Question 44.4: Determining star temperature and star size. Suppose that the......
Determining star temperature and star size. Suppose that the distances from Earth to two nearby stars can be reasonably estimated, and that their measured apparent brightnesses suggest the two stars have about the same luminosity, L. The spectrum of one of the stars peaks at about 700 \mathrm{~nm} (so it is reddish). The spectrum of the other peaks at about 350 \mathrm{~nm} (bluish). Use Wien’s law (Eq. 37-1) and the Stefan-Boltzmann equation (Section 19-10) to determine (a) the surface temperature of each star, and (b) how much larger one star is than the other.
\lambda_{\mathrm{P}} T=2.90 \times 10^{-3} \mathrm{~m} \cdot \mathrm{K} (37-1)
APPROACH We determine the surface temperature T for each star using Wien’s law and each star’s peak wavelength. Then, using the Stefan-Boltzmann equation (power output or luminosity \propto A T^{4} where A= surface area of emitter), we can find the surface area ratio and relative sizes of the two stars.
(a) Wien’s law (Eq. 37-1) states that \lambda_{\mathrm{P}} T=2.90 \times 10^{-3} \mathrm{~m} \cdot \mathrm{K}. So the temperature of the reddish star is
T_{\mathrm{r}}=\frac{2.90 \times 10^{-3} \mathrm{~m} \cdot \mathrm{K}}{700 \times 10^{-9} \mathrm{~m}}=4140 \mathrm{~K}
The temperature of the bluish star will be double this since its peak wavelength is half (350 \mathrm{~nm} vs. 700 \mathrm{~nm}) :
T_{\mathrm{b}}=8280 \mathrm{~K}
(b) The Stefan-Boltzmann equation, Eq. 19-17, states that the power radiated per unit area of surface from a blackbody is proportional to the fourth power of the Kelvin temperature, T^{4}. The temperature of the bluish star is double that of the reddish star, so the bluish one must radiate \left(2^{4}\right)=16 times as much energy per unit area. But we are given that they have the same luminosity (the same total power output); so the surface area of the blue star must be \frac{1}{16} that of the red one. The surface area of a sphere is 4 \pi r^{2}, so the radius of the reddish star is \sqrt{16}=4 times larger than the radius of the bluish star (or 4^{3}=64 times the volume).
\frac{\Delta Q}{\Delta t}=\epsilon \sigma A T^{4} (19-17)