Added to Favorites

Related Searches

Nearby Words

solar constant, the average amount of radiant energy received by the earth's atmosphere from the sun; its value is about 2 calories per min incident on each square centimeter of the upper atmosphere. The actual value of the energy varies with several factors; the most important factor is the earth's distance from the sun, which changes because of the earth's elliptical orbit. For computing the value of the solar constant, the astronomical unit, or average earth-sun distance, is used.

The Columbia Electronic Encyclopedia Copyright © 2004.

Licensed from Columbia University Press

Licensed from Columbia University Press

The solar luminosity, $L\_odot$, is a unit of luminosity (power emitted in the form of photons) conventionally used by astronomers to give the luminosities of stars.
It is equal to the current accepted luminosity of the Sun, which is 3.839 × 10^{26} W, or 3.839 × 10^{33}erg/s.
Note that the Sun is a weak variable star and its luminosity therefore fluctuates.
## Calculating with this constant

## References

You can calculate how much solar power hits the Earth by comparing a cross sectional area of the Earth and the total surface area of a sphere with a radius equal to the distance of the earth from the sun.

- The Earth's radius is 3963 miles (6,378 km).
- The Earth's cross sectional area = π×radius
^{2}= 49.3 million square miles (128,000,000 km²). - The Sun's average distance is about 93 million miles (150,000,000 km).
- The surface area of a sphere = 4×π×radius
^{2}= 1.09×10^{17}square miles (2.82×10^{17}km²). - Power reaching the Earth = P(total) × Area(earth)/Area(sphere) = 1.77×10
^{17}W. - The power hitting a square meter of area on Earth: (square meter = 1/1609
^{2}square miles) - Power over square meter = P(total)(1/1609
^{2})/area(sphere) = 1387 W (the solar constant) - Estimates have been made that humans use about 12×10
^{12}W. - How much land area would be needed to power that?
- The best solar cells can produce about 33% efficiency.
- Area needed = 12×10
^{12}/(1387×0.33) = 26×10^{9}m^{2}= 10122 square miles ~100×100 mile square. (More is needed since the sun is not always straight over head, and because some fraction of the radiation does not reach the surface due to clouds and atmospheric scattering.)

- I.-J. Sackmann, A. I. Boothroyd (2003). "Our Sun. V. A Bright Young Sun Consistent with Helioseismology and Warm Temperatures on Ancient Earth and Mars".
*The Astrophysical Journal*583 (2): 1024–1039.

*

Wikipedia, the free encyclopedia © 2001-2006 Wikipedia contributors (Disclaimer)

This article is licensed under the GNU Free Documentation License.

Last updated on Saturday August 30, 2008 at 16:11:31 PDT (GMT -0700)

View this article at Wikipedia.org - Edit this article at Wikipedia.org - Donate to the Wikimedia Foundation

This article is licensed under the GNU Free Documentation License.

Last updated on Saturday August 30, 2008 at 16:11:31 PDT (GMT -0700)

View this article at Wikipedia.org - Edit this article at Wikipedia.org - Donate to the Wikimedia Foundation

Copyright © 2015 Dictionary.com, LLC. All rights reserved.