Calculating the solar power hitting Earth

I'm trying to calculate the amount of power from the sun hitting the earth, but I am getting a number which is off by a factor of ~4.

I calculate the "area" of the earth, as seen from the sun, and then divide that by the surface area of a sphere of radius 1 AU, to get the portion of the sun's rays we absorb, and then I multiply that by the solar luminosity.

Mathematica gives me this: But when I put it into Wolfram|alpha, I get this result: Am I doing something wrong? where could that big an error be introduced? Is "mean power intercepted by earth" different than what I'm calculating?

• you made a simple mistake calculating cross section of the earth as piD². should be pir². this error accounts for exactly 4x. – gregsan Dec 4 '13 at 15:32

The area of a circle is calculated using its radius instead of its diameter: $$A = \pi r^2 = \pi \left(\frac{d}{2}\right)^2 = \pi \frac{d^2}{4}$$ which is where your missing factor of 4 reappears.

As pointed out in the comments, the method is correct aside from a careless error of using the diameter, when radius was clearly intended to be used by the argument.

I calculate the "area" of the earth, as seen from the sun, and then divide that by the surface area of a sphere of radius 1 AU, to get the portion of the sun's rays we absorb, and then I multiply that by the solar luminosity.

For a sanity check, I will propose another way to get the same thing. The sun is roughly a black body, but you didn't need this info since you started with the Luminosity. You can recalculate that value approximately as follows.

$$L_{\circ} = \sigma T^4 A_s$$

Take out the area from this expression. Now you have $W/m^2$, but at the surface of the sun. Convert this to the average solar insolation at Earth's location, you can simply multiply by the ratio of areas of those two spheres. That is, multiply the sun's $W/m^2$ value by the area of the sun's surface divided by the area of the 1 AU sphere to get the $W/m^2$ at 1 AU. The ratio of areas of two shapes will be the ratios of linear dimensions, squared. It doesn't matter which.

Returning to your objective quantity, that is simply the area of Earth presented to the sun times the $W/m^2$ at 1 AU.

$$P = \frac{ L_{\circ} }{ A_{AU} } A_{Earth} = \sigma T^4 \left( \frac{ 1 AU }{ R_{s} } \right)^2 A_{Earth}$$

I can't say this would have caught your problem, but it's a useful sanity check in general. It avoids comparing $4 \pi r^2$s against $\pi r^2$s, which gets a little bit dizzying. Here is the calculation on Wolram Alpha. It produces $1.74 \times 10^{17} W$, which might be different only within the significant figures. It's probably close enough to validate that this is another correct calculation.

• Actually, I didn't bother checking against the OP's calculation. Dividing by 4, that number gives 1.74, so these match perfectly. I was correct that our sun is a blackbody! – Alan Rominger Dec 4 '13 at 16:15

Let's go about this from a different angle and compare answers. The radius of the earth is about 6.4 Mm, so the area of its disk is 130 x 1012 m2. Figure about 1.2 kW/m2 of incident sunlight power at earth's distance, so that yields 1.5 x 1017 Watts. That's close enough for such a quick back of the envelope calculation to the 2 x 1017 value you show that it can be considered the same answer. The 1.2 kW/m2 figure I used is probably low.

However, what this answer really means is much harder to decide. You certainly can't assume all that power is heating the planet, since much of that will be reflected, especially at glancing angles.

• I think the coefficient you are looking for is the bond albedo which is equal to 0.29 for Earth according to this source, but according to the source linked by wikipedia it is equal to 0.306. – fibonatic Dec 4 '13 at 15:52