Treat a star as a spherical black body. Its spectral peak is at 557.692 nm and its luminosity is 1.75252e+26 W. Use λmax T = 2.90 × 10⁻³ m K and L = 4πr²σT⁴ with σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴. (a) Find surface temperature and radius. (b) Find received flux at distance 1.00 × 10¹⁷ m using F = L/(4πd²). (c) Explain the effect of doubling distance on flux and inferred temperature if there is no absorption.

Question

Accepted Answer

Temperature 5200 K; radius 580000000 m; flux 1.395e-9 W/m²; doubled distance quarters flux but does not change spectral temperature. Convert the peak wavelength from nanometres to metres before applying Wien's law: T = 2.90 × 10⁻³/(557.692 × 10⁻⁹) = 5200 K. The peak position describes temperature, not the star's total power. Rearrange the luminosity relation to r = √[L/(4πσT⁴)]. Substituting L = 1.75252e+26 W and the unrounded temperature gives r = 580000000 m. The square root applies to the whole ratio. At distance d the luminosity spreads over a sphere of area 4πd², so F = 1.75252e+26/[4π(1.00 × 10¹⁷)²] = 1.395e-9 W/m². This received flux is not the surface emission flux. Doubling distance multiplies the spreading area by four, so the measured flux becomes one-quarter. With no absorption or spectral shift, the peak wavelength is unchanged and the temperature inferred from it remains the same. The black-body treatment is an explicit approximation to a real stellar atmosphere.

A star's spectrum and radius 7

What a complete response should do

How did this one feel?