# Calculate temperature of the earth through blackbody radiation

I don't understand the solutions to a problem about blackbody radiation and was wondering if anybody could help me out.

Here is the question:

The sun can be considered as a blackbody radiation source at temperature T = 5778 K. Radiation from the sun which is incident on the earth is reflected by the atmosphere such that the intensity hitting the earth's surface is reduced by a factor R. Some of the radiation emitted from the earth's surface is reflected by the atmosphere such that only a fraction A leaves the atmosphere. If A = R = 0:1, what temperature would the earth be?

Then in the solutions they state: This is obtained by first trying to find the power from the sun which passes through unit area at the earth's radial distance. This is given by:

$\frac{4 \pi r_s^2}{4 \pi d_e^2}\sigma T_s^4$

where $r_s$ is the radius of the sun, $d_e$ is the distance between the earth and sun and $T_s^4 = 5778K$ is the temperature of the sun.

I know that $\sigma T_s^4$ is from the Stefan-Boltzmann law, and that $4 \pi r_s^2$ is the surface of the sun. What I don't understand is why the distance to the earth is important. Thanks in advance!

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If the Earth is farther from the sun it receives less radiation from the sun! Try sketching the geometry. (Hints: conservation of energy (flux); inverse square law) –  Michael Brown Jan 9 '13 at 13:15
@MichaelBrown So the $\frac{1}{d_e^2}$ comes from the inverse square law? The thing that confuses me is that the earth's radius isn't needed anywhere. –  Longeyes Jan 9 '13 at 13:35

## 2 Answers

The standard procedure (or at least how I think about it) for getting the temperature of a planet from that of the star consists of alternating between power and power per unit area.

• You start with $\sigma T^4$, the total power per unit area of the star.
• To get the total power, multiply by the area of the star.
• To get the power per unit area at the distance of the planet, you have to divide by the area of the sphere over which this energy is (by assumption evenly) distributed. This gets you to the quantity you have.
• Now you can find the total power absorbed by the planet, given the power per unit area it gets and an area1.
• Then you can divide by an area1 to get the power per unit area given off by the planet, which is a quantity you can set equal to $\sigma T_\text{planet}^4$ to find its temperature.

So there is a lot of multiplying and dividing by areas, and lots of factors of $\pi$ will cancel when you string it all together. Note one can modify these steps to take additional complexities into account (the $R$ and $A$ of your problem, for example).

Further, note that I used the planet's radius - twice in fact. And if you go through the arithmetic, you should find that it cancels itself. Intuitively, you might expect that an object's temperature (an average thermal energy) only depends on the strength of the heat source and its distance, but not on the object's size. Both you and your pet hamster2 get to about the same temperature sitting at equal distances from the fireplace.

1 Be careful about these two areas. You have to think about what area is appropriate where.

2 I have no idea how I came up with this example.

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Thanks! I think I understand how they got their results now :) By the first area...do you mean the "receiving" planet's surface? –  Longeyes Jan 11 '13 at 10:40
@Longeyes Yes, but there's a difference between the cross-sectional area $\pi r^2$ and the total surface area $4 \pi r^2$. –  Chris White Jan 11 '13 at 15:57

The $T^4$ term gives you the power per unit area emitted at the surface of the sun. The total area must then be multiplied by the total surface area of the sun, which is the numerator. This total power is then spread across a sphere with radius $r_e$, whose area is now $4\pi r_e$. This then gives you the total power per unit area received on Earth's orbit.

Now, you need to multiply that power density, or flux, but the effective solar collecting area of the Earth, which will be somewhat proportional to $\pi r_e^2$, which is the area of the Earth as if it were a flat circle. The effective area will be less than that because the of the spherical shape of the Earth and the atmosphere through which the solar load must pass.

As stated, some power is reflected and some is absorbed. The "solar constant" at the Earth's orbit is ~$1360 W/m^2$ using the equation in the original post. After traveling through the atmosphere one generally accepted value for solar load at the surface is ~$1120 W/m^2$. This is what causes the actual surface to warm up.

The Earth can now be treated as a graybody radiator, since that's the only way the Earth can balance out the incoming solar load. However, T will then be the effective greybody temperature of the Earth, which is a long ways from the temperature of the Sun. When the total incoming radiation is balanced against the total radiated loss and the Earth surface temperature is solved for, you get ~300 kelvins.

In order to make that calculation easier, you can simply look at a flat plate blackbody with $1m^2$ area receiving its $1360 W/m^2$ from the sun, but radiating that energy back to a 2.7 kelvin space background. The solution to this simplified scenario is ~330 kelvins, which is close to the nominal answer. Note that radiating area is doubled for this case, since the flat plate can radiate from its two sides, but can collect only from one side.

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