacer:

LOL. Basically this is like saying "Let's consider a scenario which requires disregarding our understanding of physics. What does our understanding of physics imply about what will happen when we do this?" smile

Date: Monday, 10.10.2016, 21:01 | Message # 834

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quarior14, this seems problematic. If you only know the sidereal rotation rate, and neither the length of solar day nor the orbital period nor the orbital distance, then you have an under-determined system. Basically, how long the solar day will be will depend on how far the planet is from the star, so if you don't know that, then you're stuck. (And orbital distance should be converted to meters when used in Kepler's Third Law).

Do you have any other conditions about the planet to help determine it? Do you know how bright the star is and how warm the planet is, for instance?

Date: Monday, 10.10.2016, 21:09 | Message # 835

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Watsisname, I updated my post, I have the solar period too, but I get a negative orbital period, the planet is retrograde it ?

Quote quarior14 (

)

Watsisname, thanks, but I don't have semimajor axis (it is AU ?), I juste sidereal periods, so I have P_sidereal = 1.052165173 years = 313.1561 solar days = 9216.96691325 hours
=> 1/P_sidereal = 1/P_solar + 1/P_orbital <=> 1/P_solar = 1/9216.96691325 - 1/((4π²a³)/6.6738480*10^-11*0.8*1.9891*10³⁰))

Edit : I found this for P_orbital :

Quote Watsisname (

)

Do you have any other conditions about the planet to help determine it? Do you know how bright the star is and how warm the planet is, for instance?

No, I have'nt other informations, juste gravity, solar days, sidereal years and solar years in the book Songs of a Distant Earth.

Edited by quarior14 - Monday, 10.10.2016, 23:10

Date: Monday, 10.10.2016, 21:48 | Message # 836

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Hmmm, no, orbital period should not be coming out negative. Let's go through a few examples:

Computing Earth's sidereal day:
The Earth's solar day is 24 hours, and its orbital period is 365 days (not exactly, but close enough). Then to compute Earth's sidereal day,
1/24+1/(24*365))^-1 = 23.9344 hours. Which makes sense; the solar day is a bit longer than the sidereal day.

Computing Earth's orbital period:
1/23.9344-1/24)^-1 = 8760 hours = 365 days.

Solar period of a retrograde Earth:
Use orbital period of 365 days and sidereal period of 23.9344 hours:
1/23.9344+1/(365*24))^-1 = 23.8692 hours. Makes sense; solar day should be shorter than sidereal in this case.

Orbital period of a retrograde Earth:
We started with 365 days. Let's see if we get it back.
1/23.8692-1/23.9344)^-1 = 8760 hours = 365 days.

EDIT:
Ugh, the forum is messing up the parentheses, but I think you know what I mean.

Quote quarior14 (

)

No, I have'nt other informations, juste gravity, solar days, sidereal years and solar years in the book Songs of a Distant Earth.

Oh, then what's the problem? You have the orbital period measured in two different ways. smile

The ratio of the two will tell you the ratio of sidereal day to solar day, as well.

steeljaw354

Date: Monday, 10.10.2016, 22:58 | Message # 837

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Watsisname, What if Mercury was our moon instead of Theia or the moon we have? Would they lock?

Date: Monday, 10.10.2016, 23:51 | Message # 838

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Watsisname, so unit it is hours.
1/P_orbital = 1/P_sidereal - 1/P_solar

Quote Watsisname (

)

Earth's sidereal day, 1/24+1/(24*365)^-1 = 23.9344 hours. Which makes sense; the solar day is a bit longer than the sidereal day.

If I apply for Thalassa, Thalassa's sidereal day : (1/29.4325)+(1/(29.4325*x))^-1 = (1/29.4325)*(1+(1/x))
x it is the orbital period but I don't know. I can put the solar year or sidereal year in place ?

Quote Watsisname (

)

Oh, then what's the problem? You have the orbital period measured in two different ways. smile

The ratio of the two will tell you the ratio of sidereal day to solar day, as well.

Not really understand, if I recup :
D_sidereal = (1/D_solar) + (1/Y_solar)^-1
P_orbital = Y_orbital = (1/D_sidereal) - (1/D_solar)^-1
It is correct ?

I just redo your calculation for the Earth, I do not find at all the same thing
D_sidereal = (1/24) + (1/(24*365))^-1 = (1/24) + (1/(24*365)^-1) = 8760.041667 =/= 23.9344 hours
And if I did D_sidereal = (1/24) + (1/(24*(365)^-1)) = 15.25 =/= 23.9344 hours
By cons it's okay if I did D_sidereal = ((1/24) + (1/(24*365)))^-1 = 23.9344 hours
I deduce that :
D_sidereal = ((1/D_solar) + (1/Y_solar))^-1
P_orbital = Y_orbital = ((1/D_sidereal) - (1/D_solar))^-1

For Thalassa :
D_sidereal = ((1/29.4325) + (1/(29.4325*312.5116793))^-1 = 29.33861992 hours
P_orbital = Y_orbital = ((1/29.33861992) - (1/29.432))^-1 = 9198.000001 hours = 312.5116793 solar days = 383.25 Earth's solar days = 1.05 Earth's solar years
It is correct ?

And the sidereal day it is the rotation period ?

Edited by quarior14 - Monday, 10.10.2016, 23:56

Date: Tuesday, 11.10.2016, 00:05 | Message # 839

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steeljaw:

Short answer: definitely. smile

It is pretty hard to not have a planet and moon become tidally locked over millions to billion year timescales. They have the be very widely separated. (Recall that 6th power dependence on the orbital distance). This is why most close to intermediate-range moons in the solar system are locked.

Slightly longer answer:
An easy way to compare numbers for tidal locking time -- if you already have one computed for another body, as we have done -- is as follows: The locking time is proportional to the density of the satellite, its initial spin rate, and the sixth power of its orbital distance. It is inversely proportional to the square of the mass of the primary body. The 6th power of orbital distance tends to be the most important factor. y=x⁶ grows very fast. smile

But in this case if we just replace Theia with Mercury (about half the mass, a bit more dense, and a bit smaller), then the result doesn't change very much -- not by more than an order of magnitude. So it still gets locked in about the same timescale -- millions of years.

quarior14, sorry, I have to run -- let me get back to you later with a complete answer. For now, the unit you use is whatever you want it to be, as long as you are consistent. Hours is convenient sometimes.

Date: Tuesday, 11.10.2016, 08:47 | Message # 840

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Quote Watsisname (

)

quarior14, sorry, I have to run -- let me get back to you later with a complete answer. For now, the unit you use is whatever you want it to be, as long as you are consistent. Hours is convenient sometimes.

OK, and it's true that hours are practices. Do not complicated with the hours, it's the same here : 1 hour = 3600 seconds

Looks like it's okay, in Space Engine, I get for solar day 29h25m56.77s.
I convert my solar day : 29.4325 h = 29 h 25.95 m = 29 h 25 m 57 s, it just a shift of 57-56.77 = 0.23 s, perhaps due to rounding for the period. Thank you for your formula smile

Attachments: 4340892.jpg (125.6 Kb)