# Luminosity of Sun

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1. Nov 9, 2014

### mokeejoe5

I'm trying to work out the luminosity of the sun.

1630 Watts/m2 apparent magnitude reduces with distance squared so
1/(1.496*1011)2 = 4.468*10-23m2 has the same brightness 1630 Watts from earth

Surface area of the sun divided by 4.468*10-23m2 = # of 1630 Watt sections
*1630 = 2.21*1044 Watts

• #

Can someone please explain to me where I'm going wrong?

2. Nov 9, 2014

### sophiecentaur

It is not clear what you are trying to show. I think you are choosing a 'strange path' through your calculations. Try starting again.
Given the luminosity, measured on Earth, you can work out the total Power radiated. (Area of a sphere of radius 1.496*1011 times the Watts per m2 on Earth)
You can then work out the flux for a sphere of any radius (even the radius of the Sun)

3. Nov 9, 2014

### phyzguy

Why 1/(1.496*1011)2 ? When you do this, you are calculating how much the flux increases from the radius of the Earth's orbit to a radius of 1 meter. What you want to do instead of calculating 1/Rearth^2, is calculate Rsun^2/Rearth^2. If the flux at Earth's orbit is given by Fearth (=1630 W/m^2), then the flux at the sun's radius is given by 4 * pi * Rsun^2 * Fearth / (Rsun^2/Rearth^2) = 4 * pi * Rearth^2 * Fearth. This is the same answer that SophieCentaur gave you. Do you see?

4. Nov 9, 2014

### Staff: Mentor

Another example where proper units everywhere would have made the mistake obvious.
This is 1/(1.496*1011m)2 = 4.468*10-23m-2
It does not make sense to divide the surface of sun by this value, the result would be meters to the 4th power.