Built on Facts

Reader Timothy writes in with a question:

It seems that the north pole would get more daylight during the summer solstice than the equator does during an equinox. During the summer solstice, the north pole would get the equivalent of (24hrs)sin(23.5 ° ) = 9.57 hours of sunlight from directly overhead (the zenith). During the equinox, the equator would get the equivalent of (12 hrs / pi radians) ∫ sin θ d θ (integrating from zero to pi) = 7.64 hours of sunlight from the zenith. We only integrate from zero to pi, when the sun is above the horizon.

I am curious as the reasons why the north pole would be colder even around the summer solstice than the equator around an equinox. I presume factors would include heat capacity of water and ice, albedo of ice, wind patterns, attenuation of light in the atmosphere, etc.

In other words, in the perpetual daylight of an Alaskan (or polar) summer, the total energy being pumped into each square meter is pretty hefty considering there’s no night to reduce the average. But on the equator with the sun directly overhead at noon, the sunlight will be diluted by the fact that not only is the sun’s angle constantly changing but for fully half of the 24-hour day there’s no light at all.

But pretty clearly the Equatorial spring is hotter than the polar summer. What gives?

Comments

#1 beebeeo
April 21, 2009

I can up with 2 reasons.

- reflected sunlight by the white ice

- light from the sun has to pass through more atmosphere and is dispersed
#2 JBL
April 21, 2009

Albedo? The heat capacity of the Arctic Ocean? (Just guessing here — the first certainly contributes, but probably not enough.)
#3 rob
April 21, 2009

beebeeo is correct, however those are secondary effects.

the main effect is that the net energy flux from the sun is less due to the angle of the earths tilt.

Timothy has the right idea that the earth’s tilt will effect the temperatures at the pole and equator, but he has made a mathematical error by multiplying the length of the day by the sine of an angle.

since time is a scalar, is does not make mathematical sense to multiply it by sin(23.5). this is treating time as a vector quantity by finding it’s component perpendicular to the earths surface.

what he really needs to look at is the solar constant, which is the average energy flux reaching earth from the sun. since flux is a vector, you can find the component that contributes to heating the earth.

you should find that the total energy reaching the earths surface at the pole during the summer is less than the amount reaching the equator, even though the sun is up 24 hours.
#4 Michael Vacirca
April 21, 2009

How about surface area? I would believe the equator would have a much larger amount of surface area directly under sunlight at any time do to the angle of the equator.

http://www.hko.gov.hk/education/edu06nature/ele_srad_e.htm
#5 Joe Shelby
April 21, 2009

Another reason: it is already so cold at the poles that it takes a LOT more heat than one might think to really warm it up. When combined with the reflecting ice, it makes it take a LOT more energy to really make a difference. It’s the same reason why, on a smaller scale, cities on bays and lakes are more constant in temperature than more inland cities: the cold water makes a difference, especially when it really can get cold during the other end, the season when there’s no sun at all.

At the equator, things are generally constant until an outside disturbance (like an el nino from the south pole) gets involved. The sun may only be shining 12 hours on the equinoxes, but it’s not that much less than that at the solstices.
#6 Eric Lund
April 21, 2009

Finite heat capacity makes a difference. The equator gets 12 hours of sunlight a day (modified for solar zenith angle) throughout the year, so it’s already pretty hot when the equinox comes around. The poles get zero solar input for half of the year, so they get quite cold by the time the spring equinox arrives, and only then do they warm up. That’s why the hottest days of summer at northern midlatitudes typically come in July and August, not June.

There is still an effect. In New Hampshire, where I live now, we sometimes (though thankfully not very often) see temperatures above 100F, especially away from the coast. Miami, where I grew up, will often see temperatures in the upper 90s, but I don’t think they’ve ever gone above 100F–at least, they never did in the time from when they started keeping records to when I left to go to college. Temperatures sometimes reach the 90s even at arctic or subarctic latitudes: I personally have experienced 95F in Uppsala, Sweden (latitude ~59N) in August, and I know that Fairbanks, Alaska (latitude ~65N) sees summer temperatures that high.
#7 Matt Springer
April 21, 2009

“since time is a scalar, is does not make mathematical sense to multiply it by sin(23.5). this is treating time as a vector quantity by finding it’s component perpendicular to the earths surface.”

He’s taking time to be a shorthand for time-integrated flux equivalent. I.e., the total absolbed by a solar panel on the ground over the course of a day would be the same as if the sun were directly overhead for (say) 9.57 hours.
#8 rob
April 21, 2009

ah, i see what you mean Matt about the average.

then, in that case, i think he is mixing up the angles in his average.
#9 Tim Gaede
April 21, 2009

Thanks, Matt. I thought this was a curious back-of-the-envelope demonstration and therefore perhaps appropriate for your blog.

I actually wrote up a computer program that performs such “integrations” (using rotation of vectors in 3D space and such) and the results were consistent with the two calculations I presented. The program indicates that even at 70 degrees latitude the sunlight exposure would be about 9 zenith-hours during the solstice. This of course doesn’t take into account atmospheric attenuation, which I suspect plays a significant role. I’m guessing that solar cells laid flat would not absorb that much energy during the day.

I think climate modelers have their hands full.
#10 Eric Lund
April 21, 2009

It was an interesting question, Tim. Thanks for suggesting it.

I think what you are overlooking here is that you want to integrate over an entire year, and compare that result with the annual average temperature. Let ψ be the tilt angle of the planet (23.5 degrees for Earth). For the poles the number is proportional to the integral from 0 to π of dθ sin θ sin ψ. For the equator the modification would be a bit more complicated but should be straightforward to derive with a pencil and paper. I know some people have done the calculation (though I have not done so myself). IIRC for Earth the equator is indeed a global maximum. But that is not true for all values of ψ in particular the maximum moves off the equator by the time ψ reaches 35 degrees.

The comparison will not be perfect, since there are other factors which can affect climate; e.g., due to the Gulf Stream, England is much warmer than Labrador even though both are at the same latitude. But for a given longitude, and holding elevation constant, you should find that the annual average temperature is correlated with the integrated annual insolation. As explained above, the temperature variations during the year will tend to lag the variations in insolation, especially at high latitudes, due to finite heat capacity.
#11 dWj
April 21, 2009

Note that the day/night temperature difference on the moon is a lot larger than that on earth. One reason is the length of the day, but the atmosphere is a big deal, too. It’s not so much dispersion at heat capacity, as I understand it. But essentially the answer does in fact lie with smoothing over seasons. (Not that albedo is insignificant.)
#12 percy
April 21, 2009

This question takes me back to my weather classes and global atmospheric circulation. The polar regions are high pressure. That is, the air at the polar surface is decending from the upper atmosphere and is extremely cold and very dry. It is tough to heat things up when the air around you was at -40 degrees C yesterday.
#13 Uncle Al
April 21, 2009

The (northern hemisphere) summer solstice is 21 June but the hottest days are in August (the only month without holidays – everybody is on vacation). The winter solstice is 21 December (sorry, Christianity) but the coldest days are in February. Factor in phase lag, ground solar constant, and where the heat goes (e.g., 0.2 cal/g-K for warming soil, 1 cal/gram-K for warming water, 80 cal/gram-K for melting ice to 0 C water).

However… if you grow crops, a greenhouse above the Arctic Circle is magic. The long insolation grows purely huge fruits and vegetables.
#14 Ahcuah
April 21, 2009

I’d say it’s the latent heat of all the soil and water.

Don’t forget that, even though the sun is highest at local noon (around 1:30pm here in Central Ohio on EDT), the hottest part of the day is closer to 3:30-5:00. That’s because at 1:30
the surrounding environment is still sucking up the heat. By
2:30 or so, some of that heat is being radiated back out, and it takes until 3:30 or later for the sum to start decreasing.

Also, note that the hottest day of the year falls pretty
close to July 21, a month after the solstice (northern
hemisphere). Again, it’s the latent heat in the soil
and water.
#15 InkRose
April 21, 2009

Dunno how much of a difference this would make in the calculations, but the Earth is at its farthest distance from the sun during the northern hemisphere’s summer, though iirc tilted ‘towards’ the Sun, while in winter the Earth is closest to the Sun on its orbit, but the North is tilted ‘away’ (also don’t know if there’s any deviation from the ecliptic to affect these things in some way).

A quick search suggests the effects of perihelion/aphelion are pretty well negligible in comparison to those brought on by the angle of tilt. So nevermind, then. I just know the main reason the Isle of Man is much, much more temperate (and much, much more rainy) than my native Finland is the Gulf Stream.
#16 CS
April 22, 2009

I think beebeeo is correct: attenuation by the atmosphere. Radiation from the Sun has to travel through more than twice the amount of atmosphere at 23° from the horizon versus straight overhead. The atmosphere does a fairly good job of absorbing some of that radiation, which is why you can stare at a setting Sun, but not at the Sun when it is overhead. Based on this article, only about 1/3 as much light makes it through at 23° versus 90°. This is on top of the sin term from the flux angle.
#17 Jane
April 22, 2009

The sun’s rays hit the equator almost perpendicularly and the poles obliquely, so sunlight is more intense at the equator.
#18 Bob Sykes
April 23, 2009

Michael Vacirca is real close. It’s the cosine latitude thingy. What counts is the angle between the sun’s rays and the land. At the equator, on average the rays are perpendicular to the land, and the area intercepting the rays is at its smallest. At the poles, on average the rays are parallel to the land surface, and the effective area intercepting the rays is infinite: no heating.
#19 Tim Gaede
April 23, 2009

Apparently some posters did not see that the angle of the sun’s rays is integral to the statement of the problem. In the absence of atmospheric attenuation, it is quickly shown that the north pole would receive more daily solar flux at the solstice than the equator at the equinox.

Thanks for posting, Matt.
#20 opony szczecin
April 26, 2009

oh yes,thats right