Homework 3 Solutions
Parallax + Surface Flux & Radiation Inference
Student note: These are model solutions written to show reasoning, units, and checks. Your solutions can be shorter if your setup and logic are correct.
Grade memo note: Use these to write your reflection (what you got right, what broke, and what you will do differently next time).
Part A — Distance, Parallax, and Luminosity
Problem 1 — Parallax Is Not Angular Size
Restatement: Critique the claim that a star with a large parallax must have a large angular size.
Key insight: Parallax is a position shift from changing viewpoint; angular size is apparent object diameter on the sky. They measure completely different things.
Answer:
- The statement is incorrect because large parallax means the star is nearby, not physically large in angular diameter. Parallax tells us about distance, not about the star’s physical or apparent size.
- Parallax angle: the apparent shift in a star’s sky position between two viewpoints (e.g., opposite sides of Earth’s orbit). Angular size: the apparent diameter of an object’s disk on the sky.
- Parallax reverses direction over six months as Earth moves around the Sun; angular size does not show that periodic sign flip. This six-month periodicity is the observational fingerprint of parallax.
Sanity check: Alpha Centauri has the largest parallax of any star (\(p \approx 0.76''\)), yet its angular diameter is \(\sim 0.007''\) — a hundred times smaller. Large parallax, tiny angular size. The two quantities are clearly independent. \(\checkmark\)
Common misconception: Conflating parallax (a periodic positional shift due to Earth’s orbital motion) with angular diameter (a fixed property of the object’s apparent disk). A nearby point source has large parallax but negligible angular size.
Problem 2 — Parallax to Distance
Given:
- Star A: \(p_A = 0.50''\)
- Star B: \(p_B = 0.020''\)
- \(1\ \text{pc} = 3.086\times10^{18}\ \text{cm}\)
Find: Distance to each star in pc and cm; which star is farther and by what factor.
Equation:
\[d(\text{pc}) = \frac{1}{p('')}\]
The parsec is defined so that a star with a parallax of exactly \(1''\) is at \(1\ \text{pc}\). Smaller parallax means the star’s apparent shift is smaller, so it must be farther away. The inverse relationship makes distance grow rapidly as parallax shrinks.
Steps:
(a) Star A: \(p=0.50''\) \[d_A = \frac{1}{0.50\ \text{arcsec}} = 2.0\ \text{pc}.\] \[d_A = 2.0\ \text{pc}\times 3.086\times10^{18}\ \frac{\text{cm}}{\text{pc}} = 6.17\times10^{18}\ \text{cm}.\]
(b) Star B: \(p=0.020''\) \[d_B = \frac{1}{0.020\ \text{arcsec}} = 50\ \text{pc}.\] \[d_B = 50\ \text{pc}\times 3.086\times10^{18}\ \frac{\text{cm}}{\text{pc}} = 1.54\times10^{20}\ \text{cm}.\]
(c) Farther by factor \[\frac{d_B}{d_A} = \frac{50\ \text{pc}}{2.0\ \text{pc}} = 25.\]
\[ \boxed{d_A = 2.0\ \text{pc} = 6.17\times10^{18}\ \text{cm}},\quad \boxed{d_B = 50\ \text{pc} = 1.54\times10^{20}\ \text{cm}},\quad \boxed{\text{Star B is }25\times\text{ farther.}} \]
Unit check: \(d = 1/p\) with \(p\) in arcseconds gives \(d\) in parsecs by definition. The cm conversion multiplies pc by \(3.086 \times 10^{18}\ \text{cm/pc}\), giving cm. \(\checkmark\)
Sanity check: Star B’s parallax is \(25\times\) smaller than Star A’s (\(0.020''/0.50'' = 1/25\)), so it should be \(25\times\) farther. Indeed \(d_B/d_A = 50\ \text{pc}/2.0\ \text{pc} = 25\). Star A at \(2.0\ \text{pc}\) is comparable to the nearest stars (Alpha Centauri is at \(1.3\ \text{pc}\)), so the numbers are astrophysically reasonable. \(\checkmark\)
Problem 3 — Inverse-Square Flux Scaling
Given: Two stars with equal luminosity \(L\). Star X at \(d_X = 20\ \text{pc}\); Star Y at \(d_Y = 50\ \text{pc}\). No extinction.
Find: Flux ratio \(F_X/F_Y\); which star appears brighter; why equal \(L\) does not mean equal \(F\).
Equation:
\[F = \frac{L}{4\pi d^2} \quad\Rightarrow\quad F \propto d^{-2}\ \text{at fixed}\ L\]
At fixed luminosity, flux falls as the square of the distance. The \(4\pi d^2\) factor is the surface area of the sphere over which the luminosity is spread — doubling the distance quadruples the area and quarters the flux.
Steps:
\[\frac{F_X}{F_Y}=\left(\frac{d_Y}{d_X}\right)^2 =\left(\frac{50\ \text{pc}}{20\ \text{pc}}\right)^2 =2.5^2 =6.25.\]
\[ \boxed{\frac{F_X}{F_Y}=6.25} \]
- Star X appears brighter — it is closer, so more of its luminosity reaches us per unit area.
- Equal luminosity does not imply equal observed flux because flux depends on distance through inverse-square spreading. Two identical candles look very different in brightness when one is across the room and the other is across the street.
Unit check: The ratio \(F_X/F_Y = (d_Y/d_X)^2\) — both distances are in the same units (pc), which cancel. The result is a dimensionless ratio. \(\checkmark\)
Sanity check: Star X is closer (\(20\ \text{pc} < 50\ \text{pc}\)), so it should appear brighter. \(F_X/F_Y = 6.25 > 1\) confirms this. The factor \(2.5^2 = 6.25\) also matches the distance ratio of \(50\ \text{pc}/20\ \text{pc} = 2.5\) squared — consistent with inverse-square scaling. \(\checkmark\)
Problem 4 — Luminosity from Flux + Distance
Given:
- Parallax: \(p=0.10''\)
- Flux ratio: \(F_\star/F_\odot=2.0\times10^{-11}\) (where \(F_\odot\) is the Sun’s bolometric flux at \(1\ \text{AU}\))
- \(1\ \text{pc} = 206{,}265\ \text{AU}\)
Find: Distance in pc and AU; luminosity ratio \(L_\star/L_\odot\).
Equation:
\[d(\text{pc}) = \frac{1}{p('')},\qquad \frac{L_\star}{L_\odot}=\frac{F_\star}{F_\odot}\left(\frac{d_\star}{1\ \text{AU}}\right)^2\]
The luminosity formula works because \(F_\odot\) is the Sun’s flux measured at \(1\ \text{AU}\), making it the natural reference. The \(d^2\) factor undoes the geometrical dilution: a star’s flux is its luminosity spread over a sphere of radius \(d\), so to recover \(L\) from \(F\) we multiply back by the area \(\propto d^2\).
Steps:
Distance: \[d_\star=\frac{1}{0.10\ \text{arcsec}}=10\ \text{pc}.\] \[d_\star = 10\ \text{pc}\times 206{,}265\ \frac{\text{AU}}{\text{pc}}=2.06265\times10^6\ \text{AU}.\]
Luminosity ratio (both factors are dimensionless — flux ratio and distance ratio): \[\frac{L_\star}{L_\odot} =\frac{F_\star}{F_\odot}\left(\frac{d_\star}{1\ \text{AU}}\right)^2 =\left(2.0\times10^{-11}\right)\left(\frac{2.06265\times10^6\ \text{AU}}{1\ \text{AU}}\right)^2 =\left(2.0\times10^{-11}\right)\left(4.2545\times10^{12}\right) =85.1.\]
\[ \boxed{d_\star=10\ \text{pc}=2.06265\times10^6\ \text{AU}},\qquad \boxed{\frac{L_\star}{L_\odot}\approx85} \]
The star is about \(85\times\) more luminous than the Sun.
Unit check: \(L_\star/L_\odot = (F_\star/F_\odot)(d_\star/1\ \text{AU})^2\) — the flux ratio is dimensionless and the distance ratio (AU/AU) is dimensionless, so the product is a dimensionless luminosity ratio. \(\checkmark\)
Sanity check: A star \(85\times\) the Sun’s luminosity is in the range of a luminous A-type or early F-type main-sequence star, or a subgiant — astrophysically plausible. At \(10\ \text{pc}\) its flux is tiny (\(2 \times 10^{-11}\ F_\odot\)), which makes sense: even a bright star is extremely faint compared to the Sun seen from \(1\ \text{AU}\). \(\checkmark\)
Problem 5 — Same Flux, Different Parallax
Restatement: Two stars have identical observed bolometric flux at Earth but different parallaxes (\(p_A = 0.100''\), \(p_B = 0.025''\)). Determine which is farther, by what factor, and infer the luminosity ratio.
Key elements (a complete answer must address):
- Distance ratio derived from inverted parallax ratio
- Luminosity scaling \(L \propto F d^2\) at fixed flux
- Correct numerical factor: \(L_B/L_A = (d_B/d_A)^2\)
- Physical explanation of why equal flux does not imply equal luminosity
Sample response:
(a) Given same observed flux: \[F_A = F_B.\]
Distances from parallax (\(d = 1/p\), so the star with smaller parallax is farther): \[\frac{d_B}{d_A}=\frac{p_A}{p_B}=\frac{0.100''}{0.025''}=4.\]
Star B is \(4\times\) farther.
(b) At fixed flux, luminosity scales as: \[L = 4\pi d^2 F \quad\Rightarrow\quad L\propto d^2\ \text{(at fixed}\ F\text{)} \quad\Rightarrow\quad \frac{L_B}{L_A}=\left(\frac{d_B}{d_A}\right)^2=4^2=16.\]
\[ \boxed{\frac{d_B}{d_A}=4},\qquad \boxed{\frac{L_B}{L_A}=16} \]
(c) Equal observed flux means only that \(L/d^2\) is the same for both stars. A farther star must be intrinsically more luminous to produce the same flux at Earth — it compensates for the greater distance by emitting more total power. Conversely, a nearby faint star can appear just as bright as a distant luminous one. Flux alone cannot distinguish a nearby dim star from a far-away bright one; parallax breaks the degeneracy.
(d) The scaling at fixed observed flux: \[\frac{L_B}{L_A} = \left(\frac{d_B}{d_A}\right)^2\quad\text{when}\ F_A = F_B.\]
Unit check: \(d_B/d_A\) is a dimensionless distance ratio; \((d_B/d_A)^2\) is also dimensionless. \(L_B/L_A\) is a dimensionless luminosity ratio. Both sides are dimensionless. \(\checkmark\)
Sanity check: Star B is farther (\(4\times\)) yet equally bright, so it must be intrinsically more luminous. \(L_B/L_A = 16 > 1\) confirms this, and the factor \(4^2 = 16\) is consistent with inverse-square scaling. \(\checkmark\)
Part B — Radiation, Temperature, and Radius
Problem 6 — Same Luminosity, Different Color
Restatement: Two stars have equal luminosity but different colors (one blue, one red). Determine which is hotter, which has the larger radius, and justify using Stefan-Boltzmann scaling.
Key insight: At fixed luminosity, the Stefan-Boltzmann relation \(L \propto R^2 T^4\) forces a tradeoff between temperature and radius — hotter stars must be smaller.
Answer:
At fixed luminosity: \[L \propto R^2T^4 \quad\Rightarrow\quad R\propto T^{-2}.\]
- The blue star is hotter. Bluer color means shorter peak wavelength, which Wien’s law (\(\lambda_{\text{peak}} = b/T\)) tells us corresponds to higher temperature.
- The hotter (blue) star must have the smaller radius at the same luminosity.
- From \(R \propto T^{-2}\) at fixed \(L\): if temperature increases by a factor \(f\), the emitted power per unit area rises as \(f^4\) (Stefan-Boltzmann). To keep total luminosity \(L = 4\pi R^2 \sigma T^4\) constant, the surface area \(R^2\) must decrease by \(f^4\), so \(R\) decreases by \(f^2\). The hotter star radiates so much more intensely per unit area that it needs less surface to produce the same total power.
Common misconception: Assuming the brighter-colored or hotter star must be physically larger. Temperature and radius are not independently linked — it is their combination \(R^2 T^4\) that determines luminosity. A hot star can be tiny (white dwarf) or huge (blue supergiant), depending on its luminosity.
Sanity check: Blue stars are hotter (Wien), and at fixed \(L\) the \(R \propto T^{-2}\) scaling gives smaller \(R\) for higher \(T\). The red star, being cooler, must compensate with a larger radiating area. This is consistent with observed stellar populations: red giants are large and cool, blue dwarfs are small and hot. \(\checkmark\)
Problem 7 — Temperature from Wien’s Law
Given:
- \(\lambda_{\text{peak}}=725\ \text{nm}\)
- Wien’s constant: \(b=2.898\times10^6\ \text{nm·K}\)
- \(T_\odot = 5800\ \text{K}\)
Find: Effective temperature \(T\); ratio \(T/T_\odot\); whether hotter or cooler than the Sun.
Equation:
\[T=\frac{b}{\lambda_{\text{peak}}}\]
Wien’s displacement law: the peak wavelength of a blackbody spectrum shifts to shorter wavelengths at higher temperatures. A star peaking in the red/infrared is cooler; one peaking in the blue/UV is hotter.
Steps:
Temperature: \[T=\frac{b}{\lambda_{\text{peak}}} =\frac{2.898\times10^6\ \text{nm}\cdot\text{K}}{725\ \text{nm}} =3997\ \text{K}\approx4.00\times10^3\ \text{K}.\]
Compare to Sun: \[\frac{T}{T_\odot}=\frac{3997\ \text{K}}{5800\ \text{K}}=0.689\approx0.69.\]
\[ \boxed{T\approx4.0\times10^3\ \text{K}},\qquad \boxed{T/T_\odot\approx0.69} \]
So the star is cooler than the Sun.
Unit check: \([b/\lambda_{\text{peak}}] = [\text{nm·K}]/[\text{nm}] = [\text{K}]\). The wavelength units cancel, leaving temperature. \(\checkmark\)
Sanity check: A peak at \(725\ \text{nm}\) is at the red edge of the visible spectrum (red light is \(\sim 620\text{–}750\ \text{nm}\)), suggesting a cool, reddish star. A temperature of \(\sim 4000\ \text{K}\) corresponds to a K-type star — consistent with a red/orange appearance. The Sun peaks at \(\sim 500\ \text{nm}\) and has \(T = 5800\ \text{K}\), so a star peaking at longer wavelength should indeed be cooler. \(\checkmark\)
Problem 8 — Planck Limits and Physical Meaning
Restatement: Derive the long-wavelength (Rayleigh-Jeans) and short-wavelength (Wien) limits of the Planck function, explain how the exponential suppression resolves the ultraviolet catastrophe, and connect to the statement that “photons are expensive” at high frequency.
Key elements (a complete answer must address):
- Taylor expansion \(e^x \approx 1 + x\) for \(x \ll 1\) and correct simplification to \(B_\lambda \propto T\lambda^{-4}\)
- Identification of the exponential suppression factor \(e^{-hc/\lambda k_B T}\) for \(x \gg 1\)
- Physical explanation of UV catastrophe resolution (not just restating the math)
- Connection between \(h\nu \gg k_B T\) and low photon occupation probability
Given: \[B_\lambda(T)=\frac{2hc^2}{\lambda^5}\frac{1}{e^{hc/\lambda k_B T}-1},\qquad x\equiv\frac{hc}{\lambda k_B T}.\]
(a) Long-wavelength limit (\(x\ll1\))
Use \(e^x\approx1+x\), so \(e^x-1\approx x\): \[B_\lambda \approx \frac{2hc^2}{\lambda^5}\frac{1}{x} =\frac{2hc^2}{\lambda^5}\frac{\lambda k_B T}{hc} =\frac{2ck_B T}{\lambda^4}.\]
Thus \[\boxed{B_\lambda \approx \frac{2ck_B T}{\lambda^4}}\quad\text{(Rayleigh-Jeans limit).}\]
(b) Short-wavelength limit (\(x\gg1\))
Use \(e^x-1\approx e^x\): \[\boxed{B_\lambda \approx \frac{2hc^2}{\lambda^5}\exp\!\left(-\frac{hc}{\lambda k_B T}\right)}\]
Emission is suppressed by the exponential factor \(e^{-x}\).
(c) UV catastrophe resolution
Classical Rayleigh-Jeans behavior diverges toward short \(\lambda\), but Planck’s exponential cutoff suppresses high-frequency emission, preventing the divergence.
(d) “Photons are expensive”
At high frequency, photon energy \(h\nu\) is much larger than available thermal energy scale \(k_B T\). Occupying those photon states is unlikely, so short-wavelength emission is exponentially rare. The Boltzmann factor \(e^{-h\nu/k_BT}\) quantifies exactly how rare: when \(h\nu = 10\,k_BT\), the probability is \(e^{-10} \approx 5 \times 10^{-5}\) — essentially negligible.
Unit check (Rayleigh-Jeans): \([2ck_BT/\lambda^4] = [LT^{-1}][ML^2T^{-2}\Theta^{-1}][\Theta]/[L^4] = [ML^{-1}T^{-3}]\). Spectral radiance has dimensions of power per area per steradian per wavelength interval, which is \([MT^{-3}][L^{-1}] = [ML^{-1}T^{-3}]\). \(\checkmark\)
Sanity check: In the Rayleigh-Jeans limit, \(B_\lambda \propto T\lambda^{-4}\) diverges as \(\lambda \to 0\) — this is exactly the ultraviolet catastrophe that classical physics predicted. Planck’s exponential factor kills the divergence at short wavelengths, so the total integrated intensity converges. The crossover between the two regimes occurs near \(x \sim 1\), i.e., \(\lambda \sim hc/k_BT\), which is the scale set by the peak wavelength. This is self-consistent with Wien’s law. \(\checkmark\)
Problem 9 — Radius from Luminosity and Temperature
Given:
- \(L/L_\odot = 16\)
- \(T/T_\odot = 2\)
Find: \((R/R_\odot)^2\); \(R/R_\odot\); physical interpretation (larger or smaller than the Sun?).
Equation:
\[\frac{L}{L_\odot}=\left(\frac{R}{R_\odot}\right)^2\left(\frac{T}{T_\odot}\right)^4\]
Stefan-Boltzmann in ratio form: luminosity scales as surface area (\(\propto R^2\)) times emitted power per unit area (\(\propto T^4\)). Using solar units on both sides makes the equation dimensionless and eliminates all physical constants.
Steps:
Then \[\left(\frac{R}{R_\odot}\right)^2 =\frac{16}{2^4} =\frac{16}{16} =1,\] so \[\frac{R}{R_\odot}=1.\]
\[ \boxed{\left(\frac{R}{R_\odot}\right)^2=1},\qquad \boxed{\frac{R}{R_\odot}=1} \]
The star has the same radius as the Sun.
Unit check: All quantities are dimensionless ratios (\(L/L_\odot\), \(R/R_\odot\), \(T/T_\odot\)), so both sides of the equation are dimensionless. The algebra is purely numerical. \(\checkmark\)
Sanity check: The star is \(16\times\) more luminous but \(2\times\) hotter. Since luminosity per unit area goes as \(T^4 = 2^4 = 16\), the star radiates \(16\times\) more power per unit area than the Sun. To produce \(16\times\) total luminosity with \(16\times\) the surface flux, it needs exactly the same surface area — hence \(R = R_\odot\). The higher temperature entirely accounts for the higher luminosity without requiring a larger star. \(\checkmark\)
Problem 10 — Full Chain: Parallax + Flux + Color
Restatement: Given a star’s parallax, flux ratio, and peak wavelength, execute the full observable → model → inference chain to determine distance, luminosity, temperature, and radius, then propagate uncertainty from the parallax measurement.
Key elements (a complete answer must address):
- \(d = 1/p\) in pc, converted to AU for the luminosity formula
- \(L/L_\odot = (F_\star/F_\odot)(d/1\ \text{AU})^2\) using the flux ratio method
- \(T = b/\lambda_{\text{peak}}\) from Wien’s law
- \(R/R_\odot\) from Stefan-Boltzmann ratio form
- Fractional uncertainty chain: \(\delta d/d \approx \delta p/p\), then \(\delta L/L \approx 2\,\delta d/d\)
- Explicit O→M→I summary connecting observables to inferred properties
Given: \[p=0.050'',\qquad \frac{F_\star}{F_\odot}=1.5\times10^{-12},\qquad \lambda_{\text{peak}}=500\ \text{nm}.\]
(a) Distance \[d=\frac{1}{p}=\frac{1}{0.050\ \text{arcsec}}=20\ \text{pc}.\] \[d=20\ \text{pc}\times 206{,}265\ \frac{\text{AU}}{\text{pc}}=4.1253\times10^6\ \text{AU}.\]
(b) Luminosity ratio
Both factors are dimensionless (flux ratio and distance ratio), so the result is a dimensionless luminosity ratio: \[\frac{L_\star}{L_\odot} =\frac{F_\star}{F_\odot}\left(\frac{d_\star}{1\ \text{AU}}\right)^2 =\left(1.5\times10^{-12}\right)\left(\frac{4.1253\times10^6\ \text{AU}}{1\ \text{AU}}\right)^2 =\left(1.5\times10^{-12}\right)\left(1.7018\times10^{13}\right) \approx25.5.\]
(c) Temperature \[T=\frac{b}{\lambda_{\text{peak}}} =\frac{2.898\times10^6\ \text{nm}\cdot\text{K}}{500\ \text{nm}} =5796\ \text{K}\approx5800\ \text{K}.\]
So \[\frac{T}{T_\odot}\approx\frac{5796\ \text{K}}{5800\ \text{K}}\approx1.00.\]
(d) Radius ratio \[\frac{R}{R_\odot} =\left[\frac{L_\star/L_\odot}{(T/T_\odot)^4}\right]^{1/2} =\left[\frac{25.5}{(1.00)^4}\right]^{1/2} =\left[25.5\right]^{1/2} \approx5.05.\]
(e) Uncertainty propagation
Given \(p=0.050''\pm0.003''\): \[\frac{\delta p}{p}=\frac{0.003''}{0.050''}=0.06=6\%.\]
For \(d=1/p\) (power-law error propagation: if \(d \propto p^{-1}\), then \(\delta d/d = |\!-\!1|\,\delta p/p\)): \[\frac{\delta d}{d}\approx\frac{\delta p}{p}\approx6\%.\]
For \(L\propto d^2\) at fixed flux (power-law propagation: exponent 2 doubles the fractional uncertainty): \[\frac{\delta L}{L}\approx2\,\frac{\delta d}{d}\approx2\times6\%=12\%.\]
(f) O→M→I summary
We measured observables (parallax, flux ratio, and peak wavelength). We applied models (parallax geometry, inverse-square flux spreading, and thermal-radiation scaling). From those, we inferred intrinsic properties: distance, luminosity, temperature, and radius.
Unit check: Each step produces dimensionless ratios — \(d\) in pc or AU (defined units), \(L/L_\odot\) (dimensionless), \(T\) in K (from nm·K/nm), \(R/R_\odot\) (dimensionless). Fractional uncertainties \(\delta d/d\) and \(\delta L/L\) are also dimensionless. Every intermediate result carries the expected units or is properly dimensionless. \(\checkmark\)
\[ \boxed{d=20\ \text{pc}=4.1253\times10^6\ \text{AU}},\quad \boxed{L_\star/L_\odot\approx25.5},\quad \boxed{T\approx5800\ \text{K}},\quad \boxed{R/R_\odot\approx5.05},\quad \boxed{\delta d/d\approx6\%,\ \delta L/L\approx12\%} \]
Sanity check: \(T \approx 5800\ \text{K} \approx T_\odot\) (solar temperature) with \(R \approx 5\,R_\odot\) suggests a subgiant or early red giant — a star that has expanded off the main sequence while maintaining roughly solar surface temperature. Its luminosity \(L \approx 25\,L_\odot\) is self-consistent: \(L \propto R^2 T^4\) gives \(25 \approx 5^2 \times 1^4 = 25\). \(\checkmark\) The 6% distance uncertainty propagating to 12% luminosity uncertainty also makes physical sense: luminosity depends on \(d^2\), so uncertainty doubles. \(\checkmark\)