How much energy would it take to illuminate a universe of this size?

[Bossbrosish]

How much energy would it take to illuminate a universe 8.7998051e27 m in radius for 1 second?

 

[Bossbrosish]

bump

 

[HighTempShrimp]

formula for the volume of a sphere
V = (4/3) * π * r^3
plugging in the radius of 8.7998051e27 m, i got a volume of 4.0284084e83 m^3

lets assume we need 1 lumen per cubic meter to properly illuminate the volume. we need 4.0284084e83 lumens.

since a watt is equivalent to 1/683 lumens. 4.0284084e83 lumens equates to 5.8893647e89 watts.

scaling this by the illumination time of 1 second, the total energy required is 5.8893647e89 joules

 

[Bossbrosish]

formula for the volume of a sphere
V = (4/3) * π * r^3
plugging in the radius of 8.7998051e27 m, i got a volume of 4.0284084e83 m^3

lets assume we need 1 lumen per cubic meter to properly illuminate the volume. we need 4.0284084e83 lumens.

since a watt is equivalent to 1/683 lumens. 4.0284084e83 lumens equates to 5.8893647e89 watts.

scaling this by the illumination time of 1 second, the total energy required is 5.8893647e89 joules

Damn, I wasn't expecting it to be that high. Would the results be different if calculated like same way like the one from the references for common feats?

 

[HighTempShrimp]

yes, the blog calculated based on the apparent magnitude and luminosity of the brightest star in the universe. mines assumed a uniform luminous intensity across the entire universe.

the blog used the inverse square law of light, i didnt. it scaled the luminosity of the star to the light year distance of the universe, i just used the volume of the universe.

i think you should just use the blog, seems more accurate.

 

[Bossbrosish]

yes, the blog calculated based on the apparent magnitude and luminosity of the brightest star in the universe. mines assumed a uniform luminous intensity across the entire universe.

the blog used the inverse square law of light, i didnt. it scaled the luminosity of the star to the light year distance of the universe, i just used the volume of the universe.

i think you should just use the blog, seems more accurate.

I see, that's interesting to know. Also I did try to use the blog at first but I never figured out what the Log part is in the calc.

 

[SeijiSetto]

I see, that's interesting to know. Also I did try to use the blog at first but I never figured out what the Log part is in the calc.

log is short for logarithmic function, you need the base and the number itself. If you're not given a base, assume it's base 10. Logarithm is the inverse of exponents. 10^2 is 100 (10 times itself 2 times is 100), log(100) is 2 (how many times do you have to multiply 10 by itself to get 100?)

 

[Bossbrosish]

log is short for logarithmic function, you need the base and the number itself. If you're not given a base, assume it's base 10. Logarithm is the inverse of exponents. 10^2 is 100 (10 times itself 2 times is 100), log(100) is 2 (how many times do you have to multiply 10 by itself to get 100?)

Alright, I think I get it and L in the calc is the Sun's luminosity wich is 3.486*10^26, correct or is it a different number since it would use 3.486*10^26 twice?

 

[Bossbrosish]

Alright, I think I get it and L in the calc is the Sun's luminosity wich is 3.486*10^26, correct or is it a different number since it would use 3.486*10^26 twice?

Bumping in hopes to get an answer to this

 

[Bossbrosish]

yes L is the luminosity of the sun

Thanks!