Question

A star has a surface temperature of 10,000 Kelvin and a radius 3 times that of the Sun. If the apparent magnitude of the star is 11.5, what is the distance to it?

If someone could show me step by step how to do this, I would be very appreciative.

Answer 1

A star's luminosity, or total power given off, is related to two of its properties: its temperature and surface area. If two stars have the same surface area, the hotter one will give off more radiation. If two stars have the same temperature, the one with more surface area will give off more radiation. The surface area of a star is directly related to the square of its radius (assuming a spherical star).

The luminosity of a star is given by the equation

L = 4*pi*R^2*s*T^4,

Where L is the luminosity in Watts, R is the radius in meters, s is the Stefan-Boltzmann constant

(5.67 x 10-8 Wm-2K-4), and T is the star's surface temperature in Kelvin.

We can then write the ratio of their luminosities as

L/Ls = (4pi*R^2*s*T^4)/(4pi*Rs^2*s*Ts^4) = (R/Rs)^2(T/Ts)^4

Thesurface temperature of Sun is approximately Ts=5778 K

Solving for the ratio L/Ls = yields

L/Ls = 3^2*(10000/5778)^4=80.7

The apparent magnitude m of a star depends on its absolute magnitude M. The Sun has an absolute magnitude of around 4.83. A star would have an absolute magnitude of M = 4.83 - (2.5)( log L star compared to sun)

M=4.83 - (2.5)*(log(80.7))=-6.15

m = M -5 + 5*log(d) where d is the distance of a star in parsec.

log(d)=(m-M+5)/5=(11.5+6.15+5)/5=4.53

d = e^(453/100) =~ 92.8 parsecs=302.5 light years.