Module 1 Practice Problems
Multi-Concept Synthesis Problems
How to Use These Problems
These problems go beyond single-lecture practice. They require you to: - Identify which tools apply to a given situation - Combine multiple concepts in a single problem - Check whether your answer is physically reasonable
Difficulty Legend: - ★ Warm-Up — Single concept review - ★★ Standard — Two concepts; exam-level - ★★★ Challenge — Multiple concepts; beyond exam level
- Read carefully — What are you given? What are you asked to find?
- Identify the physics — Which equations/concepts apply?
- Set up before calculating — Write the equation with symbols first
- Check units — Do they work out correctly?
- Sanity check — Is your answer physically reasonable?
Warm-Up Problems (★)
These review single concepts. If you struggle here, revisit the relevant lecture.
Concept: L8 (Blackbody Radiation)
A star has a peak emission wavelength of 350 nm.
- What is the star’s surface temperature?
- Is this star hotter or cooler than the Sun (peak at 500 nm)?
(a) Wien’s Law: \(\lambda_{peak} = \frac{b}{T}\) where \(b = 2.9 \times 10^6\) nm·K
\[T = \frac{b}{\lambda_{peak}} = \frac{2.9 \times 10^6 \text{ nm·K}}{350 \text{ nm}} = 8300 \text{ K}\]
(b) This star is hotter than the Sun. Shorter peak wavelength → higher temperature.
Concept: L10 (Doppler Effect)
A star’s calcium H line (rest wavelength 396.8 nm) is observed at 396.6 nm.
- Is the star moving toward or away from us?
- Calculate the star’s radial velocity.
(a) The observed wavelength is shorter than the rest wavelength (396.6 < 396.8), so this is a blueshift. The star is moving toward us.
(b) Doppler formula: \(\frac{\Delta\lambda}{\lambda_0} = \frac{v}{c}\)
\[\Delta\lambda = 396.6 - 396.8 = -0.2 \text{ nm}\]
\[v = c \cdot \frac{\Delta\lambda}{\lambda_0} = (3 \times 10^5 \text{ km/s}) \cdot \frac{-0.2}{396.8} = -151 \text{ km/s}\]
The negative sign confirms approach. The star is moving toward us at about 150 km/s.
Concept: L5 (Kepler’s Laws)
Planet X orbits its star at 4 AU. Planet Y orbits the same star at 1 AU.
How many times longer is Planet X’s orbital period compared to Planet Y’s?
Kepler’s Third Law (ratio form): \(\frac{P_X^2}{P_Y^2} = \frac{a_X^3}{a_Y^3}\)
\[\frac{P_X^2}{P_Y^2} = \frac{4^3}{1^3} = \frac{64}{1}\]
\[\frac{P_X}{P_Y} = \sqrt{64} = 8\]
Planet X’s period is 8 times longer than Planet Y’s.
Concept: L8 (Blackbody Radiation)
Two stars have identical radii, but Star A is twice as hot as Star B.
How many times more luminous is Star A compared to Star B?
Stefan-Boltzmann Law: \(L = 4\pi R^2 \sigma T^4\)
If radii are equal, luminosity scales as \(T^4\):
\[\frac{L_A}{L_B} = \frac{T_A^4}{T_B^4} = \left(\frac{T_A}{T_B}\right)^4 = 2^4 = 16\]
Star A is 16 times more luminous than Star B. Temperature has a huge effect!
Concept: L7 (Light as Information)
Star A and Star B have identical luminosities. Star A is 10 pc away; Star B is 100 pc away.
How many times brighter does Star A appear than Star B?
Inverse-square law: \(F = \frac{L}{4\pi d^2}\)
With equal luminosity, brightness (flux) scales as \(1/d^2\):
\[\frac{F_A}{F_B} = \frac{d_B^2}{d_A^2} = \frac{100^2}{10^2} = \frac{10000}{100} = 100\]
Star A appears 100 times brighter than Star B.
Concept: L4 (Moon Geometry)
You observe the Moon rising at midnight. What phase is the Moon in?
At midnight, you’re facing away from the Sun (Sun is on the opposite side of Earth). If the Moon is rising at that moment, it’s 180° from the Sun in the sky.
A Moon that is opposite the Sun is fully illuminated from our perspective.
The Moon is in the full phase.
Memory trick: Full Moon rises at sunset, is highest at midnight, and sets at sunrise.
Concept: L9 (Spectral Lines)
A star shows very strong hydrogen Balmer lines. Another star shows no hydrogen lines but strong titanium oxide (TiO) molecular bands.
Which star is hotter? Explain.
The star with strong Balmer lines is hotter (~10,000 K, A-type).
The star with TiO bands is cooler (~3,000 K, M-type).
Reasoning:
- Balmer lines require electrons in the n=2 state, which needs moderate temperature (~10,000 K optimal)
- TiO molecules are destroyed at high temperatures; they only survive in cool stellar atmospheres
- The presence of molecular bands is a signature of cool stars (K and M types)
Standard Problems (★★)
These combine two concepts. This is exam-level difficulty.
Concepts: L8 (Wien + Stefan-Boltzmann)
Star A has a peak wavelength of 290 nm and radius 2 R☉. Star B has a peak wavelength of 580 nm and radius 8 R☉.
Which star is more luminous, and by what factor?
Step 1: Find temperatures using Wien’s Law
\[T_A = \frac{2.9 \times 10^6}{290} = 10{,}000 \text{ K}\] \[T_B = \frac{2.9 \times 10^6}{580} = 5{,}000 \text{ K}\]
Star A is twice as hot as Star B.
Step 2: Compare luminosities using Stefan-Boltzmann
\[\frac{L_A}{L_B} = \frac{R_A^2 T_A^4}{R_B^2 T_B^4} = \frac{2^2 \cdot (10{,}000)^4}{8^2 \cdot (5{,}000)^4}\]
\[= \frac{4 \cdot 10^{16}}{64 \cdot 6.25 \times 10^{14}} = \frac{4 \times 10^{16}}{4 \times 10^{16}} = 1\]
The stars have equal luminosity! Star A is smaller but hotter; Star B is larger but cooler. The effects cancel.
Concepts: L7 (Inverse-Square) + L8 (Stefan-Boltzmann)
Two stars appear equally bright in the sky. Star A has temperature 10,000 K and radius 2 R☉. Star B has temperature 5,000 K and radius 4 R☉.
If Star A is at a distance of 50 pc, how far away is Star B?
Step 1: Compare intrinsic luminosities
\[\frac{L_A}{L_B} = \frac{R_A^2 T_A^4}{R_B^2 T_B^4} = \frac{2^2 \cdot (10{,}000)^4}{4^2 \cdot (5{,}000)^4} = \frac{4 \cdot 10^{16}}{16 \cdot 6.25 \times 10^{14}} = \frac{4 \times 10^{16}}{10^{16}} = 4\]
Star A is 4 times more luminous than Star B.
Step 2: If they appear equally bright, the more luminous star must be farther away
\[\frac{F_A}{F_B} = 1 = \frac{L_A/d_A^2}{L_B/d_B^2} = \frac{L_A}{L_B} \cdot \frac{d_B^2}{d_A^2}\]
\[1 = 4 \cdot \frac{d_B^2}{50^2}\]
\[d_B^2 = \frac{50^2}{4} = 625\]
\[d_B = 25 \text{ pc}\]
Star B is at 25 pc — closer because it’s intrinsically fainter.
Concepts: L10 (Doppler) + L12 (Exoplanet Detection)
An exoplanet transits its Sun-like star with a depth of 0.25%. The star’s radial velocity varies with amplitude 20 m/s.
- What is the planet’s radius (in Earth radii)?
- Is this likely a rocky planet or a gas giant? Explain your reasoning.
(a) Transit depth = \((R_p/R_*)^2 = 0.0025\)
\[\frac{R_p}{R_*} = \sqrt{0.0025} = 0.05\]
For a Sun-like star (\(R_* = R_☉ = 109 R_⊕\)):
\[R_p = 0.05 \times 109 R_⊕ = 5.5 R_⊕\]
The planet has radius ~5.5 Earth radii.
(b) This is likely a gas-dominated planet (mini-Neptune or Neptune-like).
Reasoning:
- Rocky planets (super-Earths) typically have R < 1.5–2 R⊕
- Planets with R > 4 R⊕ almost always have thick H/He envelopes
- The 20 m/s RV signal is moderate — consistent with a low-density planet
- To confirm, we’d need the mass from the full RV solution to calculate density
Concepts: L6 (Newton’s Gravity) + L5 (Kepler’s Laws)
A satellite orbits Mars at a distance of 9,380 km from Mars’s center with a period of 7.65 hours.
- Use Newton’s form of Kepler’s Third Law to calculate Mars’s mass.
- Compare to Earth’s mass (6.0 × 10²⁴ kg). How many times more massive is Earth than Mars?
Hint: G = 6.67 × 10⁻¹¹ N·m²/kg², convert units carefully.
(a) Newton’s form: \(M = \frac{4\pi^2 a^3}{GP^2}\)
Convert to SI units:
- \(a = 9380 \text{ km} = 9.38 \times 10^6 \text{ m}\)
- \(P = 7.65 \text{ hr} = 7.65 \times 3600 = 27{,}540 \text{ s}\)
\[M = \frac{4\pi^2 (9.38 \times 10^6)^3}{(6.67 \times 10^{-11})(27{,}540)^2}\]
\[M = \frac{4\pi^2 \times 8.26 \times 10^{20}}{6.67 \times 10^{-11} \times 7.58 \times 10^{8}}\]
\[M = \frac{3.26 \times 10^{22}}{5.06 \times 10^{-2}} = 6.4 \times 10^{23} \text{ kg}\]
(b) \(\frac{M_{Earth}}{M_{Mars}} = \frac{6.0 \times 10^{24}}{6.4 \times 10^{23}} \approx 9.4\)
Earth is about 9-10 times more massive than Mars.
Concepts: L10 (Doppler) + L6 (Newton’s Gravity)
Astronomers observe a star with periodic Doppler shifts. The star’s spectral lines shift by ±0.01 nm around a rest wavelength of 500 nm, with a period of 10 days.
- What is the star’s radial velocity amplitude?
- Is this more likely caused by a planet or a stellar companion? Explain.
(a) Doppler formula:
\[v = c \cdot \frac{\Delta\lambda}{\lambda_0} = (3 \times 10^5 \text{ km/s}) \cdot \frac{0.01}{500} = 6 \text{ km/s}\]
(b) This is likely a stellar companion (binary star), not a planet.
Reasoning: - Planets cause stellar wobbles of ~1-100 m/s (Jupiter causes ~13 m/s on the Sun) - This signal is 6 km/s = 6000 m/s — much too large for a planet - A stellar companion with comparable mass would cause km/s-scale wobbles
The 10-day period is also consistent with a close binary star system.
Concepts: L9 (Spectral Lines) + L8 (Blackbody)
A star’s spectrum shows very strong hydrogen Balmer absorption lines. Its peak emission is at 290 nm.
- What is the star’s approximate spectral type?
- Is this consistent with the strong Balmer lines? Explain.
(a) From Wien’s Law:
\[T = \frac{2.9 \times 10^6}{290} = 10{,}000 \text{ K}\]
This corresponds to spectral type A (A stars have T ~ 7,500–10,000 K).
(b) Yes, this is consistent. Balmer lines are strongest around 10,000 K because: - Hot enough that many hydrogen atoms have electrons in the n=2 level (ready to absorb Balmer photons) - Not so hot that hydrogen is ionized (which would eliminate Balmer absorption)
A-type stars are famous for having the strongest hydrogen lines in the OBAFGKM sequence.
Challenge Problems (★★★)
These require multiple concepts and careful reasoning. They go beyond typical exam questions.
Concepts: L8 + L9 + L10 + L6
A binary star system is observed: - Star A’s spectrum peaks at 250 nm and shows weak hydrogen lines - Star B’s spectrum peaks at 725 nm and shows strong TiO molecular bands - The orbital period is 50 years and the semi-major axis is 20 AU - Star A’s Doppler shift amplitude is 1/4 of Star B’s amplitude
- Estimate the temperatures of both stars.
- What spectral types are these stars?
- Calculate the total system mass.
- What is the mass ratio \(M_A/M_B\)?
(a) Temperatures from Wien’s Law:
\[T_A = \frac{2.9 \times 10^6}{250} = 11{,}600 \text{ K}\] \[T_B = \frac{2.9 \times 10^6}{725} = 4{,}000 \text{ K}\]
(b) Spectral types: - Star A (~11,600 K, weak H lines): B-type (or late A) - Star B (~4,000 K, TiO bands): M-type (TiO appears in cool stars)
(c) Total mass from Newton-Kepler:
\[M_{total} = \frac{4\pi^2 a^3}{GP^2}\]
Using solar units where \(M\) in \(M_☉\), \(a\) in AU, \(P\) in years:
\[M_{total} = \frac{a^3}{P^2} = \frac{20^3}{50^2} = \frac{8000}{2500} = 3.2 \, M_☉\]
(d) Mass ratio from velocity ratio:
Both stars orbit the center of mass. The less massive star moves faster:
\[\frac{v_A}{v_B} = \frac{M_B}{M_A} = \frac{1}{4}\]
Therefore \(M_A/M_B = 4\).
With \(M_A + M_B = 3.2 \, M_☉\) and \(M_A = 4 M_B\): \[5 M_B = 3.2 \implies M_B = 0.64 \, M_☉, \quad M_A = 2.56 \, M_☉\]
Concepts: L8 + L12 + L11
A star currently has luminosity L = 1 L☉ and hosts a planet at 1 AU (in the habitable zone). As the star ages, it will evolve into a red giant with L = 100 L☉.
When the star becomes a red giant, at what distance would a planet need to orbit to receive the same flux Earth receives from the Sun today?
The original planet at 1 AU — by what factor does the flux it receives increase?
What happens to the original habitable zone planet as the star evolves?
(a) For Earth-like flux: \(F = L/(4\pi d^2)\)
Setting \(F_{new} = F_{Earth}\):
\[\frac{L_{RG}}{d_{new}^2} = \frac{L_☉}{(1 \text{ AU})^2}\]
\[d_{new}^2 = \frac{L_{RG}}{L_☉} \times (1 \text{ AU})^2 = 100 \times 1 = 100 \text{ AU}^2\]
\[d_{new} = 10 \text{ AU}\]
The habitable zone moves out to ~10 AU (Saturn’s distance!).
(b) Flux at 1 AU when L = 100 L☉:
\[\frac{F_{new}}{F_{old}} = \frac{L_{new}}{L_{old}} = \frac{100 L_☉}{1 L_☉} = 100\]
The planet receives 100 times more flux.
(c) The planet at 1 AU becomes uninhabitable — it would experience extreme heating, likely losing any atmosphere and oceans. For rocky planets, surface temperatures would rise dramatically. The planet that was Earth-like becomes more Venus-like or worse.
This is relevant to Earth’s future: in ~5 billion years, the Sun will become a red giant and Earth will be scorched.
Concepts: L5 + L6 + L12
You want to design an exoplanetary system where:
- The host star has mass 0.5 M☉
- Planet A is in the habitable zone (receives Earth-like flux)
- Planet B is at the frost line (~3 AU equivalent for our Sun)
For a 0.5 M☉ star, the habitable zone is at ~0.5 AU (since L ∝ M³·⁵ approximately for main-sequence stars, so L ~ 0.18 L☉, and HZ distance scales as √L).
- What is Planet A’s orbital period?
- If the frost line scales with stellar luminosity the same way the HZ does, where is this star’s frost line?
- What is the orbital period at the frost line?
(a) Newton’s form of Kepler III in solar units:
\[P^2 = \frac{a^3}{M} = \frac{(0.5)^3}{0.5} = \frac{0.125}{0.5} = 0.25\]
\[P = 0.5 \text{ years} = 6 \text{ months}\]
(b) The frost line for the Sun is at ~3 AU. For a star with L ~ 0.18 L☉:
\[d_{frost} = 3 \text{ AU} \times \sqrt{\frac{L}{L_☉}} = 3 \times \sqrt{0.18} \approx 3 \times 0.42 = 1.3 \text{ AU}\]
(c) At 1.3 AU:
\[P^2 = \frac{(1.3)^3}{0.5} = \frac{2.2}{0.5} = 4.4\]
\[P = 2.1 \text{ years}\]
Summary: This small star has a compact planetary system — habitable zone at 0.5 AU (6-month year), frost line at 1.3 AU (2-year period). Gas giants would form closer than Jupiter does in our system!
(a) Temperatures from Wien’s Law:
\[T_A = \frac{2.9 \times 10^6}{250} = 11{,}600 \text{ K}\] \[T_B = \frac{2.9 \times 10^6}{725} = 4{,}000 \text{ K}\]
(b) Spectral types: - Star A (~11,600 K, weak H lines): B-type (or late A) - Star B (~4,000 K, TiO bands): M-type (TiO appears in cool stars)
(c) Total mass from Newton-Kepler:
\[M_{total} = \frac{4\pi^2 a^3}{GP^2}\]
Using solar units where \(M\) in \(M_☉\), \(a\) in AU, \(P\) in years:
\[M_{total} = \frac{a^3}{P^2} = \frac{20^3}{50^2} = \frac{8000}{2500} = 3.2 \, M_☉\]
(d) Mass ratio from velocity ratio:
Both stars orbit the center of mass. The less massive star moves faster:
\[\frac{v_A}{v_B} = \frac{M_B}{M_A} = \frac{1}{4}\]
Therefore \(M_A/M_B = 4\).
With \(M_A + M_B = 3.2 \, M_☉\) and \(M_A = 4 M_B\): \[5 M_B = 3.2 \implies M_B = 0.64 \, M_☉, \quad M_A = 2.56 \, M_☉\]
Concepts: L8 + L11 + L12
A star with luminosity \(L = 4 L_☉\) has a planet orbiting at 2 AU.
- How does the flux received by this planet compare to Earth’s solar flux?
- Would this planet be inside, outside, or near the habitable zone?
- If the star’s luminosity increases to \(16 L_☉\) as it evolves, at what orbital distance would a planet receive Earth-like flux?
(a) Flux comparison:
Flux scales as \(F = L/(4\pi d^2)\). Using ratios:
\[\frac{F_{planet}}{F_{Earth}} = \frac{L_{star}}{L_☉} \cdot \frac{d_{Earth}^2}{d_{planet}^2} = \frac{4}{1} \cdot \frac{1^2}{2^2} = \frac{4}{4} = 1\]
The planet receives the same flux as Earth!
(b) Habitable zone:
Since the flux equals Earth’s, this planet is likely in or near the habitable zone — assuming similar atmospheric properties, it could potentially support liquid water.
(c) New habitable distance:
For Earth-like flux with \(L = 16 L_☉\):
\[\frac{F_{Earth}}{F_{new}} = 1 = \frac{16}{1} \cdot \frac{1^2}{d^2}\]
\[d^2 = 16 \implies d = 4 \text{ AU}\]
The habitable zone moves outward to 4 AU when the star’s luminosity quadruples.
Capstone: Real Data Problem
All Module 1 Concepts
You observe a star and collect the following data:
| Measurement | Value |
|---|---|
| Peak wavelength | 483 nm |
| Hα line (rest 656.3 nm) | Observed at 656.1 nm |
| Strong absorption lines | H, He I, Mg II |
| Apparent brightness | 100× fainter than a similar star at 10 pc |
Using only Module 1 tools, determine:
- The star’s surface temperature
- The star’s radial velocity (toward or away? how fast?)
- The star’s approximate spectral type
- The star’s distance
Show your work and explain your reasoning.
(a) Temperature from Wien’s Law:
\[T = \frac{2.9 \times 10^6 \text{ nm·K}}{483 \text{ nm}} = 6{,}000 \text{ K}\]
(b) Radial velocity from Doppler:
\[\Delta\lambda = 656.1 - 656.3 = -0.2 \text{ nm}\]
Blueshift → approaching.
\[v = c \cdot \frac{\Delta\lambda}{\lambda_0} = (3 \times 10^5) \cdot \frac{0.2}{656.3} = 91 \text{ km/s toward us}\]
(c) Spectral type:
- T ~ 6,000 K is consistent with late F or early G type
- Strong H lines + He I + Mg II supports F/G classification
- (The Sun is G2 at 5,800 K, so this is slightly hotter)
(d) Distance from inverse-square law:
If a star is 100× fainter but intrinsically identical:
\[\frac{F_1}{F_2} = \frac{d_2^2}{d_1^2} = 100\]
\[\frac{d_2}{d_1} = 10\]
If the comparison star is at 10 pc, this star is at 100 pc.
Answer Key Summary
| Problem | Key Answer |
|---|---|
| 1 | T = 8,300 K (hotter than Sun) |
| 2 | v = 150 km/s toward us |
| 3 | 8× longer period |
| 3b | 16× more luminous |
| 3c | 100× brighter |
| 3d | Full Moon |
| 3e | Balmer star is hotter (~10,000 K) |
| 4 | Equal luminosity |
| 4b | Star B at 25 pc |
| 4c | R ~ 5.5 R⊕; likely gas-dominated |
| 4d | Mars mass = 6.4 × 10²³ kg; Earth ~10× more massive |
| 5 | 6 km/s; stellar companion |
| 6 | A-type; yes, consistent |
| 7 | \(M_A\) = 2.56 M☉, \(M_B\) = 0.64 M☉ |
| 7b | HZ moves to 10 AU; planet at 1 AU gets 100× flux |
| 7c | HZ at 0.5 AU (P=6 mo); frost line at 1.3 AU (P=2.1 yr) |
| 8 | Same flux as Earth; 4 AU for evolved star |
| 9 | 6,000 K; 91 km/s toward; F/G type; 100 pc |
What’s Next?
If you’ve worked through these problems successfully, you’re ready for the Exam Prep materials and the Module 1 Exam.