Show that the angular radius of the star is given by....

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SUMMARY

The angular radius of a star, as observed from Earth, is defined by the equation θ = (2.06 × 105 / Teff2) √(F / σ), where Teff is the effective temperature, F is the flux, and σ is the Stefan-Boltzmann constant. The derivation involves rearranging the flux equation F = L / (4πd2) to express luminosity L in terms of flux and distance. The relationship between radius R and distance d is established, leading to the final angular radius formula. This discussion clarifies the connection between radians and arcseconds through the factor 2.06 × 105.

PREREQUISITES
  • Understanding of stellar luminosity and flux equations
  • Familiarity with effective temperature (Teff) concepts
  • Knowledge of the Stefan-Boltzmann law
  • Basic trigonometry related to angular measurements
NEXT STEPS
  • Study the derivation of the Stefan-Boltzmann law in detail
  • Learn about the relationship between luminosity, flux, and distance in astrophysics
  • Explore angular measurements in astronomy, specifically radians and arcseconds
  • Investigate the implications of effective temperature on stellar characteristics
USEFUL FOR

Astronomy students, astrophysicists, and educators seeking to understand the relationship between a star's effective temperature, flux, and angular radius as observed from Earth.

FaraDazed
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Homework Statement


A star has an effective temperature [itex]T_{eff}[/itex] and is observed to have a flux [itex]F[/itex]. Show that the angular radius in arcseconds of the star (as seen from Earth) is given by

[itex] \theta = (\frac{2.06 \times 10^5}{T_{eff}^2}) \sqrt{\frac{F}{\sigma}}[/itex]

Homework Equations


[itex] L = 4 \pi R^2 \sigma T_{eff}^4 \\<br /> T_{eff} = (\frac{L}{4 \pi R^2 \sigma})^{\frac{1}{4}} \\<br /> F = \frac{L}{4 \pi d^2}[/itex]

And probably..
[tex] \omega = \frac{A}{R^2}<br /> [/itex] ?<br /> <br /> <h2>The Attempt at a Solution</h2> <br /> [/B]<br /> I am a bit lost here! I missed several of the lecture containing this material, but from looking at what I have go already I should be able to do it?<br /> <br /> First I started by rearranging the eqn for Flux for L<br /> [itex] F = \frac{L}{4 \pi d^2} \\<br /> L = 4 \pi d^2 F[/itex]<br /> And then subsituted that into the eqn for T_eff<br /> [itex] T_{eff} = (\frac{L}{4 \pi R^2 \sigma})^{\frac{1}{4}} \\<br /> T_{eff} = (\frac{4 \pi d^2 F}{4 \pi R^2 \sigma})^{\frac{1}{4}} \\<br /> T_{eff}^4 = (\frac{ d^2 F}{R^2 \sigma}) = (\frac{d}{R})^2 (\frac{F}{\sigma}) \\<br /> (\frac{R}{d})^2 = \frac{1}{T_{eff}^4}(\frac{F}{\sigma}) \\<br /> (\frac{R}{d}) = \sqrt{\frac{1}{T_{eff}^4}(\frac{F}{\sigma})} \\<br /> \frac{R}{d} = \frac{1}{T_{eff}^2} \sqrt{\frac{F}{\sigma}} \\[/itex]<br /> <br /> That is as far as I have got. I assume in mine, R/d is the angle in radians, so I hope that it relates to the angle in arcsecond by that 2.06 * 10^5 factor? Or I am way off. I'd appreciate some help/advie with this please.<br /> <br /> EDIT: yes It does and I can see how now. My bad.[/tex]
 
Last edited:
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Yep. Looks good.
 
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