A star Has twice the luminosity of the sun but has 1/3 the suns temperature, calculate the stars surface area

Answer 1

The luminosity of a star is the total energy emitted from the surface of the star. It is related to the temperature and radius (size/surface area) of the star. L = σ T⁴ 4π R², where σ is the Stefan-Boltzmann constant, T is the temperature in K and R is the radius in m.

However, this equation can be simplified to L = T⁴ R², if L, T, and R are given in terms of Solar units.

Given: L = 2 L_s and T = (1/3) T_s (since the correct symbol isn't available, the subscript "s" will indicate solar values)

Want to find R in terms of R_s. (or more accurately, 4π R² )

Using the equation: L = T⁴ R², it can be shown that R_s² = L_s ⁄ T_s⁴ and for the given star:

2 L_s = [ (1/3) T_s ]⁴ R² solve for R² : R² = (3⁴ * 2 * L_s ) ⁄ T_s⁴ = 162 * (L_s ⁄ T_s⁴) = 162 R_s²

Now that you have the radius in terms of solar radii, you can find the surface area, either in terms of the Sun's surface area or by solving for the surface area of the Sun then multiplying by 162.

Surface area = 4π R²,