Here’s a somewhat mathematical look at why it is hotter at the equator that at the poles. This is high school or basic college level science, using trigonometry (pre calculus), a slight step beyond the basic statement that the sun hits down more directly at the equator than at the poles. That’s the kid’s explanation, but we can understand better if we add a little math.

Lets use the diagram at right and trigonometry to compare the amount of sun-energy that falls on a square meter of land at the equator (0 latitude) and in a city at 42.5 N latitude (Detroit, Boston, and Rome are at this latitude). In each case, let’s consider high-noon on March 21 or September 20. These are the two equinox days, the only days each year when the day and night are equal length, and the only times when it is easy to calculate the angle of the sun as it deviates from the vertical by exactly the latitude on the days and times.

More specifically the equator is zero latitude, so on the equator at high-noon on the equinox, the sun will shine from directly overhead, or 0° from the vertical. Since the sun’s power in space is 1050 W/m2, every square meter of equator can expect to receive 1050 W of sun-energy, less the amount reflected off clouds and dust, or scattered off or air molecules (air scattering is what makes the sky blue). Further north, Detroit, Boston, and Rome sit at 42.5 latitude. At noon on March 31 the sun will strike earth at 42.5° from the vertical as shown in the lower figure above. From trigonometry, you can see that each square meter of these cities will receive cos 42.5 as much power as a square meter at the equator, except for any difference in clouds, dust, etc. Without clouds etc. that would be 1050 cos 42.5 = 774 W. Less sun power hits per square meter because each square meter is tilted. Earlier and later in the day, each spot will get less sunlight than at noon, but the proportion is the same, at least on one of the equinox days.

To calculate the likely temperature in Detroit, Boston, or Rome, I will use a simple energy balance. Ignoring heat storage in the earth for now, we will say that the heat in equals the heat out. We now ignore heat transfer by way of winds and rain, and approximate to say that the heat out leaves by black-body radiation alone, radiating into the extreme cold of space. This is not a very bad approximation since Black body radiation is the main temperature removal mechanism in most situations where large distances are involved. I’ve discussed black body radiation previously; the amount of energy radiated is proportional to luminosity, and to T^{4,} where T is the temperature as measured in an absolute temperature scale, Kelvin or Rankin. Based on this, and assuming that the luminosity of the earth is the same in Detroit as at the equator,

T_{ Detroit }/ T_{equator } = √√ cos 42.5 = .927

I’ll now calculate the actual temperatures. For American convenience, I’ll choose to calculation in the Rankin Temperature scale, the absolute Fahrenheit scale. In this scale, 100°F = 560°R, 0°F = 460°R, and the temperature of space is 0°R as a good approximation. If the average temperature of the equator = 100°F = 38°C = 560°R, we calculate that the average temperature of Detroit, Boston, or Rome will be about .927 x 560 = 519°R = 59°F (15°C). This is not a bad prediction, given the assumptions. We can expect the temperature will be somewhat lower at night as there is no light, but the temperature will not fall to zero as there is retained heat from the day. The same reason, retained heat, explains why it is warmer will be warmer in these cities on September 20 than on March 31.

In the summer, these cities will be warmer because they are in the northern hemisphere, and the north pole is tilted 23°. At the height of summer (June 21) at high noon, the sun will shine on Detroit at an angle of 42.5 – 23° = 19.5° from the vertical. The difference in angle is why these cities are warmer on that day than on March 21. The equator will be cooler on that day (June 21) than on March 21 since the sun’s rays will strike the equator at 23° from the vertical on that day. These temperature differences are behind the formation of tornadoes and hurricanes, with a tornado season in the US centering on May to July.

When looking at the poles, we find a curious problem in guessing what the average temperature will be. At noon on the equinox, the sun comes in horizontally, or at 90°from the vertical. We thus expect there is no warming power at all this day, and none for the six months of winter either. At first glance, you’d think the temperature at the poles would be zero, at least for six months of the year. It isn’t zero because there is retained heat from the summer, but still it makes for a more-difficult calculation.

To figure an average temperature of the poles, lets remember that during the 6 month summer the sun shines for 24 hours per day, and that the angle of the sun will be as high as 23° from the horizon, or 67° from the vertical for all 24 hours. Let’s assume that the retained heat from the summer is what keeps the temperature from falling too low in the winter and calculate the temperature at an .

Let’s assume that the sun comes in at the equivalent of 25° for the sun during the 6 month “day” of the polar summer. I don’t look at equinox, but rather the solar day, and note that the heating angle stays fixed through each 24 hour day during the summer, and does not decrease in the morning or as the afternoon wears on. Based on this angle, we expect that

T_{Pole }/ T_{equator } = √√ cos 65° = .806

T_{Pole }= .806 x 560°R = 452°R = -8°F (-22°C).

This, as it happens is 4° colder than the average temperature at the north pole, but not bad, given the assumptions. Maybe winds and water currents account for the difference. Of course there is a large temperature difference at the pole between the fall equinox and the spring equinox, but that’s to be expected. The average is, -4°F, about the temperature at night in Detroit in the winter.

One last thing, one that might be unexpected, is that temperature at the south pole is lower than at the north pole, on average -44°F. The main reason for this is that the snow on south pole is quite deep — more than 1 1/2 miles deep, with some rock underneath. As I showed elsewhere, we expect that, temperatures are lower at high altitude. Data collected from cores through the 1 1/2 mile deep snow suggest (to me) chaotic temperature change, with long ice ages, and brief (6000 year) periods of warm. The ice ages seem far worse than global warming.

Dr. Robert Buxbaum, December 30, 2017