Thread: What is the range of distances in metric that a planet orbiting a red dwarf could have Liquid Ammonia?

Assuming that the star it orbits has a surface temperature of about 4000 degrees kelvin, and its surface pressure is about one atmosphere. What is the range of distances that a planet like this could have liquid ammonia on its surface?
Also assume that the radius of the parent stare is about half the radius of the sun.

I could derive the formula for planet temperature estimate, but why bother since I've done that enough?

The formula is:
T = Ts*sqrt(Rs/(2*d))

where:
Ts is the temperature of the parent star
Rs is the radius of the parent star
d is the distance from the parent star

This formula assumes:
1. The star radiates as a black body
2. The planet is a body of either infinite conductivity of very fast sidereal spin speed.
3. The planet is a gray body, such that its emissivity is frequency independent
4. The planet is thermally dormant producing no heat energy of its own
5. The atmosphere of the planet does not have any selective transmission or Raleigh attenuation.

Re-arrange for d:
d = Rs*Ts^2/(2*T^2)


We desire that T be somewhere between the Earth atmospheric pressure melting temperature of anhydrous ammonia and the Earth atmospheric pressure boiling temperature of anhydrous ammonia.

Thus, construct one version for melting and one for boiling:
dfar = Rs*Ts^2/(2*Tmelt^2)
dnear = Rs*Ts^2/(2*Tboil^2)


You need to give the radius of the parent star for me to actually answer you.

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Edit:

Data:
Rs:= 0.002326 AU; Ts:=4000 Kelvin; Tmelt:=195 Kelvin; Tboil:=240 Kelvin;

Results in AU:
dfar = 0.4893 AU
dnear = 0.3231 AU

I didn't get the same result as Gintable, so here is my back of the envelop reasoning.

The star's luminosity L=(R/R_Sun)^2 (T/T_Sun)^4 = 0.5^2 (4000/5780)^4 = 0.057 L_Sun.
So we will find back Earth conditions at sqrt(0.057) = 0.24 AU.

The surface temperature of the Earth without greenhouse is 288 Kelvin. Ammonia is liquid between 200 and 240 Kelvin, say 220 Kelvin with a 10% error.
To go from 288 to 220 Kelvin we have to move our planet outwards by a factor sqrt(288/220) = 1.14, so that puts it at 0.273 AU +- 5%.
So my limits are 0.271-0.275 AU, or about 40 +- 0.4 million km.