Surface Flux & Colors of Stars
Useful constants: nm·K cm·K; erg cm⁻² s⁻¹ K⁻⁴; pc cm; erg/s; K; cm km; km cm.
Show explicit units, and for each result run a sanity check. Worked solutions are released after the homework due date.
Conceptual
Problem
⭐ Two stars, same luminosity, different colors. Star A is blue; Star B is red; both have the same luminosity.
- (a) Which has the higher effective temperature?
- (b) Which must have the larger radius? Explain using .
- (c) Which surface radiates more power per cm²? How do you know?
Color tracks temperature: blue light peaks at shorter wavelength than red, so by Wien’s law the bluer star is hotter. With held fixed in , must compensate for .
Surface flux is — it depends only on temperature, not on size. Decide which star wins on alone for part (c), then let settle part (b).
Problem
⭐ Why Betelgeuse is huge. Betelgeuse has and K (~60% of the Sun’s). Use Stefan-Boltzmann to explain qualitatively why it must be so large despite being cooler than the Sun. Reference and surface area.
Split the luminosity into two factors: . A cooler shrinks the second factor, so the first factor must grow to reach .
Per cm², Betelgeuse radiates only of the Sun’s flux. To still emit the Sun’s total power, its surface area — and hence — must be enormous.
Problem
⭐⭐ Peak wavelength, temperature, and radius. A star peaks at nm; the Sun peaks at 500 nm.
- (a) Hotter or cooler than the Sun, and by what factor? (Wien as a ratio.)
- (b) At the same luminosity, which has the larger radius, and by what factor? (.)
- (c) Which has the greater surface flux , and by what factor?
Use Wien’s law as a ratio: . Temperature is inversely proportional to peak wavelength, so a longer peak means a cooler star.
Once you have , the radius follows from at fixed , and the surface flux from . Carry the single factor through both.
Problem
⭐⭐ Surface brightness independence. Explain why you cannot determine a star’s distance from how bright its disk appears per unit solid angle, given that is distance-independent. Does this apply most directly to unresolved point sources or to resolved objects?
Both flux and solid angle fall off as — flux because of the inverse-square law, solid angle because the disk subtends a smaller angle when farther away. Form the ratio and watch the cancel.
“Per unit solid angle” only means something if the object’s disk is resolved — you must spread the flux across measured angular area. Ask whether a point source has any measurable at all.
Calculation
Problem
⭐ Sun’s temperature from Wien’s law. The Sun peaks at nm.
- (a) Convert 500 nm to cm.
- (b) Use with cm·K; show the unit cancellation.
- (c) Compare to the known 5800 K.
For (a), nm cm, so nm cm. Wien’s law is — use the cm·K value of so the centimeters cancel and leave kelvin.
Divide cm·K by cm. The cm cancel, leaving K. Then check your result against the stated 5800 K.
Problem
⭐ Stellar temperature from color. A star peaks at 700 nm (red).
- (a) Calculate its effective temperature.
- (b) Hotter or cooler than the Sun, and by what factor?
Convert nm to cm ( nm cm), then apply with cm·K.
For (b), skip a second calculation — form the ratio . A peak at longer wavelength means a cooler star.
Problem
⭐⭐ From received flux to luminosity. A star at pc has bolometric flux erg s⁻¹ cm⁻².
- (a) Convert to cm.
- (b) Use ; show units at every step.
- (c) Express in . Does it remind you of a worked-example star?
The inverse-square law runs the flux back up to total power. First convert pc to cm using pc cm.
Compute in cm², multiply by — the units work out to (cm²)(erg s⁻¹ cm⁻²) = erg s⁻¹. Then divide by erg/s for part (c).
Problem
⭐⭐ Stellar radius (ratio method). A star has , .
- (a) Show .
- (b) Raise to the power:
- (c) Larger or smaller than the Sun? Convert to km.
Rearrange into the dimensionless ratio — the constant cancels. Substitute and .
With , take the square root for . For (c), multiply that ratio by km.
Problem
⭐⭐ Rigel’s radius (full calculation). Rigel has and K.
- (a) Compute (verify dimensionless).
- (b) Raise to the fourth power.
- (c) Compute , then the power for .
- (d) Express in km.
- (e) Compare to Betelgeuse () — why is one so much larger?
Form — the kelvin cancel, leaving a pure number. Then with .
Take the fourth power of first, divide by it, then square-root for . For (e), Betelgeuse is much cooler, so at comparable luminosity inflates its radius far more.
Synthesis
Problem
⭐⭐ Complete inference chain. A red giant: erg s⁻¹ cm⁻², nm, pc.
- (a) Temperature from Wien (show units), then .
- (b) Convert to cm; use ; express as .
- (c) Radius in solar units: then the power.
- (d) Convert to km; compare to the Sun and to Betelgeuse. Does cool + large make sense?
Chain three tools in order: Wien () for temperature, inverse-square () for luminosity, then for radius.
Convert nm to cm and pc to cm before substituting. A cool raised to the fourth power in the denominator is what drives the large radius — that’s the “cool + large” giant signature.
Problem
⭐⭐ HR diagram reasoning. Three stars: A ( K, ); B ( K, ); C ( K, ).
- (a) Radius of each in solar radii ( first, then power).
- (b) Largest? Smallest?
- (c) Sketch on an HR diagram ( increasing leftward) with lines of constant radius. What do they tell you about giants and white dwarfs?
Apply to each star, using K. Build first for each, then raise to the fourth power.
A cool, luminous star (B) lands at large radius (upper right); a hot, faint star (C) lands at small radius (lower left). On the HR diagram, lines of constant radius run diagonally — giants sit above them, white dwarfs below.
Problem
⭐⭐⭐ Extinction’s effect on temperature inference. Dust reddens starlight. An observer finds nm; the true (dust-corrected) peak is nm.
- (a) Apparent temperature from the reddened color (show units).
- (b) True temperature from the corrected color.
- (c) Bias factor .
- (d) Using the reddened temperature at the correct luminosity, would the inferred radius be too large or too small, and by what factor? (.)
Reddening pushes the apparent peak to longer wavelength, so by the apparent temperature comes out cooler than the truth. Compute both temperatures with Wien’s law before comparing.
The bias factor is just . Since at fixed , an under-estimated inflates the inferred radius by .