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Thread: Designing a fictional extrasolar planet

  1. #1

    Designing a fictional extrasolar planet

    I'm trying to to invent a fictional extrasolar planet which is the setting for a novel I'm writing. I would like it to to be scientifically believable, so I'd be very grateful if someone could help me work out the parameters the planet could have, or link relevant sites/threads that might help.

    =======UPDATE======

    Ok, I think I have all the major parameters sorted. I'm going to go away and have a think about moons, landmasses and tides and things - I may come back for more help once I've done that, thanks to everyone who contributed

    Latest planetary model
    I'll keep editing this based on feedback

    Semi major axis: 0.4 AU
    Orbital period*: ~100 earth days
    Rotation period*: ~120 hours (range 3-5 days)
    Sun: 0.88 solar masses (KIV/K2V)
    Average surface temp of planet*: 15°C
    Gravity*: same as earth

    *These are the features I wish the planet to have. All other variables can be amended.

    ===================


    Quote Originally Posted by Veiligo - rest of original post

    I have prayed for several hours to Google, and this is the vision I have been granted so far...

    Design parameters:
    • Semi-major axis <1AU (e.g. 0.5-0.7AU?)
    • Shortest possible orbital period without being tidally locked (e.g. ~100 earth-days?)
    • Cooler sun (e.g. K-type dwarf)
    • Longer rotation period (e.g. 3-6 earth days)
    • Same surface gravity and atmospheric density to earth
    • Similar average surface temperature (liquid water present on surface)
    • Same or higher oxygen concentration (between 23-30%, as life on planet is oxygen breathing)
    • Small, split up landmasses


    In summary, what I want to achieve is a planet where days are longer, years are shorter, and extremes of temperature are less seasonal or the seasons themselves are shorter in duration. The longer days would presumably create higher extremes of temperature between day and night (although this would be mititaged slightly by small, spread out landmasses).

    What I need to work out:
    1. How close can the planet be to the sun without it being tidally locked? (NB There is a formula for estimating the time it takes for a body to become tidally locked, can I use that to work out a range of distances at which my planet probably won't be tidally locked?)
    2. How slow can the planet's rotation be yet still produce a magnetosphere? Also, what explanation could I give for the slower-than-normal rotation period - a collision with it's own moon?
    3. Can the planet retain a moon? (This can be artificially introduced if necessary. I would prefer to have a moon, but can manage without)
    4. Will the planet be techtonically active?
    5. What is the mass and size of the planet? (Probably the same as Earth's, if the gravity is the same)
    6. What is the mass and size of the planet's star? (Whatever will provide the right temperature conditions on the planet's surface.)
    7. Does it matter what the axial tilt is?
    8. What would the sun look like from the surface of the planet in terms of both size and colour? (NB Other threads on this forum suggest that the colour of the light or the sky under a K or M-type dwarf star would not be noticeably different from Earth's. But it would be nice if I could have a big orangey sun for the concept art. I mean, what kind of self-respecting alien world doesn't have a big red sun?)


    The planet does not have to have evolved naturally into its current state, if necessary it can have been terraformed at some point. If some of the answers are not possible to theorise exactly, that's ok - guestimates are fine. If I can get this information sorted, I can start thinking more about how the differences in day/night and seasonal cycles affect the weather and lifeforms that live there.

    If I have got some of the terminology wrong, please let me know and I will edit the OP.
    Last edited by Veiligo; 2012-Jun-06 at 07:15 PM.

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    Since you're writing a story, you should figure out what sort of planet you want first. Then we'll help you figure out the parameters. There are many ways to explain any number of common or rare combinations of physical characteristics. If you want a fast rotation, claim an impact from one direction, if you want a slower rotation, claim an impact from another direction. If you want a similar gravity, keep it like earth, or make it bigger but less dense, or smaller but denser. If you want seasons give it a tilt or, less likely, give it an eccentric orbit. If you don't want seasons, give it little tilt, and then you may have static latitudinal variations. If you want moderate climate with/and/or/without seasons, give it lots of surface water.
    Et tu BAUT? Quantum mutatus ab illo.

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    Observationally, Venus, at 225 day orbit, is not tidally locked. Mercury, at 88 days, is.

    Generally, tidal dissipation is hard to predict. Tide-raising acceleration is not. 100 day orbit means tides 13 times higher than the solar tides of Earth - meaning 6 times higher than lunar tides on Earth, or 4x spring tides on Earth. Tides equivalent to Earth spring tides mean about 200 day period.

    Semimajor axis and requirement of Earth equivalent bolometric illumination gives the star bolometric magnitude - range thus 0,25...0,49. Nearby star examples are, from http://www.gdnordley.com/_files/Near...20Summary.html, epsilon Eridani (0,379, K2, mass assumed 0,85 - then period 191 days), epsilon Indi A (0,262, K5, mass 0,67 - then period 163 days), Keid A (0,467, K1, mass assumed 0,89 - then period 219 days)

    The planetary rotation period is rather free parametre - compare Earth and Venus. No particular explanation needed.

    If you need specifically Earth equal gravitational acceleration, you could have the planet slightly smaller than Earth - with bigger iron core, like Mercury, or slightly bigger, with smaller core.

    No objection to a moon. Note that 6 days is the orbital period of Charon - if the moon is massive enough, you can have mutual tidal lock.

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    Quote Originally Posted by chornedsnorkack View Post
    Observationally, Venus, at 225 day orbit, is not tidally locked. Mercury, at 88 days, is.
    Actually, Mercury is not tidally locked. It's in a 3:2 spin-orbit resonance. Over a couple orbits, all of Mercury's longitudes are exposed to the sun a few times.
    Et tu BAUT? Quantum mutatus ab illo.

  5. #5
    Quote Originally Posted by Ara Pacis View Post
    Since you're writing a story, you should figure out what sort of planet you want first. Then we'll help you figure out the parameters.
    Thanks I'll try and refine the parameters a bit more.

    Quote Originally Posted by chornedsnorkack View Post
    Generally, tidal dissipation is hard to predict.
    Ok, I'll just stick with 100 days then.

    Quote Originally Posted by chornedsnorkack View Post
    100 day orbit means tides 13 times higher than the solar tides of Earth - meaning 6 times higher than lunar tides on Earth, or 4x spring tides on Earth. [...]

    Semimajor axis and requirement of Earth equivalent bolometric illumination gives the star bolometric magnitude - range thus 0,25...0,49. Nearby star examples are [...]

    The planetary rotation period is rather free parametre - compare Earth and Venus. No particular explanation needed.

    If you need specifically Earth equal gravitational acceleration, you could have the planet slightly smaller than Earth - with bigger iron core, like Mercury, or slightly bigger, with smaller core.

    No objection to a moon. Note that 6 days is the orbital period of Charon - if the moon is massive enough, you can have mutual tidal lock.
    Thanks for all this, that helps a lot. I hadn't thought too deeply about tides yet, but that would make a huge difference.

    So the revised model for the planet/system is as follows...

    Semi major axis: 0.6AU (range 0.5-0.7)
    Orbital period: ~100 days
    Rotation period: ~5 days (range 3-5 days)
    Sun: K2 dwarf (probably luminosity 0.35-0.4, using Epsi Eridani as a base)
    Average surface temp: 15°C (which is what I think earth is?)
    Gravity: same as earth

    The story starts in a temperate zone, where daily temperatures fluctuate between 0-30°C. Cant decide right now whether the planet is bigger or the same size as earth, but I think the next thing I need to think about is how the land mass is divided up, in order to acheive the desired temperature range, and also consider how higher tides will affect coastal areas.

    And I will have a small moon after all

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    Quote Originally Posted by Veiligo View Post
    The story starts in a temperate zone, where daily temperatures fluctuate between 0-30°C. Cant decide right now whether the planet is bigger or the same size as earth, but I think the next thing I need to think about is how the land mass is divided up, in order to acheive the desired temperature range, and also consider how higher tides will affect coastal areas.

    And I will have a small moon after all
    A large component of the tides will depend on the bathymetry, not just the gravitational gradients. Shallow and sheltered gulfs and isolated bodies of water will experience lower tides but wide and deep ocean basins will tend to magnify the effects, and certain shaped bays may magnify them even more (Bay of Fundy), hence the difference between the Gulf of Mexico tides and tides along the Atlantic seaboard of the US. Part of that is due to a basin-wide rhythmic bulge that moves around the entire basin.

    0-30°C sounds extreme, so I doubt there'd be a large heat reservoir such as a body of water or thick, moist atmosphere at that location. You need to take into account the effect of dew point on temperature, so this extreme suggests to me an inland desert, perhaps at high altitude. Or perhaps there is some sort of standing High pressure due to the climate and geography.
    Et tu BAUT? Quantum mutatus ab illo.

  7. #7
    Quote Originally Posted by Ara Pacis View Post
    A large component of the tides will depend on the bathymetry, not just the gravitational gradients. Shallow and sheltered gulfs and isolated bodies of water will experience lower tides but wide and deep ocean basins will tend to magnify the effects, and certain shaped bays may magnify them even more (Bay of Fundy), hence the difference between the Gulf of Mexico tides and tides along the Atlantic seaboard of the US. Part of that is due to a basin-wide rhythmic bulge that moves around the entire basin.

    0-30°C sounds extreme, so I doubt there'd be a large heat reservoir such as a body of water or thick, moist atmosphere at that location. You need to take into account the effect of dew point on temperature, so this extreme suggests to me an inland desert, perhaps at high altitude. Or perhaps there is some sort of standing High pressure due to the climate and geography.
    Sounds like I have a lot of reading to do. And some things to rethink. This is all really useful, 'preciate it.

    (NB The temperature range doesn't have to be that high - it's not a plot requirement - so I'll reduce it your opinion is that it's unrealistic.)

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    A useful calculator that doesn't exist yet (as far as I know) would give a reasonable range of densities for Earth-like worlds; the density of a rocky world is dependent on how big the iron core is, and how much compression due to gravity has occurred - this value is greater in larger worlds, so a larger world will be denser. Maybe someone would like to try their hand at making one someday.

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    Quote Originally Posted by Veiligo View Post
    Semi major axis: 0.6AU (range 0.5-0.7)
    Orbital period: ~100 days
    Sun: K2 dwarf (probably luminosity 0.35-0.4, using Epsi Eridani as a base)
    That doesn't work out. Kepler's third law says that P(years)2 = a(AU)3 * M(solar masses). With a = 0.6 AU and P = 100 days ~ 0.3 years, M = 0.25 solar masses, which would be an M star. A hot K dwarf is about 3 times as massive.

  11. #11
    Useful tool, bookmarked ^^

    Quote Originally Posted by Tobin Dax View Post
    That doesn't work out. Kepler's third law says that P(years)2 = a(AU)3 * M(solar masses). With a = 0.6 AU and P = 100 days ~ 0.3 years, M = 0.25 solar masses, which would be an M star. A hot K dwarf is about 3 times as massive.
    Thanks, that's the kind of info I need.

    I found a tool that helps calculate surface temperature for a planet based on inputting the solar mass, the semi major axis, and the albedo and greenhouse effects so I added this to the model. After playing with the variables I get this:

    Solar mass: 0.43
    Axis: 0.7
    Orbital period: 0.3

    To get a surface temperature on the planet of 15°C, that assumes the same albedo as earth, and a greenhouse effect that's 18x stronger. So now I need to go and read up on greenhouse effects as well as bathymetry ^^

    Updated the OP with this info.

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    Quote Originally Posted by eburacum45 View Post
    A useful calculator that doesn't exist yet (as far as I know) would give a reasonable range of densities for Earth-like worlds; the density of a rocky world is dependent on how big the iron core is, and how much compression due to gravity has occurred - this value is greater in larger worlds, so a larger world will be denser. Maybe someone would like to try their hand at making one someday.
    I did have a brief look a while ago but the formula seemed a bit daunting to code. However what I am thinking is that I could just store key points on the curve in a table and the density of any mass planet lying between 2 points in the table could be inferred by averaging - which should give a reasonable approximation.

    The referenced formula does appear to give a clear minimum and maximum value for planet density of a given mass for both solid planets and gas giants (see graph on page 6) - which takes into account density increase with mass. Solid planets would fall between 100% iron and 100% water. Gas giants would fall between 100% water and 100% Hydrogen.

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    Quote Originally Posted by Veiligo View Post
    Sounds like I have a lot of reading to do. And some things to rethink. This is all really useful, 'preciate it.

    (NB The temperature range doesn't have to be that high - it's not a plot requirement - so I'll reduce it your opinion is that it's unrealistic.)
    It's not extreme in the sense of unlikely, but extreme in the sense of fitting with what many people think of as temperate. A friend of mine just got back from Afghanistan and he mentioned normal diurnal temperture fluctuations close to what you mention. So, if the setting you want is an high altitude desert, you're good to go. If you want it to be something with more life and water, then you may be able to have that, but you'll have to play around with geography and rainfall patterns and plant biologies. but I'm looking at it as similar to Earth. If you have a slower rotation, which I just remembered, then you have more time for temperatures to fluctuate to more extreme positions, so your design sounds more doable now that I think about it.

    And you don't have to design the bathymetry, as long as you have something that sounds like a normal pattern given the gravitational fluctuations.
    Et tu BAUT? Quantum mutatus ab illo.

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    Quote Originally Posted by Veiligo View Post

    I found a tool that helps calculate surface temperature for a planet based on inputting the solar mass, the semi major axis, and the albedo and greenhouse effects so I added this to the model. After playing with the variables I get this:

    Solar mass: 0.43
    Axis: 0.7
    Not likely on main sequence.

    If what you need is orbital period 0,3 years, you are looking for a more massive but dimmer star.

  15. #15
    Quote Originally Posted by JohnBStone View Post
    I did have a brief look a while ago but the formula seemed a bit daunting to code. However what I am thinking is that I could just store key points on the curve in a table and the density of any mass planet lying between 2 points in the table could be inferred by averaging - which should give a reasonable approximation.
    That sounds a cool thing to have. I think in this case I will just stick with same density, size and gravity as earth for simplicity's sake, but if you did ever put something up I'd like to see it.

    Quote Originally Posted by Ara Pacis View Post
    If you have a slower rotation, which I just remembered, then you have more time for temperatures to fluctuate to more extreme positions, so your design sounds more doable now that I think about it.

    And you don't have to design the bathymetry, as long as you have something that sounds like a normal pattern given the gravitational fluctuations.
    Ok thanks. And yeah, I'm supposing what I need to do is find somewhere that has similar geography on earth, and then decide how having a different distance/solar mass would affect the tides.

    Quote Originally Posted by Tobin Dax View Post
    That doesn't work out. Kepler's third law says that P(years)2 = a(AU)3 * M(solar masses). With a = 0.6 AU and P = 100 days ~ 0.3 years, M = 0.25 solar masses, which would be an M star. A hot K dwarf is about 3 times as massive.
    Actually, I just realised I'm having difficulty with this equation. I don't get 0.25 solar masses from 0.6AU and 0.3 days.

    0.32 / 0.63 = 0.42 solar masses.

    What am I doing wrong? Also, if I make a table of AU vs solar masses it looks... weird:

    Code:
    Axis	Per.	Solar masses
    0.10	0.27	72.90
    0.20	0.27	9.11
    0.30	0.27	2.70
    0.40	0.27	1.14
    0.50	0.27	0.58
    0.60	0.27	0.34
    0.70	0.27	0.21
    0.80	0.27	0.14
    0.90	0.27	0.10
    0.95	0.27	0.09
    1.00	0.27	0.07
    2.00	0.27	0.01

    Quote Originally Posted by chornedsnorkack View Post
    Not likely on main sequence.

    If what you need is orbital period 0,3 years, you are looking for a more massive but dimmer star.
    In the table you linked it provided a column showing the AU at which the insolation would equal earth's (a bit slow realising but I got there in the end ><). So what I should do is: for each star in that table, calculate the AU at which a planet has a period of 0.27 yrs*, and then match this against the AU figure that provides equivalent to earth insolation. If there are any correlations, I can use that star as a model.

    *As I said above, this is the bit I'm having difficulty with.

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    Could the formula given by Tobin Dax contain an error in the power of the mass of the primary?

    As I pointed out, 0,5 AU is too far (consider epsilon Indi - 163 day orbit, not 100).

    For a less massive star example, consider 61 Cygni A.
    Luminosity 0,150 solar. Mass 0,70 solar.

    Distance square root 0,150 is 0,387 AU.

    Cube of distance is 0,058.
    Divided (sic) by the mass the square of period is 0,083
    So year length turns out as 0,288 years - 105 days.

    Other examples from similar luminosity range:

    Lac 8760 - 0,130 solar luminosity, 0,60 solar mass
    Grb 1618 - 0,120 solar lumonosity, 0,64 solar mass.

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    I use this calculator to create planets around stars;
    http://web.archive.org/web/200302181...1/mseqstar.htm
    I must admit that I tend to ignore the stellar class it gives, as it seems to be slightly askew. Enter a luminosity for the star and a luminosity as seen at the planet and you get your year length and sem-imajor axis values.
    Then go to this site to get a more accurate stellar class
    http://web.archive.org/web/200802131...tarTables.html

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    One confusing thing about the relationship between mass and luminosity is that an older star will be brighter than a younger star of the same mass. Red dwarfs will slowly get brighter over time until some of them are as bright as our Sun is today.

  19. #19
    Quote Originally Posted by chornedsnorkack View Post
    For a less massive star example, consider 61 Cygni A.

    [...]

    Lac 8760 - 0,130 solar luminosity, 0,60 solar mass
    Grb 1618 - 0,120 solar lumonosity, 0,64 solar mass.
    Quote Originally Posted by eburacum45 View Post
    I use this calculator to create planets around stars;
    http://web.archive.org/web/200302181...1/mseqstar.htm
    Ok, so this link from eburacum and the examples Chornedsnorckack is giving correlate - I'll have a look at those stars in more detail (after my brain recovers from being exposed to all these numbers).

    I get this from the link...

    Star
    Mass: 0.6299
    Lum: 0.1282
    Eff temp: 3965 K
    Diameter: 0.7122
    Stellar classification: K

    Planet - (which receives equivalent radiation from star as earth)
    Period: 0.27
    Axis: 0.3581 AU
    Apparent size of star viewed from planet: 1.9890

    I'm getting there

    Quote Originally Posted by eburacum45 View Post
    One confusing thing about the relationship between mass and luminosity...
    Heh, ok, will bear that in mind.

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    Quote Originally Posted by eburacum45 View Post
    A useful calculator that doesn't exist yet (as far as I know) would give a reasonable range of densities for Earth-like worlds; the density of a rocky world is dependent on how big the iron core is, and how much compression due to gravity has occurred - this value is greater in larger worlds, so a larger world will be denser. Maybe someone would like to try their hand at making one someday.
    I got my head around the formulas from that paper and added it to the planet designer web page. It now shows Minimum, Maximum and typical density for a terrestrial planet of a given mass. See the paper for fine details of accurate ranges for the formula (varies by composition).

    Minimum density is for a planet that is pure water.

    Maximum density is for a planet that is pure iron.

    Web page to calculate lots of useful planetary parameters given density, mass (relative to Earth) and day length

    It seems to give reasonable results across a wide range. Let me know if you spot any issues.

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    Quote Originally Posted by Veiligo View Post
    Actually, I just realised I'm having difficulty with this equation. I don't get 0.25 solar masses from 0.6AU and 0.3 days.

    0.32 / 0.63 = 0.42 solar masses.

    What am I doing wrong? Also, if I make a table of AU vs solar masses it looks... weird:

    Code:
    Axis	Per.	Solar masses
    0.10	0.27	72.90
    0.20	0.27	9.11
    0.30	0.27	2.70
    0.40	0.27	1.14
    0.50	0.27	0.58
    0.60	0.27	0.34
    0.70	0.27	0.21
    0.80	0.27	0.14
    0.90	0.27	0.10
    0.95	0.27	0.09
    1.00	0.27	0.07
    2.00	0.27	0.01
    Oops, I made a typo. It should be P2 = a3/M, or M = a3/P2. I also made a mistake in my arithmetic. Below is an updated table. a = 0.6 AU still will not work for a 100-day orbit around a K star, but a = 0.4 AU will.

    Code:
    Axis	Per.	Solar masses
    0.3	0.27	0.37037037
    0.4	0.27	0.877914952
    0.5	0.27	1.714677641
    0.6	0.27	2.962962963
    Edit: 0.88 solar masses falls nicely in the range of a K1V or K2V star, as you had originally hoped for.

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    Is the 100 day orbit measured in 24-hour Earth Days or in 120-hour Local Days?
    Et tu BAUT? Quantum mutatus ab illo.

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    Quote Originally Posted by JohnBStone View Post
    I got my head around the formulas from that paper and added it to the planet designer web page. It now shows Minimum, Maximum and typical density for a terrestrial planet of a given mass. See the paper for fine details of accurate ranges for the formula (varies by composition).

    Minimum density is for a planet that is pure water.

    Maximum density is for a planet that is pure iron.

    Web page to calculate lots of useful planetary parameters given density, mass (relative to Earth) and day length

    It seems to give reasonable results across a wide range. Let me know if you spot any issues.
    That's really useful; thanks! I'll try a few planets out and see how it works.
    I wouldn't expect to see many, or indeed any, natural planets made of pure water or pure iron, but they are useful as extreme limits when worldbuilding.

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    Quote Originally Posted by eburacum45 View Post
    That's really useful; thanks! I'll try a few planets out and see how it works.
    I wouldn't expect to see many, or indeed any, natural planets made of pure water or pure iron, but they are useful as extreme limits when worldbuilding.
    I can imagine a small planet loosing a lot of it's lighter elements in a glancing collision. I can imagine a small ice body melting as it's sun got warmer.
    Et tu BAUT? Quantum mutatus ab illo.

  25. #25
    Quote Originally Posted by Tobin Dax View Post
    Oops, I made a typo. It should be P2 = a3/M, or M = a3/P2. I also made a mistake in my arithmetic. Below is an updated table. [/code]
    Ah thanks, for the correction.

    Quote Originally Posted by JohnBStone View Post
    I got my head around the formulas from that paper and added it to the planet designer web page. It now shows Minimum, Maximum and typical density for a terrestrial planet of a given mass. See the paper for fine details of accurate ranges for the formula (varies by composition). [/URL]
    This is going to come in very handy if my writing career ever gets anywhere. Thanks. ^^

    Quote Originally Posted by Ara Pacis View Post
    Is the 100 day orbit measured in 24-hour Earth Days or in 120-hour Local Days?
    Earth days. Updating the OP to reflect this.

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