Question

A random star has a surface temperature of 7300 K and a radius 100 times that of the Sun. Calculate the luminosity of the star compared to the Sun. Estimate what it’s apparent magnitude would be if it was only ∼ 1 pc away like the nearest stars. Compare this to other night sky objects.

Answer 1

`(F')/(F) = 5.07*10^(-12)`

Step-by-step explanation:

given data:

surface temperature 7300 K

`R_(star) = 100R_(sun)`

we know that

`L = \sigma A T^4`

Where

`\sigma = 5.67*10^(-8) wm^(-2) k^(-4)`

A area of illuminated surface
`= 4 \pi R^2`

T = temperature of surface

WE KNOW THAT

`(L)/(L_O) = [(R)/(R_(sun))]^2+[(T)/(T_(sun))]^4`

T = 7300 K

`T_(sun) = 5778 K`

`(L)/(L_O) = 100^2 * [(7300)/(5778)]^4`

L = L_O * 2.54*10^4

WE KNOW THAT

r = 1 parsec, pc = 3.085*10^{16] m

luminosity decrease when move away from the core

`F' = (L)/(4 \pi r^2)`

`= (L)/(4 \pi R^2) *[(R)/(r)]^2`

`F' = F  *[(R)/(r)]^2`

`(F')/(F) = [(6.95)/(3.085)]^2*10^(-12)`

`(F')/(F) = 5.07*10^(-12)`