Sirius_Alpha Admin
Number of posts : 4320 Location : Earth Registration date : 2008-04-06
| Subject: Planet Insolation 2nd May 2009, 10:27 pm | |
| Can someone give me a planetary insolation equation that uses the luminosity of the star, and the distance of the planet? I can't find any that make sense. L / d^2 doesn't seem to work either.
Also, what about an equation that takes into account the radius of the star? _________________ Caps Lock: Cruise control for 'Cool'!
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Edasich dK star
Number of posts : 2291 Location : Tau Ceti g - Mid Latitudes Registration date : 2008-06-02
| Subject: Re: Planet Insolation 3rd May 2009, 7:00 am | |
| I'm not sure if it's what you're asking but a radius relation for the star is that I know for luminosity:
LStar=(TeffStar/TeffSun) 4 x (RStar/RSun)2
Where Teff is effective temperature of the star (TeffSun=5870 K) and R is radius (RSun=1).
For example, you wanna get Sirius' luminosity:
LSirius=(TeffSirius/TeffSun) 4 x (RSirius/RSun)2
LSirius=LStar=(9940/5780) 4 x (1.71)2=25.6 LSun
Then, if you wanna also get the habitable zone for the star, you just get the square root of the luminosity and that's all:
HZSirius=√LSirius
HZSirius=√25.6= 5 AUs | |
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Sirius_Alpha Admin
Number of posts : 4320 Location : Earth Registration date : 2008-04-06
| Subject: Re: Planet Insolation 3rd May 2009, 11:42 am | |
| I'm looking for an equation for planetary insolation (i.e. a measure of how much energy a planet receives from its star, relative to Earth, or measured in watts per square metre). _________________ Caps Lock: Cruise control for 'Cool'!
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jbjerk Micrometeorite
Number of posts : 13 Registration date : 2009-11-19
| Subject: Re: Planet Insolation 19th November 2009, 4:01 pm | |
| - Sirius_Alpha wrote:
- I'm looking for an equation for planetary insolation (i.e. a measure of how much energy a planet receives from its star, relative to Earth, or measured in watts per square metre).
I = L / R^2 (inverse-square law) I = the insolation. This is equivalent to the sun apparent brightness, i.e. how bright it appears when viewed from the planet. R = the distance between the planet and the sun. L = the sun's luminosity, i.e. how much light it gives out. from Creating an Earthlike Planet: http://www.cix.co.uk/~morven/worldkit/index.html#astro-sun | |
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Lazarus dF star
Number of posts : 3337 Registration date : 2008-06-12
| Subject: Re: Planet Insolation 19th November 2009, 5:03 pm | |
| If the planet is on an eccentric orbit, the orbit-averaged flux is given by f = L / (4πa 2) * 1/√(1-e 2) where L is luminosity of the star, a is semimajor axis, e is eccentricity, given in SI units - if you are working in units scaled to the Earth/Sun system (i.e. solar luminosity, AU, flux relative to Earth's), you can drop the 4π. Derivation of the factor involving eccentricity is fairly straightforward but requires somewhat better mathematics rendering than is available here
Last edited by Lazarus on 20th November 2009, 6:44 pm; edited 2 times in total (Reason for editing : oops dropped a factor of 4pi) | |
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Sedna Planetary Embryo
Number of posts : 87 Registration date : 2008-08-21
| Subject: Re: Planet Insolation 20th November 2009, 4:57 pm | |
| This formula works too: I = (r² * sigma * T^4)/d²; where I is the insolation in W/m², r is the radius of the star in meters, T is the temperature of the star in kelvin and d is the semi-major axis in meters.
Bye
Sedna | |
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| Subject: Re: Planet Insolation | |
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