Homework 5 Solutions
Masses from Motion + Magnitudes/HR Synthesis (Midterm 1 Review)
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 — Masses from Motion + Magnitudes
Problem 1 — Why “Weighing a Star” Is an Inference
Restatement: Explain how stellar masses are determined without direct weighing.
Key insight: Mass is inferred from dynamical observables (positions, periods, radial velocities) through Newton/Kepler models.
Answer:
- Example direct observable: orbital period, angular separation, or radial-velocity amplitude from spectral-line Doppler shifts.
- Physical model: Newtonian gravity + Kepler’s laws.
- This is still measurement because we directly measure observables with instruments and use a tested physical law to map those observables to mass. In science, “measured” often means model-mediated inference, not literal direct contact. The critical check is whether the model is valid for the system and whether uncertainties are tracked.
Sanity check: If independent observables (imaging orbit + RV curve) give consistent masses, the inference is behaving like a reliable measurement. \(\checkmark\)
Common misconception: “Inferred” means “guessed.” In reality, inferred quantities can be tightly constrained and highly reliable.
Problem 2 — Total Mass from a Visual Binary
Given: \(P = 8.0\,\text{yr}\), \(a = 4.0\,\text{AU}\).
Find: \(M_1 + M_2\), then each mass if equal.
Equation: \[ \frac{M_1 + M_2}{M_\odot} = \frac{\left(a/\text{AU}\right)^3}{\left(P/\text{yr}\right)^2} \]
Steps:
(a) \[ \frac{M_1 + M_2}{M_\odot} = \frac{4.0^3}{8.0^2} = \frac{64}{64} = 1.0 \]
\[ \boxed{M_1 + M_2 = 1.0\,M_\odot} \]
(b) Equal masses: \[ M_1 = M_2 = \frac{1.0\,M_\odot}{2} = 0.50\,M_\odot \]
\[ \boxed{M_1 = M_2 = 0.50\,M_\odot} \]
(c) The total mass is equal to the Sun’s mass.
Unit check: \((a/\text{AU})\) and \((P/\text{yr})\) are dimensionless; result is in solar masses. \(\checkmark\)
Sanity check: \(a\) and \(P\) were chosen so \(a^3 = P^2\); a total mass near \(1\,M_\odot\) is expected. \(\checkmark\)
Problem 3 — Mass Ratio from SB2 Velocities
Given: \(K_1 = 30\,\text{km/s}\), \(K_2 = 45\,\text{km/s}\), \(M_1 + M_2 = 3.0\,M_\odot\).
Find: \(M_2/M_1\), then \(M_1\), \(M_2\).
Equation: \[ \frac{M_2}{M_1} = \frac{K_1}{K_2} \]
Steps:
(a) \[ \frac{M_2}{M_1} = \frac{30}{45} = \frac{2}{3} = 0.667 \]
\[ \boxed{\frac{M_2}{M_1} = 0.667} \]
(b) Let \(M_2 = 0.667\,M_1\). Then: \[ M_1 + M_2 = M_1 + 0.667M_1 = 1.667M_1 = 3.0\,M_\odot \] \[ M_1 = \frac{3.0}{1.667}M_\odot = 1.8\,M_\odot \] \[ M_2 = 3.0 - 1.8 = 1.2\,M_\odot \]
\[ \boxed{M_1 = 1.8\,M_\odot,\quad M_2 = 1.2\,M_\odot} \]
(c) Star 1 is more massive. The more massive component has the smaller orbital speed/amplitude.
Unit check: \(K_1/K_2\) is dimensionless; masses remain in \(M_\odot\). \(\checkmark\)
Sanity check: Since \(K_1 < K_2\), star 1 should be more massive; \(1.8 > 1.2\) is consistent. \(\checkmark\)
Problem 5 — Required Capstone: Eclipsing SB2 Full Inference Chain
Restatement: Use \(P\), \(a\), and SB2 velocity amplitudes to infer component masses and relative luminosities.
Key elements:
- Use Kepler solar-unit form for total mass
- Use \(K\) ratio for mass ratio
- Solve the two-equation system for \(M_A\), \(M_B\)
- Convert mass ratio to luminosity ratio with \(L \propto M^{3.5}\)
- Clearly separate measured observables from inferred quantities
Sample response:
(a) Total mass \[ \frac{M_A + M_B}{M_\odot} = \frac{\left(2.00\,\text{AU}/1\,\text{AU}\right)^3}{\left(2.00\,\text{yr}/1\,\text{yr}\right)^2} = \frac{(2.00)^3}{(2.00)^2} = \frac{8}{4} = 2.00 \]
\[ \boxed{M_A + M_B = 2.00\,M_\odot} \]
(b) Mass ratio and individual masses \[ \frac{M_B}{M_A} = \frac{K_A}{K_B} = \frac{40\,\text{km/s}}{20\,\text{km/s}} = 2.0 \] So \(M_B = 2M_A\). Then: \[ M_A + M_B = M_A + 2M_A = 3M_A = 2.00\,M_\odot \] \[ M_A = \frac{2.00}{3}\,M_\odot = 0.667\,M_\odot, \qquad M_B = 1.333\,M_\odot \]
\[ \boxed{M_A = 0.667\,M_\odot,\quad M_B = 1.333\,M_\odot} \]
(c) Star B is more massive, and star A moves faster (\(K_A > K_B\)).
(d) Luminosity ratio \[ \frac{L_B}{L_A} \approx \left(\frac{M_B}{M_A}\right)^{3.5} = 2.0^{3.5} \approx 11.3 \]
\[ \boxed{\frac{L_B}{L_A} \approx 11} \]
So B should dominate the total luminosity.
(e) Observable → model → inference chain: We directly observe \(P\), \(a\), and Doppler amplitudes \(K_A\), \(K_B\). Kepler’s law maps \((P,a)\) to total mass, and center-of-mass dynamics maps \(K_A/K_B\) to mass ratio. Combining those gives individual masses. Then the main-sequence scaling \(L\propto M^{3.5}\) gives luminosity ratio. Key assumptions include Newtonian two-body dynamics, reliable orbital parameter extraction, and that both stars are main-sequence stars so the adopted mass-luminosity scaling applies.
Unit check: \((a/\text{AU})\), \((P/\text{yr})\), and \((K_A/K_B)\) are dimensionless ratios, so the Kepler and mass-ratio outputs are dimensionless until we multiply by \(M_\odot\). The luminosity ratio \(L_B/L_A\) is dimensionless. \(\checkmark\)
Sanity check: The more massive star should move more slowly, and the ratio \(K_A/K_B=2\) implies \(M_B/M_A=2\), which matches the solved masses. \(\checkmark\)
Grading guidance: Full credit requires all links in the chain (total mass, mass ratio, component masses, luminosity ratio, and one explicit assumption).
Problem 6 — Apparent vs. Absolute Magnitude Misconception
Restatement: Correct a misuse of apparent magnitude for intrinsic brightness claims.
Key insight: Apparent magnitude depends on both luminosity and distance; intrinsic brightness requires absolute magnitude or luminosity.
Answer:
- The statement is not always valid because \(m\) measures how bright a star appears from Earth, not how much power it emits.
- You need distance (or directly absolute magnitude \(M\)) to compare intrinsic brightness.
- Apparent magnitude \(m\) is observed brightness at Earth; absolute magnitude \(M\) is the magnitude a source would have at \(10\,\text{pc}\).
Sanity check: A nearby low-luminosity star can appear brighter than a distant high-luminosity star, so apparent brightness alone cannot rank intrinsic power. \(\checkmark\)
Common misconception: Brighter-looking stars are always intrinsically brighter.
Problem 7 — Distance Modulus Practice
Given:
- \(m = 11.2\), \(M = 1.2\)
- \(d = 250\,\text{pc}\), \(m = 14.0\)
- Magnitude difference \(\Delta m = 5.0\)
Find: distance, absolute magnitude, and flux ratio.
Equation: \[ m - M = 5\log_{10}\!\left(\frac{d}{10\,\text{pc}}\right) \]
Steps:
(a) \[ 11.2 - 1.2 = 10.0 = 5\log_{10}\!\left(\frac{d}{10\,\text{pc}}\right) \] \[ \log_{10}\!\left(\frac{d}{10\,\text{pc}}\right)=2.0 \Rightarrow \frac{d}{10\,\text{pc}}=10^2=100 \Rightarrow d=1000\,\text{pc} \]
\[ \boxed{d = 1.0\times10^3\,\text{pc}} \]
(b) \[ m - M = 5\log_{10}\!\left(\frac{250}{10}\right)=5\log_{10}(25)=5(1.39794)=6.99 \] \[ M = m - (m-M)=14.0-6.99=7.01 \]
\[ \boxed{M \approx 7.0\,\text{mag}} \]
(c) Magnitude-flux relation: \[ \Delta m = -2.5\log_{10}\!\left(\frac{F_2}{F_1}\right) \] For \(\Delta m=5.0\): \[ \frac{F_1}{F_2} = 10^{5/2.5}=10^2=100 \]
\[ \boxed{\text{A 5-mag difference corresponds to a factor of }100\text{ in observed flux; the brighter (smaller-}m\text{) star has }100\times\text{ more flux.}} \]
Unit check: Distance modulus argument uses a distance ratio (dimensionless). \(\checkmark\)
Sanity check: Larger \(m-M\) means farther distance; \(m-M=10\) giving \(1000\,\text{pc}\) is consistent. \(\checkmark\)
Part B — HR Synthesis + Cumulative Midterm Bridge
Problem 8 — Main-Sequence Fitting by Magnitude Offset
Restatement: Use a vertical magnitude shift in matched main sequences to infer distance ratio and distance.
Key elements:
- Use \(\Delta m = 5\log_{10}(d_t/d_r)\) when intrinsic sequence is assumed same
- Convert ratio to distance
- Explain extinction bias direction
Sample response:
(a) \[ \Delta m = 5\log_{10}\!\left(\frac{d_{\text{target}}}{d_{\text{ref}}}\right) \Rightarrow 3.0 = 5\log_{10}\!\left(\frac{d_{\text{target}}}{d_{\text{ref}}}\right) \] \[ \log_{10}\!\left(\frac{d_{\text{target}}}{d_{\text{ref}}}\right)=0.6 \Rightarrow \frac{d_{\text{target}}}{d_{\text{ref}}}=10^{0.6}=3.98 \]
\[ \boxed{\frac{d_{\text{target}}}{d_{\text{ref}}}\approx3.98} \]
(b) \[ d_{\text{target}} = 3.98\times100\,\text{pc}=398\,\text{pc} \]
\[ \boxed{d_{\text{target}}\approx4.0\times10^2\,\text{pc}} \]
(c) Extinction makes stars appear fainter, increasing the observed magnitude offset. If uncorrected, this mimics larger distance, so the inferred distance is too large.
Unit check: \(\Delta m\) (mag) maps to \(\log_{10}(d_{\text{target}}/d_{\text{ref}})\), so only a dimensionless distance ratio can appear inside the logarithm. Multiplying that ratio by \(d_{\text{ref}}\) returns parsecs. \(\checkmark\)
Sanity check: Positive magnitude shift should mean farther target cluster; ratio \(>1\) confirms this. \(\checkmark\)
Grading guidance: Full credit requires correct ratio math and correct extinction-bias direction.
Problem 9 — Bridge to Module 3: Mass, Luminosity, and Lifetime
Given: \[ \frac{t_{\text{MS}}}{t_\odot} \approx \left(\frac{M}{M_\odot}\right)^{-2.5}, \qquad t_\odot=10\,\text{Gyr} \]
Find: Lifetimes for \(2.0\,M_\odot\) and \(5.0\,M_\odot\), plus age interpretation.
Steps:
(a) For \(2.0\,M_\odot\): \[ \frac{t_{\text{MS}}}{10\,\text{Gyr}} = 2.0^{-2.5}=0.1768 \Rightarrow t_{\text{MS}}=1.77\,\text{Gyr} \]
For \(5.0\,M_\odot\): \[ \frac{t_{\text{MS}}}{10\,\text{Gyr}} = 5.0^{-2.5}=0.0179 \Rightarrow t_{\text{MS}}=0.179\,\text{Gyr} \]
\[ \boxed{t_{\text{MS}}(2M_\odot)\approx1.8\,\text{Gyr},\quad t_{\text{MS}}(5M_\odot)\approx0.18\,\text{Gyr}} \]
(b) If turnoff mass is about \(2.0\,M_\odot\), cluster age is about the \(2.0\,M_\odot\) main-sequence lifetime:
\[ \boxed{\text{Age}\approx1.8\,\text{Gyr}} \]
(c) Massive stars burn fuel much faster and leave the main sequence quickly, so old clusters no longer contain many high-mass main-sequence stars.
Unit check: Lifetime ratio is dimensionless; multiplying by \(10\,\text{Gyr}\) returns time units. \(\checkmark\)
Sanity check: Higher mass gives shorter lifetime; \(0.18\,\text{Gyr} < 1.8\,\text{Gyr} < 10\,\text{Gyr}\) is correct ordering. \(\checkmark\)
Problem 10 — Cumulative Inference: One System, Multiple Models
Restatement: Combine orbital, RV, and photometric information to infer masses and intrinsic brightness, and identify uncertainty-reducing follow-up.
Key elements:
- Kepler total mass from \(P,a\)
- Mass ratio from \(K_1,K_2\)
- Absolute magnitude from distance modulus
- Explicit measured vs inferred separation
- Reasonable follow-up observation with justification
Sample response:
(a) Total and individual masses \[ \frac{M_1+M_2}{M_\odot} =\frac{(4.0\,\text{AU}/1\,\text{AU})^3}{(4.0\,\text{yr}/1\,\text{yr})^2} =\frac{4.0^3}{4.0^2} =\frac{64}{16} =4.0 \Rightarrow M_1+M_2=4.0\,M_\odot \]
\[ \frac{M_2}{M_1} =\frac{K_1}{K_2} =\frac{25\,\text{km/s}}{20\,\text{km/s}} =1.25 \] So \(M_2=1.25M_1\) and: \[ M_1+1.25M_1=2.25M_1=4.0\,M_\odot \Rightarrow M_1=1.78\,M_\odot, \quad M_2=2.22\,M_\odot \]
\[ \boxed{M_1\approx1.78\,M_\odot,\quad M_2\approx2.22\,M_\odot} \]
(b) Combined absolute magnitude \[ M = m - (m-M) = 8.0 - 10.0 = -2.0 \]
\[ \boxed{M_{\text{combined}}=-2.0\,\text{mag}} \]
(c) Measured vs inferred: Measured directly are \(P\), \(a\), \(K_1\), \(K_2\), and apparent magnitude \(m\). The cluster distance modulus used here is an inferred intermediate from prior photometric modeling (not a raw direct observable). Using Kepler/Newton, we infer total mass from \(P\) and \(a\). Using center-of-mass dynamics, we infer mass ratio from \(K_1/K_2\), then individual masses. Using the distance modulus relation, we infer intrinsic brightness as absolute magnitude. Each step is model-dependent but testable, and each has uncertainty.
(d) Follow-up observation: A strong follow-up is eclipse photometry (or other inclination constraint), because it reduces geometric uncertainty and enables tighter component-property inference (masses/radii/luminosities) rather than only system-level estimates.
Unit check: Kepler and RV steps use dimensionless ratios to return masses in \(M_\odot\); the distance-modulus subtraction returns magnitude units (mag). \(\checkmark\)
Sanity check: The larger-mass component should have smaller velocity amplitude; \(M_2>M_1\) and \(K_2<K_1\) are consistent. \(\checkmark\)
Grading guidance: Full credit requires coherent chain logic, not just numerical answers.