A Lesson in Basic Geometry

June 28, 2013

28 Jun (PRINCIPIA SCIENTIFIC) This is the best rendering of a ball that I can do on a computer. It has a single source of light and I’ve made it slightly gibbous (more than half-lit to the eye) in order to emphasize its 3-dimensionality. What’s observable about this ball is elementary but vital.

by Alan Siddons

ball

First of all, only half of it can be illuminated at any one time. Secondly, most light falling onto it falls obliquely; at only one point is the surface perpendicular to the light source, thus receiving the maximum amount of energy.

Now, it bears repeating that a 2-dimensional disk has four times less surface area than a sphere of the same diameter. Perpendicular to a beam of light, though, a disk’s surface is able to absorb the full intensity. A sphere, by contrast, absorbs the same amount but that amount is spread over a larger area, thus diluting it. Only one spot on the sphere is perpendicular to the source.

graph

This has a direct impact on the temperature the two surfaces can reach. A blackbody temperature equation for the disk goes like this.

Kelvin = (P ÷ 5.67)0.25 × 100.

Where P is the power of the beam impinging on the disk. Let’s make it 1000 W/m² here. Ergo,

(1000 ÷ 5.67)0.25 × 100 = 364.42

So 364.42 Kelvin — or 91.27° Celsius — is the highest temperature the disk can reach. Notice another simple thing: the average and the peak temperature on a disk are identical, for the disk is receiving the same amount of energy everywhere.
The temperature equation for a sphere requires but one adjustment. Since we know that the radiant power is going to be diluted 4 times, we divide the initial 1000 W/m² by 4. Ergo,

(250 ÷ 5.67)0.25 × 100 = 257.69

So 257.69 Kelvin — or minus 15.46° Celsius — is the highest AVERAGE temperature the disk can reach. But in this case the average and the peak temperature are not the same, for the sphere is not receiving the same amount of energy everywhere.

It is crucial to understand this distinction. Only one point on a sphere can face radiant energy perpendicularly. For that reason, only this single point can reach the temperature of a perpendicular disk. There is a simple way to quantify that temperature. Once you have determined the sphere’s average temperature in Kelvin, multiply it by the square root of two. Ergo,

257.69 × SQRT2 = 364.42

In other words, the sphere’s peak temperature and the disk’s temperature are identical.

Let’s test these equations in a more real-life scenario. We will adopt NASA’s figure of a 1370 W/m² solar constant and have this fall onto the earth’s moon, a sphere whose albedo is given as 0.07, thus an absorbency of 93%. So we divide radiance by four: 342.5, and multiply 342.5 by 0.93 to correct for reflection losses, obtaining 318.53 W/m². Ergo,

Average Kelvin = (318.53 ÷ 5.67)0.25 × 100 = 273.77

Now to determine the peak temperature on that spherical surface, multiply average Kelvin by the square root of two.

273.77 × SQRT2 = 387.17 K

Alternately, going for the peak temperature alone, 1370 W/m² × 0.93 = 1274.10 W/m² absorbed. So,

Disk Kelvin = (1274.10 ÷ 5.67)0.25 × 100 = 387.17 K

The same.

Here’s how NASA handles the problem.

For slowly rotating planets like Mercury and the Moon, one must take into account that these bodies receive energy over their projected (disk) areas and emit energy, not over their full spherical surface areas but only over the same projected areas because the remaining surface area is considered to be too cold to radiate a significant amount of energy back to space. For such bodies, the thermal equilibrium is thus established when…

math

… where, as before, the Sun-Earth distance d is expressed in AU. A comparison of equations (10) and (13a) shows that for slowly rotating planets, the equilibrium temperature is higher by a factor equivalent to the fourth root of the projected area (i.e., the ratio of sphere surface-to-disk area), namely the fourth root of 4. Apply equation (13a), for d = 1 AU, ε = 1, and A = 0.07, to Earth’s moon to obtain…

math2

… which is the maximum temperature at the lunar equator at noon.

http://gltrs.grc.nasa.gov/reports/2001/TM-2001-210063.pdf

387 K. The same.

By the way, notice the 394 above? That would be the peak Kelvin temperature if the 0.07 albedo loss weren’t factored in.

Disk Kelvin = (1370÷ 5.67)0.25 × 100 = 394.26 K

Divide that by the square root of two and you have the average temperature of a perfectly-absorbing sphere, 278.78 Kelvin.

Summary

I have shown you the thermal aspects of a ball absorbing radiant energy and have done this with blackbody equations simplified to the nth degree by omitting needlessly repetitive expressions. As you see, the results of my methods correspond to those of much more elaborate mathematical exercises.

The reason for doing this is to give you a solid grounding and preparation for material to follow, because what I’ve presented here is the principle of method for testing whether planet earth shows any sign of radiative forcing from greenhouse gases.

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