formula for the Luminosity

[DenisineD]

Hi everyone,

I m doing a excel program to calculate the luminosity of a star with the Radius and the temperature in Kelvin of is surface....

I use the formula L = A X F

A is the Surface area of the sphere (lets supose its a perfect sphere...)
the radius i use is in Km 

So A = 4лR2

F = σT4

σ = 5,67E-8

It give pretty huge numbers....do i need to devise it by something to get a realistic luminosity?
Ex: My star have a radius of: 540 000 Km and a surface temperature of: 3950

Our sun have a luminosity of : 3.84E26
My star have a luminosity of : 5,05788E19

So if i divide My star with the sun to have a 1 Ratio it equal: 1,31716E-7    Wich is ridiculously low....lol

But...it seem to be realistic when i multiply by (1,0E6) wich it give a final luminosity of 0,1317

It seem kinda realistic for a star of this size and temperature...

let me know what you think....i was having some free time lol

[DenisineD]

But the problem is....when i try it at my second star wich have a radius of 240508 Km and a surface temperature of 3050 K

it give me: 0,009287955 L

Kinda too low...lol

[Darvince]

Actually it's right.

[DenisineD]

oh oki but when i try the sun it gave me 0,90....... something :/ lol

Now i did change the constant and it give me now for the sun 1,000000001

[Darvince]

The 01 at the end is because of not exact numbers used for sun's diameter and mass.

[DenisineD]

ok i ll keep it this way....maybe the difference are for the approximation of numbers and etc....i did try it with syrus and it gave me 23,4....kinda near...