* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download Brightness Luminosity and Inverse Square Law
History of astronomy wikipedia , lookup
Copernican heliocentrism wikipedia , lookup
Equation of time wikipedia , lookup
Rare Earth hypothesis wikipedia , lookup
History of Solar System formation and evolution hypotheses wikipedia , lookup
Astronomical spectroscopy wikipedia , lookup
Solar System wikipedia , lookup
Geocentric model wikipedia , lookup
Formation and evolution of the Solar System wikipedia , lookup
Observational astronomy wikipedia , lookup
Tropical year wikipedia , lookup
Perseus (constellation) wikipedia , lookup
Aquarius (constellation) wikipedia , lookup
Corvus (constellation) wikipedia , lookup
Dialogue Concerning the Two Chief World Systems wikipedia , lookup
Timeline of astronomy wikipedia , lookup
Why do the stars look as bright as they do? the nature box Why do the stars look as bright as they do? Consider two light bulbs. 30 Watts 120 Watts 30 Watts vs. 120 Watts If you wanted the bulbs to appear equally bright………how much farther away would you place the 120 Watt bulb? Nope, not 4 times further. TWO TIMES FURTHER AWAY. Brightness follows the inverse square law! The Inverse Square Law! Apparent brightness (b) changes with the inverse square of the distance (d). b α 1/d2 At 2x the distance, brightness is 1/22 or 1/4th At 3x the distance, brightness is 1/32 or 1/9th How would the brightness change if you increased your distance from a star by 10 times? brightness “b” and Luminosity “L” Luminosity is the Power output of a star in Watts (1 Watt = 1 Joule/second) brightness as viewed from a particular location is the Power received per unit area, in W/m2 b= L 4πd2 How to use the inverse square law for brightness of stars… We know that the apparent brightness “b” of our Sun as viewed from Earth is 1362 W/m2 We also know that Saturn is 9.7 times further away from our Sun than the Earth. If b α 1/d2 , then as viewed from Saturn, the sun would appear 1/(9.7)2 or 1/94th as bright. The brightness would then be 1362 W/m2(1/94) = 14.5 W/m2 Exit slip! We know our “Solar Constant” (the apparent brightness of our Sun from Earth) is 1362 W/m2 Neptune is 30 times the distance from our Sun than is the Earth. Calculate, using the inverse square law, the apparent brightness of the Sun as viewed from Neptune.