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Is the sun going to end up being called a "Dwarf Star"? Also, this means nothing to me: "With a dayside temperature of more than 7,800 degrees Fahrenheit (4,60

by nonbel 9y ago

Is the sun going to end up being called a "Dwarf Star"? Also, this means nothing to me: "With a dayside temperature of more than 7,800 degrees Fahrenheit (4,600 Kelvin)" Where in the atmosphere is this temperature begin measured? The temperature can vary wildly by altitude, latitude, and time of "day". As far as we can tell, as the atmosphere of a planet gets thicker the temperature keeps increasing indefinitely (eg check out fig 1): http://faculty.washington.edu/dcatling/Robinson2014_0.1bar_Tropopause.pdf http://faculty.washington.edu/dcatling/Robinson2014_0.1bar_T...

greglindahl 9y ago
In this case 4600K is what a black body at the planet's distance would be heated to by the star.
- nonbel 9y ago
  Thanks, actually take a look at the self-response I was writing as you posted this.
nonbel 9y ago
Looking closer: >"Given the high stellar luminosity and close orbit, the planet receives a large stellar insolation flux (Table 1). As a result, it has an extremely high equilibrium temperature: calculations, assuming zero albedo and perfect heat redistribution, give a value of approximately 4,050 K."[1] That 4,050 K value is reported as the equilibrium temperature, which I understand to be the output of the Stefan-Boltzmann law: T = (I*(1-alpha)/(epsilon*sigma))^.25 where, I = Incident Flux alpha = albedo epsilon = emissivity sigma = SB constant = 5.670373e-8 According to the paper the planet receives 61.1 x10^9 erg/s/cm^2.[1] The solar constant is ~1360 W/m^2.[2] In the same units the planet gets 6.11 x 10^7 W/m^2, which is about 45 thousand times greater than what the earth receives. Anyway, plugging into the SB law I get: (6.11e7*(1-0)/(1*5.670373e-8))^.25 = 5729.377 K So I am certain it is I rather than them with the error. Where did I go wrong? [1] http://www.nature.com/nature/journal/vaop/ncurrent/full/nature22392.html http://www.nature.com/nature/journal/vaop/ncurrent/full/natu... [2] https://en.wikipedia.org/wiki/Solar_constant https://en.wikipedia.org/wiki/Solar_constant
- nonbel 9y ago
  I see now, I was calculating the temperature at the subsolar point, they are averaging over a uniformly illuminated sphere. Since the illuminated side is a disc with area pi x R^2 and the sphere has surface area 4 x pi x R^2 we need to include a factor of 1/4: (6.11e7*(1-0)/(4*1*5.670373e-8))^.25 = 4051.281 K

by nonbel 9y ago

greglindahl 9y ago

In this case 4600K is what a black body at the planet's distance would be heated to by the star.

nonbel 9y ago
Thanks, actually take a look at the self-response I was writing as you posted this.

nonbel 9y ago

Looking closer: >"Given the high stellar luminosity and close orbit, the planet receives a large stellar insolation flux (Table 1). As a result, it has an extremely high equilibrium temperature: calculations, assuming zero albedo and perfect heat redistribution, give a value of approximately 4,050 K."[1] That 4,050 K value is reported as the equilibrium temperature, which I understand to be the output of the Stefan-Boltzmann law: T = (I*(1-alpha)/(epsilon*sigma))^.25 where, I = Incident Flux alpha = albedo epsilon = emissivity sigma = SB constant = 5.670373e-8 According to the paper the planet receives 61.1 x10^9 erg/s/cm^2.[1] The solar constant is ~1360 W/m^2.[2] In the same units the planet gets 6.11 x 10^7 W/m^2, which is about 45 thousand times greater than what the earth receives. Anyway, plugging into the SB law I get: (6.11e7*(1-0)/(1*5.670373e-8))^.25 = 5729.377 K So I am certain it is I rather than them with the error. Where did I go wrong? [1] http://www.nature.com/nature/journal/vaop/ncurrent/full/nature22392.html http://www.nature.com/nature/journal/vaop/ncurrent/full/natu... [2] https://en.wikipedia.org/wiki/Solar_constant https://en.wikipedia.org/wiki/Solar_constant

nonbel 9y ago
I see now, I was calculating the temperature at the subsolar point, they are averaging over a uniformly illuminated sphere. Since the illuminated side is a disc with area pi x R^2 and the sphere has surface area 4 x pi x R^2 we need to include a factor of 1/4: (6.11e7*(1-0)/(4*1*5.670373e-8))^.25 = 4051.281 K