Equilibrium temperature as function of latitude

[StephenPrivitera]

Find the equilibrium temperature of the moon as a function of latitude assuming that the moon is a rapid rotater with an emissivity of 1, zero obliquity, and a bond albedo of 0.07.

The only variable for this problem in the equation for equilibrium temperture is the distance from the Sun. If I say that the center of the moon is a distance d (1AU) from the Sun, then a given point on the surface is a distance d+Rcos(latitude) where R is the radius of the moon. Using this value in the equation for equilibrium temperature doesn't change the temperature much from pole to equator (about 0.01 K).

Defeated, I resort to this,

F_abs=(1-A)LpiR²/(4pid²)
d is the distance from the Sun, which I take to be about 1AU
A is the albedo, L is the solar luminosity
F_emit=4piR²esT⁴
e=1=emissivity
s=sigma=Boltzmann constant
T=temp
It appears that the first equation estimates the area of the moon using a circle (piR^2) and the second using the equation for the area of a sphere (4pir^2). I follow along these lines. The effective area of the moon is dA=2Rcos(latitude)dh in the first equation (two dimensional area). For the second, I use a cylindrical estimation of the area (3D): dA=piR²cos²(latitude)dh. I substitute these into the equations and solve for T. The answer I get indicates that the temperature of the moon is about 8K at the equator and approaches infinity near the poles.

edit: I caught a mistake. I used the volume of a cylinder rather than the area of the cylinder. In that case, the solution approximately reduces to the original method off by a small constant factor.

 

[arcnets]

F_abs=(1-A)LpiR²/(4pid²)

This refers to the total absorbed power.
I think it's better to use intensity (power per unit area).

For absorption, Intensity varies because of the inclination of the surface:
I_abs=(1-A)Lcos(latitude)/(4pid²).
For emission, we have
I_emit=esT⁴.

OK?

 

[M98Ranger]

Thank you for sharing your approach to finding the equilibrium temperature of the moon as a function of latitude. It is interesting to see how you incorporated the distance from the Sun and the moon's rotation into your calculations. However, I believe there may be a few errors in your method.

Firstly, I believe the first equation you used to estimate the area of the moon is incorrect. It should be dA = 2Rsin(latitude)dh instead of dA = 2Rcos(latitude)dh. This is because the moon is a sphere and the distance from the Sun is the radius of the sphere, which is perpendicular to the surface at all points. Therefore, the angle should be measured from the equator, not the pole.

Secondly, the equation for the effective area of the moon should be dA = piR2cos(latitude)dh, not dA = piR2cos2(latitude)dh. This is because the cosine function already takes into account the angle of incidence, so there is no need to square it.

Lastly, I believe your final answer of 8K at the equator and approaching infinity near the poles is incorrect. The actual equilibrium temperature of the moon is estimated to be around 220 K, which is significantly higher than your calculation. This is likely due to the errors in your method, as well as the simplifications made (such as assuming the moon is a perfect sphere and neglecting other factors such as thermal inertia).

In conclusion, while your approach to finding the equilibrium temperature of the moon as a function of latitude is interesting, it may not produce accurate results. It is important to carefully consider all factors and equations involved in order to get a more precise estimation.