The **solar luminosity**, , is a unit of luminosity (power emitted in the form of photons) conventionally used by astronomers to give the luminosities of stars. Luminosity has different meanings in several different fields of science. ...
In physics, power (symbol: P) is the amount of work done per unit of time. ...
In physics, the photon (from Greek Ï†Ï‰Ï‚, phÅs, meaning light) is the quantum of the electromagnetic field; for instance, light. ...
An astronomer or astrophysicist is a scientist whose area of research is astronomy or astrophysics. ...
The Pleiades star cluster A star is a massive body of plasma in outer space that is currently producing or has produced energy through nuclear fusion. ...
It is equal to the luminosity of the Sun, which is 3.827 × 10^{26} W, or 3.827 × 10^{33}ergs/s. The Sun is the spectral type G2V yellow star at the center of Earths solar system. ...
The watt (symbol: W) is the SI derived unit of power. ...
For other uses see Erg (disambiguation) An erg is the unit of energy in the centimetre-gram-second (CGS) system of units, symbol erg. The erg is a quite small unit, equal to equal to one gram·centimetre2/second2. ...
## Calculating with this constant
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.) |