Science and Astronomy Questions
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steeljaw354 | Date: Sunday, 09.10.2016, 00:32 | Message # 826 |
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| Watsisname, Why would it be smaller? Can you make this in Space Engine?
Edited by steeljaw354 - Sunday, 09.10.2016, 00:45 |
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Watsisname | Date: Sunday, 09.10.2016, 02:15 | Message # 827 |
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| I was going by the properties of the moon you showed, which is smaller and denser. It doesn't have to be that way. If you are looking at a scenario where Theia, instead of impacting the Earth to form the Moon, settles into orbit to become a moon itself, then that moon would be much larger and more massive -- something like 10% of Earth's mass and half the diameter of Earth. It would raise larger tides, be bigger on the sky and provide more light. It would still become tidally locked very quickly. But the exact numbers depend on exactly what Theia was, and this varies across different models.
This can be made in Space Engine. See the tutorial for how to make a planet.
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steeljaw354 | Date: Sunday, 09.10.2016, 02:42 | Message # 828 |
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| Watsisname, I just made it. The Barycenter and orbital period is probably dead wrong. What would it be like to have Theia as a moon?
Edited by steeljaw354 - Sunday, 09.10.2016, 02:42 |
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Watsisname | Date: Sunday, 09.10.2016, 07:23 | Message # 829 |
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| The tides would be big: five or six times larger than the tidal change now. The moon would be about 75% larger on the sky and provide three times as much light, assuming the same albedo. The barycenter would be well outside the Earth: about 21,400 miles from the center and 17,500 miles above the surface.
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steeljaw354 | Date: Sunday, 09.10.2016, 14:43 | Message # 830 |
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| Watsisname, Would Earth and Theia tidally lock to eachother like Pluto and Charon?
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Watsisname | Date: Sunday, 09.10.2016, 22:05 | Message # 831 |
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| Yes, and fairly quickly (on geologic timescales). If we set them at the current Earth-moon distance and let them both start with a rotation period of 12 hours, then Theia would become tidally locked in a few million years, and Earth would become locked in a few hundred million years.
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steeljaw354 | Date: Sunday, 09.10.2016, 22:28 | Message # 832 |
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| Watsisname, Interesting! Could life possibly form on both worlds? Would Theia stay in it's current orbit or will it move out or in?
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Watsisname | Date: Monday, 10.10.2016, 04:48 | Message # 833 |
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| I don't know offhand any reason why life could not form on either world. 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.
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quarior14 | Date: Monday, 10.10.2016, 16:51 | Message # 834 |
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| How to convert solar day to sidereal day (rotation period) ? It is : Tsolar*(Ysolar/Ysidereal) (with Hsolar : Time solar (3600 s = 1 h) in hours, Ysolar : 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 !
Quarior
Edited by quarior14 - Monday, 10.10.2016, 16:52 |
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Watsisname | Date: Monday, 10.10.2016, 18:24 | Message # 835 |
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| 1/Psidereal = 1/Psolar + 1/Porbital
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.
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quarior14 | Date: Monday, 10.10.2016, 18:59 | Message # 836 |
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| Quote Watsisname ( ) 1/Psidereal = 1/Psolar + 1/Porbital Ah, already found a solution and a problem : For me, I thought Psidereal = Porbital, a big mistake I see, and I do'nt know Psolar and Porbital, I just Psidereal.
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 ?
Quarior
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Watsisname | Date: Monday, 10.10.2016, 20:02 | Message # 837 |
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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.
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spacer | Date: Monday, 10.10.2016, 20:27 | Message # 838 |
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| try to think about that and maybe solve it Watsisname,
"we began as wanderers, and we are wanderers still" -carl sagan
-space engine photographer
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quarior14 | Date: Monday, 10.10.2016, 20:47 | Message # 839 |
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| ]Watsisname, thanks, but I don't have semimajor axis (it is AU ?), I juste sidereal periods, so I have Psidereal = 1.052165173 years = 313.1561 solar days = 9216.96691325 hours => 1/Psidereal = 1/Psolar + 1/Porbital <=> 1/Psolar = 1/9216.96691325 - 1/((4π2a3)/6.6738480*10-11*0.8*1.9891*1030))
Edit : I found this for Porbital : (Source : Wikipedia)
Edit 2 : I'm stupid... 1/Porbital = 1/Psidereal - 1/Psolar <=> 1/Porbital = 1/1.052165173 - 1/1.0500000001138413242009132420091 = -0,0019598343188704316143210359517 years Porbital = 1/(-0,0019598343188704316143210359517) = -510,24721343606184076731659026361 years I mistake or it is retrograde ?
Quarior
Edited by quarior14 - Monday, 10.10.2016, 23:11 |
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Watsisname | Date: Monday, 10.10.2016, 20:47 | Message # 840 |
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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?"
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