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. It is equal to the current accepted luminosity of the Sun, which is 3.839 × 10²⁶ W, or 3.839 × 10³³erg/s.¹ Note that the Sun is a weak variable star and its luminosity therefore fluctuates.

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×radius² = 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×radius² = 1.09×10¹⁷ square miles (2.82×10¹⁷ km²).
• Power reaching the Earth = P(total) × Area(earth)/Area(sphere) = 1.77×10¹⁷ W.
• The power hitting a square meter of area on Earth: (square meter = 1/1609² square miles)
  • Power over square meter = P(total)(1/1609²)/area(sphere) = 1387 W (the solar constant)
• Estimates have been made that humans use about 12×10¹² W.
• How much land area would be needed to power that?
  • The best solar cells can produce about 33% efficiency.
  • Area needed = 12×10¹²/(1387×0.33) = 26×10⁹ m² = 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.)

References

• 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. doi:10.1086/345408, http://adsabs.harvard.edu/abs/2003ApJ...583.1024S. 

• Carroll, Bradley W.; Dale A. Ostlie (2007). An Introduction to Modern Astrophysics. Pearson Addison Wesley, Appendix A. ISBN 0-8053-0402-9.