Off Topic — HabitableZone

Off Topic » in reply to re: E=M*(C^2) question?

An example problem
ER — (archived)

Here is a numerical application of your two equations in action. All raw data is from "Astronomical and Physical Data" in the Royal Astronomical Society of Canada Observer's Handbook, 2006 editition. All calculations are rounded off to three significant digits.

The solar intensity is 1.370 x 10^3 Watts per square meter, that is, every square meter (at the top of the atmosphere) normal to the path of light from sun to earth, passes 1370 Joules or radiant solar energy every second. The sun's distance from earth (an Astronomical Unit) is approximately 1.496 x 10^11 meters.

Plugging these numbers into your first equation and solving for the total solar power output

P = 1.370 x 10^3 x 4 x Pi x (1.496 x 10^11)^2
or...
P = 3.853 x 10^26 Watts
That is the number of Joules of energy the sun produces every second

The speed of light is approx 2.998 x 10^8 m/s
Substituting these values into the Einstein equation and solving for mass

M = (3.853 x 10^26)/( (2.998 x 10^8)^2 )
or
M = 1.285 x 10^10 kg

In other words, every single second the sun converts to energy
12,850,000 metric tonnes of matter.

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