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

Answer:

$\frac{F'}{F} = 5.07*10^{-12}$

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

$\frac{L}{L_O} = [\frac{R}{R_{sun}}]^2+[\frac{T}{T_{sun}}]^4$

T = 7300 K

$T_{sun} = 5778 K$

$\frac{L}{L_O} = 100^2 * [\frac{7300}{5778}]^4$

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

$F' = \frac{L}{4 \pi r^2}$

 $= \frac{L}{4 \pi R^2} *[\frac{R}{r}]^2$

$F' = F  *[\frac{R}{r}]^2$

$\frac{F'}{F} = [\frac{6.95}{3.085}]^2*10^{-12}$

$\frac{F'}{F} = 5.07*10^{-12}$