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StephenPrivitera

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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

d is the distance from the Sun, which I take to be about 1AU

A is the albedo, L is the solar luminosity

F

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=piR

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.

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)LpiR^{2}/(4pid^{2})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}=4piR^{2}esT^{4}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=piR

^{2}cos^{2}(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.

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