Calculating Luminosity and Temperature of the Sun from Pluto

[EzequielSeattle]

Hey all, I think that I'm doing this problem correctly but I'm getting an answer that's a couple thousand Kelvins off. Sorry if I should have posted this in the "advanced" section.

1. Homework Statement
Part of a flat region of Pluto has the Sun directly overhead. Its surface temperature is 58 K. You have determined Pluto is 33 AU away from the sun, and the region of interest has an albedo of 0.5. Use all of this information to determine the Sun's luminosity temperature. Assume the Sun and Pluto are perfect black bodies.

Homework Equations

P/A of a black body is equal to σT⁴, where σ = 5.67x10^-8 W*m^-2*K^-4.

The Attempt at a Solution

Pluto's temperature is 58 K. By the equation above, its power/area radiated outward is σ*58⁴, which is 0.642 W*m^-2. Because its albedo is 0.5, it receives double this power from the sun, so the power per area from the sun hitting the surface is 1.283 W*m^-2.

Because this power will fall off as the radius squared from the sun, I multiplied this by the square of 33 AU, which is 2.437x10²⁵ square meters. I came away with a total luminosity of 3.13 W. Then, using the same equation above, I get that the effective surface temperature of the Sun is 3193 K, which is almost 3000 K away from the true value. Am I doing something wrong, or is this because of the imprecise variables that I was given in the problem?

Thanks in advance.

 

[Bandersnatch]

Pluto's temperature is 58 K. By the equation above, its power/area radiated outward is σ*584, which is 0.642 W*m-2. Because its albedo is 0.5, it receives double this power from the sun, so the power per area from the sun hitting the surface is 1.283 W*m-2.

Why do you care for the power radiated from Pluto's surface, though?

 

[EzequielSeattle]

The power radiated from Pluto's surface is equal to half of the total power it receives from the Sun, because the albedo is 0.5. Right?

 

[Bandersnatch]

Right, sorry. I posted without thinking. Give me a moment to go through it again.
Tentatively, I'd probably look at the difference in areas that receive and reradiate the energy.

 

[Bandersnatch]

Because this power will fall off as the radius squared from the sun, I multiplied this by the square of 33 AU, which is 2.437x1025 square meters.

I think this is the issue right here. To get the total power produced by the Sun, you need to take the flux you've just calculated at Pluto's orbit, and multiply it by the area of a sphere at that distance. That is, 4π times the radius squared. This gives the correct luminosity for the Sun.

I'm not sure what method you used to get the temp. from luminosity, though. Don't you need to know the Sun's radius? In any case, with the correct luminosity you should get the correct temperature.