Luminosity of the Earth covered in bulbs

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SUMMARY

The discussion focuses on calculating the total power output of 100W light bulbs covering the Earth's surface and comparing it to the Sun's luminosity. The surface area of the Earth is calculated as approximately 4.5 x 10^20 cm², leading to a total power output of 1.5 x 10^21 W from the bulbs. This results in a luminosity ratio of the Sun to the Earth-bulb setup of approximately 7 x 10^4. To match the Sun's luminosity, each bulb would need to be 10,000W, or 100 times more luminous than the original 100W bulb.

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1. The problem statement, all variables and given/known ddata
a.) Assume that a 100W light bulb covers 30cm^2 of area, and covers the surface of the Earth (land and water) with such 100W bulbs. How would the total power output compare with the luminosity of the Sun? Note: L_sun = 10^26W and the radius of the Earth is 6*10^6 meters.

b.) Which wattage would each light bulb need to be to make the luminosity of the entire Earth equal to the luminosity of the Sun?

Homework Equations


Surface area = 4πr^2

The Attempt at a Solution


[/B]
Studying for finals and found an old test problem that I got wrong.

Initially I calculated the Surface area of the Earth
4*π*(6*10^6 m )^2 = 4.5 * 10^14 m^2 = 4.5 * 10^20 cm^2

I set the surface area of the Earth over the area of the bulb and multiplied that by the wattage of the bulb:

SA/A = (4.5*10^20 cm^2)/(30 cm^2) = 1.5 * 10^19 * 100W = 1.5 * 10^21 W

Then made a ratio of the luminosity of the Sun over the Earth-bulb:

Ls/Le = 10^26/1.5 * 10^21 ≈ 7 * 10^4 difference.

For part b, each bulb would have to be 100 times more luminous, so 100^2 W instead of just 100W.

This isn't what I did on the test, rather what I did just now. I don't quite know if this is correct, so I'm hoping someone could verify if it is or not. Thank you for your time!
 
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There are 100 cm in 1 m, so 10^14 m^2 is 10^18 cm^2, not 10^20 cm^2.
 
Ah, yes, thank you. I made a calculator error there. Does everything else seem fine to you?
 

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