

Luminosity and Sunlight
Name: Donna Schakelaar
Status: other
Age: 30s
Location: N/A
Country: N/A
Date: 1999  2000
Question:
What are the relative intensities at sunrise, sunrise +
10 minutes,
sunrise + 30 minutes, and noon, and sunset? How would altitude and
geographical
location affect this e.g. between Darwin, Australia and
Melbourne,Australia?
Are there any other factors that affect the intensity of the sunlight?
Replies:
Unfortunately, I don't know a simple mathematical formula to get these
numbers. So, I can't answer your question quantitatively, as you would
want. I can only repond in generalizations.
The intensity of sunlight during any given day is pretty close to constant:
as long as the sun is up, its intensity depends mostly on the day of the
year. It rises sharply at sunrise and drops sharply at sunset, and in
between it curves gently. A plot of the solar intensity versus time looks
a bit like an upsidedown "U", somewhat squared off. The main factor
influencing solar intensity is the length of atmosphere through which the
sunlight has to travel. If the sun were directly overhead, such as at noon
on an equinox at the equator, the sunlight is at its most intense, as it
only needs to penetrate one atmospheric thickness. At latitudes away from
the equator, the sunlight reaches the earth's surface at a more grazing
angle, so that it needs to penetrate a greater slice of atmosphere. The
angle depends on the latitude, the time of year (more overhead in summer,
closer to the horizon in winter), and the time of day (more overhead at
noon, right at the horizon at sunrise and sunset). The lower the angle
(closer to the horizon), the more atmosphere stands between you and the
sun, and the lower the intensity of the sunlight that arrives.
Richard Barrans Jr., Ph.D.
I don't know the answer, but I know how I'd try to calculate it.
You need to make some simplifying assumptions and do a little
trigonometry. Let's say the difference in intensity is entirely
attributable to the greater distance in the atmosphere the morning and
evening sunlight must travel to get to you. I think this is a very
good assumption. You can solve for the distance variation by drawing a
circle of radius 4000 miles (the Earth's surface) and a concentric
circle of 4010 miles (the top of the atmosphere) and doing the trig.
The varying air density with height is ignorable because you're
interested in relative intensity, and because we'll assume different
air densities affect all light wavelengths in the same way (not too bad
an assumption if we're only interested in light that actually makes it
down to the surface). A quick a dirty calculation says morning
sunlight travels through 283 miles of air while noon sun travels
through 10, if you're at the equator in midsummer.
So now you can calculate the distance variation. You can calibrate
this scale to an intensityvariation scale with two numbers: the
intensity at noon at the equator in midsummer, and the intensity at
the pole in midsummer. If you can't find these numbers, you could
make a reasonable stab at them by comparing the temperatures at those
places and times (in Kelvins: Fahrenheit and Centigrade have arbitrary
zeros so they're no good for this purpose).
Altitude makes no difference under the assumptions I've made.
Geographical location makes a big difference, of course. (If you're at
the pole, there's no difference between morning and noon. I'd guess
the dependence on latitude involves nothing worse than a cosine.)
Other factors that affect the relative intensity include clouds and
mountains.
Tim Mooney
Click here to return to the Astronomy Archives
 
Update: June 2012

