User:GabrielVelasquez/sandbox3

Insolation data for Extra Solar Planets
Main article: Luminosity & Climate modeling 

Luminosity can be calculated given a distance (in Lightyears) and apparent magnitude (m):
 * Lstar/$$L_{\odot}$$ = (diststar/distsun)2 &middot; 10[(msun &minus;mstar) &middot; 0.4]
 * Lstar = 0.0813 &middot; diststar2 &middot; 10(&minus;0.4 &middot; mstar) &middot; $$L_{\odot}$$

From Luminosity and distance irridance can be calculated"
 * $$L = 4\pi R^2\sigma T^4 \,$$  and,   $$L = 4\pi d^2\sigma f \,$$

Where d is in AU for the planet and f is the the irradiance.

Therefore:

$$f_p= \frac{( (R \times 6.955 \times 10^8 )^2 ) \times (5.67051 \times 10^{-8}) \times (T^4)} { ( ( d - ( d\times e) ) \times 149597876600 )^2 }$$

$$f_p= \frac{ (L \times 3.0572\times 10^{25})} { ( ( d - ( d\times e) ) \times 149597876600 )^2 }$$

HD 221287 b for example may harbour habitable moons.

Formula for radius and temperature
Here's the formula for calculating radius and temperature for extrasolar planets. You should add calculated r & T to extrasolar planet articles.

$$ R_{planet} = \sqrt{\frac {\sqrt [3]{M_{planet}} \times \sqrt {R_{star}}} {\sqrt [4]{a}}} $$

$$ T_{planet} = \frac {\sqrt [4]{L_{star}} \times 256} {\sqrt{a}} $$

where:

$$M_{planet}$$ is the mass of the planet

$$R_{planet}$$ is the radius of the planet

$$T_{planet}$$ is the temperature of the planet

$$a$$ is the semimajor axis

$$R_{star}$$ is the radius of the star

$$L_{star}$$ is the luminosity of the star

Note: After finding the radius, you should find density by formula m/r³ and gravity by formula m/r². The radius of planets can only be calculated orbiting the main-sequence stars. Giants and subgiants have grew larger in radius and does not grow planets in radius because of the expanding star's radius.

--BlueEarth 18:05, 24 July 2007 (UTC)
 * Where did you get those from?!? I fail to grasp how the radius of a planet can be dependent on the radius of its parent star or of its semimajor axis. (Not to mention that density also needs to be factored in.) The formula for the temperature of the planet looks more reasonable. However, it has no units, and the use of semimajor axis instead of (weighted) mean distance to the star is highly suspect. Ben Hocking (talk 18:20, 24 July 2007 (UTC)
 * I tried different set of radius formula for trying to find radius of the planet Gliese 581 c until I'll get exactly 1.5 RE as it was referenced. I made the temperature formula by trying to calculating the temperature of extrasolar planets with changing formula until I'll get exact value and accepting the formula from calculating effective temperature of Mars and Jupiter. BlueEarth 19:54, 24 July 2007 (UTC)
 * A general formula for the radius would be:$$R_{planet} = \sqrt [3]{\frac{3M_{planet}}{4 \pi \rho_{planet}}}$$ where $$\rho_{planet}$$ is the density of the planet. If your units are cgs, $$\rho_{planet}$$ might not differ significantly from 1 (the density of water in g/cm3). There will be no simple way to deduce the density (except for using the radius and mass) as some planets are rocky, gaseous, etc. The temperature of the planet is also a little more complicated than what you have there. (I'll have to look into that one, as I don't have it off the top of my head.) Ben Hocking (talk 22:07, 24 July 2007 (UTC)
 * A planet's density is not only affected by the material it is made of, but also by gravitational pressure which is significant for giant planets (gas giants like Jupiter, brown dwarfs and low-mass red dwarfs are all about the same size despite their highly different masses).— JyriL talk 22:52, 1 August 2007 (UTC)

This radius formula is nonsense. First off, the author claims it is based on the properties of a planet for which the radius isn't actually known, secondly it is not dimensionally consistent, thirdly it doesn't actually work for our solar system, so expecting it to work for every other planetary system is ridiculous. Look, if the radius and temperature of an extrasolar planet are unknown, just don't put them in the infobox. This is an encyclopaedia, it is a summary of what is known. If something isn't known, then we shouldn't pretend it is. Chaos syndrome 21:32, 29 July 2007 (UTC)

Theoretical surface temperature of a planet is

$$T = T_\star \sqrt[4] { \frac {1 - A} { 2 } } \sqrt { \frac {R_\star} {r} }$$

and for a rapidly rotating planet (when the amount of emitted radiation is equal across the whole surface)

$$T = T_\star \sqrt[4] { \frac {1 - A} { 4 } } \sqrt { \frac {R_\star} {r} }$$

$$A$$ is the planet's Bond albedo, which obviously is unknown for extrasolar planets, and $$r$$ is its orbital distance. $$R_\star$$ is the star's radius. When we take atmosphere into account, things like greenhouse effect and heat currents guarantee the values given by the equations become wildly off-mark. For example, based on these Venus would be slightly cooler than Earth because of its very high albedo. — JyriL talk 22:52, 1 August 2007 (UTC)