12
$\begingroup$

Recently I heard that in winter season the distance between the Earth and the Sun will be closest and it is farthest in summer. If it is true then why is it hotter in summer and cooler in winter?

$\endgroup$
  • 1
    $\begingroup$ I assume you mean for the Northern Hemisphere - here in the Southern Hemisphere, January is in the middle of the summer season. $\endgroup$ – user889 Jan 3 '15 at 13:36
  • $\begingroup$ yes im in Northen Hemisphere, here December to mid March is Winter season $\endgroup$ – Hari-pacific Jan 3 '15 at 13:39
  • 1
    $\begingroup$ True. However, this question may be more appropriate for Astronomy.SE. $\endgroup$ – Gimelist Jan 3 '15 at 14:24
  • 2
    $\begingroup$ Asides from David Hammen's excellent answer below, the following document may also be of help luc.edu/faculty/dslavsk/courses/phys478/classnotes/… $\endgroup$ – user889 Jan 3 '15 at 14:28
18
$\begingroup$

Yes, it's true (in the northern hemisphere). The small eccentricity of the Earth's orbit is not anywhere close to a key driver in the seasons. The key driver of the seasons is the Earth's obliquity. In the northern hemisphere, the axial tilt of Earth's rotation axis has the northern half of the Earth facing a bit toward the Sun in June/July/August and away from the Sun in December/January/February. The opposite is true in the southern hemisphere.

Eccentricity would be a driver of the seasons if the Earth's rotation and orbital axes were much closer in line with one another than they are. If that were the case, summer and winter would be world-wide phenomena. As it stands, when its summertime in the northern hemisphere its wintertime in the southern hemisphere, and vice versa.

Somewhat paradoxically, even though the Earth is closest to the Sun in early January and furthest from the Sun in early July, the Earth as a whole is cooler during December/January/February than it is during June/July/August. The reason is the uneven distribution of land and ocean between the northern and southern hemispheres.

In 13000 years, northern hemisphere summers will occur near perihelion passage and winters near aphelion passage. Those will be brutal times! Fortunately, I won't be around to see them.

$\endgroup$
  • $\begingroup$ @Sabre Tooth: I think that has more to do with the fact that Melbourne a) isn't really all that far south, and b) is in Australia. People in Patagonia, the South Island of New Zealand, or even Tasmania experience rather different climates. $\endgroup$ – jamesqf Jan 3 '15 at 18:18
6
$\begingroup$

David's answer is correct, but I wanted to elaborate a bit further.

The Earth's axis is tilted about $23.5^\circ$ from being perpendicular to the plane of its orbit (as David mentioned, 'obliquity' or 'axial tilt' is the name for this.) As a result of this tilt, at the solstices (the first day of winter or summer,) the sun is directly facing latitudes of either $23.5^\circ$ North (the first day of summer in the Northern hemisphere) or $23.5^\circ$ South (the first day of summer in the Southern hemisphere.) Areas between these two latitudes are known as the tropics, where the sun will be directly overhead at some point during the year.

For latitudes farther away from the equator than the tropics, while the sun will never be directly overhead, the first day of summer in a given hemisphere will be the day in which the incident angle of the sun is the most shallow (i.e. the sun is as close as it will come to being directly overhead) and the first day of winter is the day in which the incident angle will be the most steep (i.e. the sun will be the farthest from passing directly overhead.) For example, at a latitude of $40^\circ$, on the first day of summer, the sun will be only $16.5^\circ$ away from being directly overhead, while, on the first day of winter, it will be $63.5^\circ$ away from being directly overhead. Solar power density for areas directly under the sun is about $1.4\ \mathrm{kW/m^2}$, so this gives us:

Solar power density at noon on the first day of summer at 40 degrees latitude:
$$1.4\ \mathrm{kW/m^2} * \cos(16.5^\circ) = 1.4\ \mathrm{kW/m^2} * 0.959 = 1.34\ \mathrm{kW/m^2}$$

And for winter:
$$1.4\ \mathrm{kW/m^2} * \cos(63.5^\circ) = 1.4\ \mathrm{kW/m^2} * 0.446 = 0.62\ \mathrm{kW/m^2}$$

Additionally, the sun will spend much more time above the horizon on the first day of summer than on the first day of winter, again, due to the axial tilt of Earth. The combination of these two effects results in the dramatic difference in average solar heating observed between summer and winter in latitudes that are distant from the equator. This local difference in solar heating through the year is the dominant factor in determining seasons as the Earth orbits the sun, not the comparatively minor differences in actual distance between the Earth and the sun.

For a graphical comparison, take a look at the solar power density measurement graphs on these pages, from a weather station located about $36.2^\circ$ North:

December 20, 2014 (a day before the solstice because the actual solstice was cloudy)

June 22, 2014 (again, a day off the solstice, due to clouds on the solstice.)

I would copy the graphs into this post, but they're copyrighted by that site, so you'll have to visit the links and scroll down to the power density graphs. Note that apparent noise in the graphs is primarily due to clouds briefly passing over.

The areas under these curves (i.e. total solar energy per unit area per day) are roughly $2.4\ \mathrm{kW\ hr/m^2}$ for the beginning of winter and $7.3\ \mathrm{kW\ hr/m^2}$ for the beginning of summer. At least, that's about what they would have been had there been no clouds in the way.

$\endgroup$
  • $\begingroup$ With all these calculations I was surprised not to see a comparison of the size of the two effects! The eccentricity effect is about 7%, as compared to the factor of two of summer vs. winter (at 40 degrees). See e.g. space.com/26421-earth-farthest-from-sun-2014.html $\endgroup$ – Rex Kerr Jan 4 '15 at 11:08
  • $\begingroup$ I thought that 1.4kW/m2 was for edge of the Earth's atmosphere; isn't it more like 1kW/m2 at the surface? $\endgroup$ – EnergyNumbers Jan 5 '15 at 19:19
  • $\begingroup$ @EnergyNumbers I think you're right that the 1.4 kW/m^2 number is at the edge of the atmosphere. However, as you can see at the link to the June data above, it hits around 1,100 W/m^2 at the surface even at 36.2° North. Btw, I just noticed that I accidentally typed 31.2° in the answer. I'll fix that. $\endgroup$ – reirab Jan 5 '15 at 21:11
  • $\begingroup$ Excellent comparisons of the extraterrestrial irradiances! $\endgroup$ – user889 Jan 5 '15 at 21:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.