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Forum » SpaceEngine » Space Journeys » Challenge: The Quest for Earth's Twin (Find the closest possible match to our home planet)
Challenge: The Quest for Earth's Twin
Which Terra is as similar as possible to Earth?
1.earth match 90-100%[ 44 ][78.57%]
2.earth match 80-90%[ 10 ][17.86%]
3.earth match 70-80%[ 6 ][10.71%]
4.earth match 60-70%[ 2 ][3.57%]
5.earth match 50-60%[ 3 ][5.36%]
6.earth match 40-50%[ 3 ][5.36%]
7.earth match 30-40%[ 4 ][7.14%]
8.earth match 20-30%[ 2 ][3.57%]
9.earth match 10-20%[ 2 ][3.57%]
10.earth match 0-10%[ 5 ][8.93%]
Poll expiry date: Tuesday, 23.11.2021, 09:34
Answers total: 56
neutronium76Date: Saturday, 26.11.2011, 10:34 | Message # 1
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Once Talisman's Challenge finishes I am starting a new challenge: To find the closest matching terra to our home planet!
All planetary parameters will be considered and taken into account. Planet class, planet size (diameter), Mass, Gravity, Distance to parent star, Class of parent star, Orbital period, Rotaion period, inclination to the ecliptic/axial tilt (i.e seasons) And most imp[ortantly: Atmospheric conditions: Pressure, Temperature and greenhouse effect. Theses 3 will have the highest importance i.e. they will be multiplied with a factor of 2 or 3 (open to suggestions). Also gravity will have a high importance factor as well.

Certain criteria will be set for the quest (not yet determined - I am open to suggestions rolleyes ):
1. Distance from our solar system (shall it be a matching factor as well? I think not as it will complicate things a lot wacko )
2. Shall we search in our galaxy only or shall I include more galaxies?
3. Shall I choose one distant Galaxy like Talisman? And not search in our Galaxy at all?

There will be 10 positions for the final ranking. 1st place will be the closest match i.e the deviation from earth's parameters is closest to zero. 2nd place more deviation, and so on, etc..

The challenge will start when Talisman's challenge finishes.
The winner will have the honor to name that planet with his name/nickname or whatever name he wants B).
What do you people think?





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Edited by neutronium76 - Saturday, 26.11.2011, 11:58
 
SpaceEngineerDate: Saturday, 26.11.2011, 14:50 | Message # 2
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I'd like to add another criteria (but not important) - presence of one selena moon with parameters similar to our Moon: 3400 km diameter, 384000 distance from planet, etc.

*





 
neutronium76Date: Saturday, 26.11.2011, 15:17 | Message # 3
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Quote (SpaceEngineer)
I'd like to add another criteria (but not important) - presence of one selena moon with parameters similar to our Moon: 3400 km diameter, 384000 distance from planet, etc.


Definetly! I've read somewhere, a long time ago, that the presence of the moon, with its specific size, orbital parameters, distance from earth, etc.. contributes in the stabilization of earth's axial tilt and rotation and hence is responsible for the existance of seasons and climate stabilization. Don't know if this is scientifically correct though unsure .





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Edited by neutronium76 - Saturday, 26.11.2011, 15:18
 
SpaceEngineerDate: Saturday, 26.11.2011, 21:47 | Message # 4
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And finally: should this terra have a life?

*





 
neutronium76Date: Saturday, 26.11.2011, 22:32 | Message # 5
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Quote (SpaceEngineer)
And finally: should this terra have a life?


I have though about this - can't really decide.

Not necessarily IMHO... It may sound weird but there is no 100% guarantee that if a terra has almost identical to earth conditions, will definitely have life. Most likely - yes. Definetly - No cool .





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Edited by neutronium76 - Saturday, 26.11.2011, 22:34
 
TalismanDate: Saturday, 26.11.2011, 23:31 | Message # 6
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This is a great challenge idea!

You don't have to wait for mine to finish cool





 
SpaceEngineerDate: Sunday, 27.11.2011, 00:40 | Message # 7
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Quote (neutronium76)
Most likely - yes. Definetly - No

With probability 30% - yes wink

*





 
neutronium76Date: Sunday, 27.11.2011, 08:23 | Message # 8
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Ok. The search criteria will be:

1. The planet must be in our galaxy only. i.e the system must start with 8404 code. Can be in a cluster/nebula as long as the cluster/nebula is in our galaxy biggrin .
2. It doesn't count towards more matching points if the planet's system is closer to our sol i.e. it can be in the oposite side or the edge of milky way.
3. The presence of life is compulsory.

I will try to find a mathematical formula in order to calculate the matching points - help from Space Engineer and/or anybody will be greatly appreciated wink .

As a reference, the earth's matching score will be:
Code

               
                             planet diameter                    planet's mass             pl's orbital period          pl's rotation prd
Earth's MS: ---------------------- +/ x (please advice) ------------------- +/ x ----------------------- +/ x ---------------------- +/x
                             earth's diameter                    earth's mass               erth's orb.period          erth's rottn prd
                
            pl's axial tilt           pl's age               pl's Gravity                 pl's atm ps               pl's temp                pl's grhs eff
            ---------------- +/ x --------------- +/ x ---------------- (x 2) +/x ---------------(x2) +/x ------------(x2) +/x -----------------(x2) +/x
            erh's axial tlt        erth's age             erth's G                   earth's atm ps          earth's Tmp            earth's ghs eff

pl's moons +/x moon's size +/x distance from parent
------------------------------------------------------------------- = earth's ideal matching score 15 if adding / 16 if multiplying)
earth's moon +/x moon's sz +/x distance from earth



angry2 :angry2: can't allign the fractions within the code frame angry >(

PS: I did not include in the formula the planet's parent star astronomical parameters (Class, spectral type, diameter, mass, temperature, luminosity, distance from planet etc..). Shall I include them? It's going to be really really tough wacko





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Edited by neutronium76 - Sunday, 27.11.2011, 09:30
 
SpaceEngineerDate: Tuesday, 29.11.2011, 22:41 | Message # 9
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neutronium76, if you are familiar with LaTeX, you can use this online editor and attach gif image to your message.

I think formula should be like this:



(Surface gravity is unnecessary as long as it defined by mass and diameter of the planet). Here every factor is a function:



This function gives normal distribution with maximum value of 1 then parameter D is equal to Earth's one. So the similarity coefficient k will be between 0 and 1 (or 0% to 100% if you multiply it by 100%).



To take a moon into account, I think we should use second coefficient:



where parameters with <'> is a moon's parameters, related to Earth's moon.

And final similarity coefficient is a multiplication of both. The coefficients can be calculated using any math software or MS Excel.

*

Attachments: 6500507.gif(3Kb) · 6528518.gif(5Kb) · 6163009.gif(4Kb) · 7549477.gif(4Kb)





 
TalismanDate: Wednesday, 30.11.2011, 07:31 | Message # 10
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I found a pretty nice one with a single selena moon. Sadly it was not even close to the milky way so I can't submit it. cool






 
neutronium76Date: Wednesday, 30.11.2011, 20:02 | Message # 11
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Quote (SpaceEngineer)
neutronium76, if you are familiar with LaTeX, you can use this online editor and attach gif image to your message


I'll check it out for next time - thanks Space Engineer!

Quote (SpaceEngineer)
Surface gravity is unnecessary as long as it defined by mass and diameter of the planet


Hmm does this mean it is directly proportionsl to the density of the planet? The only reason I consider gravity important is because too much deviation from 1g is not going to help with the evolution of life - at least as we know it here on earth cool . The closer the planet's "g" to earth "g", the more comfortable the conditions for human colonization. If the above equation you suggest takes gravity into account - even indirectly by measuring of F(D) and F(M) - then it's perfectly fine by me. happy

Quote (SpaceEngineer)



This function gives normal distribution with maximum value of 1 then parameter D is equal to Earth's one. So the similarity coefficient k will be between 0 and 1 (or 0% to 100% if you multiply it by 100%).


So the similarity coefficient K should be as close to one as possible, in order for the planet to be as close to earth as possible? Or is it the oposite? i.e close to zero? I am confused wacko . Sorry my brain is a bit rusty with logs and e's biggrin ...





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SpaceEngineerDate: Wednesday, 30.11.2011, 22:47 | Message # 12
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Quote (neutronium76)
Hmm does this mean it is directly proportionsl to the density of the planet? The only reason I consider gravity important is because too much deviation from 1g is not going to help with the evolution of life - at least as we know it here on earth . The closer the planet's "g" to earth "g", the more comfortable the conditions for human colonization. If the above equation you suggest takes gravity into account - even indirectly by measuring of F(D) and F(M) - then it's perfectly fine by me.




Where G is a gravitational constant. If M (mass) and R (radius) of a planet is equal to Earth's, than surface gravity is equal automatically.

Quote (neutronium76)
So the similarity coefficient K should be as close to one as possible, in order for the planet to be as close to earth as possible? Or is it the oposite? i.e close to zero? I am confused . Sorry my brain is a bit rusty with logs and e's ...

Yes, for pretty Earth's twin the formula gives 1.0 result.

*

Attachments: 3835441.gif(2Kb)





 
neutronium76Date: Thursday, 01.12.2011, 07:16 | Message # 13
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Aha. Crystal clear now - thanks Space Engineer!

Let the hunting begin then ... cool

Please all submissions to include exact details of your findings i.e either a screenshot with HUD on (Ctrl+F11), or in written with coordinates in order to check your finding. Remember that screenshot quality - no matter how beatiful it may look - does not count towards matching points. This is NOT a beauty contest - it is a search for a new home. biggrin





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PC2:Core2Quad X9770@3.2GHz, 2 cores/4 threads 4GB DDR2 RAM@1GHz, GTX-285 GPU 1GB VRAM(DDR3)@1.24GHz, OS:WinVistax64SP2


Edited by neutronium76 - Thursday, 01.12.2011, 07:17
 
SimonDooDate: Thursday, 01.12.2011, 09:00 | Message # 14
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Great challenge indeed !
The one I post in the "best terra challenge" is off contest, as it's not in the Milky Way, but as an example, it has a nice similarity coeff K=38%
Its main dissimilarities : more than twice earth mass, 5 years orbit period, and a short 10 hours day only.
The same with a 1 year orbit would reach K=67%





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SpaceEngineerDate: Thursday, 01.12.2011, 18:38 | Message # 15
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If someone make an Excel document for compute k, please post it here.

*





 
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