Universe Sandbox
Universe Sandbox Legacy => Universe Sandbox Legacy | Discussion => Topic started by: DenisineD on June 30, 2011, 09:38:59 PM
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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
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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
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Actually it's right.
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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
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The 01 at the end is because of not exact numbers used for sun's diameter and mass.
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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...