# Homework Help: Luminosity of the Earth covered in bulbs

1. Dec 8, 2016

### Stendhal

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?

2. Relevant equations
Surface area = 4πr^2

3. The attempt at a solution

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!

2. Dec 8, 2016

### phyzguy

There are 100 cm in 1 m, so 10^14 m^2 is 10^18 cm^2, not 10^20 cm^2.

3. Dec 8, 2016

### Stendhal

Ah, yes, thank you. I made a calculator error there. Does everything else seem fine to you?