Lecture 19 Solutions: Stellar Evolution

Practice Problem Solutions

Solutions to the Lecture 19 practice problems.
Author

Dr. Anna Rosen

Published

March 25, 2026

Note

Student note: These are model solutions written to show setup, units, and reasoning clearly. Your own work can be shorter as long as the physics, units, and logic are sound.

Use these after you try the problems yourself. The point is to compare your reasoning to a worked example, not to skip the thinking step.

Core Problems

Problem 1: Main-Sequence Lifetime Calculation

Restatement: A star has a mass of 3 \(M_\odot\).

  1. Use \(t_{\rm MS} \approx 10~\text{Gyr}(M/M_\odot)^{-2.5}\) to estimate its main-sequence lifetime.
  2. If the star formed at the same time as the Sun (4.6 Gyr ago), would it still be on the main sequence today?
  3. If not, what broad conclusion can you draw about its present evolutionary state?

Given: \(M = 3\,M_\odot\), \(\tau_\odot \approx 10\,\text{Gyr}\)

Equation: \[ t_{\rm MS} \approx 10~\text{Gyr}\left(\frac{M}{M_\odot}\right)^{-2.5} \]

  1. Using \(t_{\rm MS} = 10 \text{ Gyr} \times (M/1 M_\odot)^{-2.5}\):

\[t_{\rm MS} = 10 \text{ Gyr} \times (3)^{-2.5} = 10 \text{ Gyr} \times 0.064 = 0.64 \text{ Gyr} \approx 640 \text{ Myr}\]

  1. No. The star’s total main-sequence lifetime is about 640 Myr, but it has already existed for 4.6 Gyr, which is much longer than its main-sequence lifetime.

  2. It has already exhausted its core hydrogen and must be in a post-main-sequence stage, most plausibly a giant or later evolved phase.

Unit check: The mass ratio is dimensionless, so the result stays in units of time.

Sanity check: A star more massive than the Sun should live far less than 10 Gyr, so a lifetime of a few hundred million years is sensible.

Answer: A \(3\,M_\odot\) star stays on the main sequence for about \(0.64\ \text{Gyr}\), so it would no longer be a main-sequence star today if it formed when the Sun did.

Problem 2: The Red-Giant Sun

Restatement: In about 5 billion years, the Sun will exhaust its core hydrogen and become a red giant.

Suppose its luminosity becomes 2,000 \(L_\odot\) and its surface temperature drops to 3,500 K.

  1. Estimate the Sun’s future radius in units of \(R_\odot\).
  2. Explain in words why the radius must become so large.

Hint: Use the Stefan-Boltzmann ratio form: \(\frac{L_{\text{future}}}{L_{\odot}} = (R_{\text{future}}/R_{\odot})^2 (T_{\text{future}}/T_{\odot})^4\)

  1. Using the Stefan-Boltzmann ratio:

\[\frac{2,000}{1} = \left(\frac{R_{\text{future}}}{R_{\odot}}\right)^2 \left(\frac{3,500}{5,800}\right)^4\]

\[2,000 = \left(\frac{R_{\text{future}}}{R_{\odot}}\right)^2 (0.603)^4\]

\[2,000 = \left(\frac{R_{\text{future}}}{R_{\odot}}\right)^2 \times 0.132\]

\[\frac{R_{\text{future}}}{R_{\odot}} = \sqrt{\frac{2,000}{0.132}} = \sqrt{15,152} \approx 123 R_{\odot}\]

  1. The radius must become so large because luminosity depends on both temperature and surface area. If the surface cools from about 5,800 K to 3,500 K, the star emits much less energy per square meter, so it needs a much larger emitting area to reach a total luminosity of 2,000 \(L_\odot\).

Unit check: The ratio form uses dimensionless quantities, so the answer naturally comes out in units of \(R_\odot\).

Sanity check: Red giants are known to reach radii tens to hundreds of times the Sun’s, so about \(10^2\,R_\odot\) is the right scale.

Problem 3: Interpreting an H-R Diagram

Restatement: You observe a star that is cooler (lower temperature, farther right) than the Sun but has the same luminosity as the Sun (~1 \(L_\odot\)).

  1. Using \(L \propto R^2 T^4\), is this star larger or smaller than the Sun?

  2. What can you infer confidently from temperature and luminosity alone?

  3. What can you not determine from the H-R diagram alone?

  4. It must be larger than the Sun. If two stars have the same luminosity but one is cooler, the cooler one needs a larger surface area to radiate the same total power.

  5. From temperature and luminosity alone, you can infer that this star is cooler, has roughly solar luminosity, and therefore has a larger radius than the Sun. That places it away from the ordinary solar-like main sequence.

  6. You cannot determine its exact mass, exact age, or exact evolutionary history from the H-R diagram alone. Different kinds of evolved stars can overlap in similar parts of the diagram, so temperature and luminosity do not tell the whole story by themselves.

Problem 4: Cluster Age From Main-Sequence Turnoff

Restatement: An open cluster shows a main-sequence turnoff near spectral type F, with temperature about \(6500\ \text{K}\) and luminosity about \(4\,L_\odot\).

  1. Estimate the mass of the turnoff star.

  2. Use \(t_{\rm MS} \approx 10~\text{Gyr}(M/M_\odot)^{-2.5}\) to estimate the cluster age.

  3. Explain in one sentence why the turnoff point acts like a clock.

  4. A turnoff star with luminosity about \(4\,L_\odot\) should be somewhat more massive than the Sun. Using the rough main-sequence scaling \(L \propto M^{3.5}\) gives

\[ \frac{M}{M_\odot} \approx 4^{1/3.5} \approx 1.5, \]

so a reasonable estimate is \(1.5\,M_\odot\).

  1. The cluster age equals the main-sequence lifetime of the turnoff mass:

\[ t_{\rm MS} \approx 10(1.5)^{-2.5}\ \text{Gyr} \approx 3.6\ \text{Gyr} \]

The cluster is approximately 3.6 billion years old. That makes it an intermediate-age open cluster, not a globular cluster.

  1. The turnoff acts like a clock because it marks the most massive stars that are still on the main sequence, and more-massive stars have shorter main-sequence lifetimes.

Sanity check: An F-type turnoff should correspond to a cluster a few billion years old, not an extremely young cluster or an ancient globular cluster.

Problem 5: Two Stars, Same Cluster

Restatement: A young cluster contains both a 50 \(M_\odot\) star and a 1 \(M_\odot\) star, formed at the same time.

  1. Which star will leave the main sequence first?

  2. About how long will the 1 \(M_\odot\) star remain on the main sequence?

  3. What does this tell you about why clusters can contain stars in very different evolutionary stages at the same time?

  4. The 50 \(M_\odot\) star will leave the main sequence first, by a huge margin, because very massive stars burn fuel vastly faster than low-mass stars.

  5. For the 1 \(M_\odot\) star:

\[t_{\rm MS} = 10~\text{Gyr}\]

So the \(1\,M_\odot\) star remains on the main sequence for about 10 billion years.

  1. This tells us why clusters can show stars in very different evolutionary stages at the same time: all the stars may have formed together, but high-mass stars evolve much faster than low-mass stars. A single-age population can therefore contain some stars that have already become giants while lower-mass stars remain on the main sequence.

Challenge Problems

Challenge 1: Gravity As An Energy Source

Restatement: A star contracts slightly under gravity.

  1. What happens to its gravitational potential energy?

  2. What happens to its thermal energy and temperature?

  3. Why does that help explain the thermostat behavior of a star?

  4. Its gravitational potential energy becomes more negative. The star is more tightly bound after it contracts.

  5. Some of that lost gravitational potential energy shows up as increased thermal energy, so the temperature rises.

  6. This helps explain the thermostat behavior because contraction naturally heats the core, which raises pressure and pushes back against further collapse.

Challenge 2: A Very Old Universe

Restatement: Imagine the universe 1 trillion years from now.

  1. Which kinds of stars would still remain on the main sequence?

  2. Where would the main-sequence turnoff lie on the H-R diagram?

  3. Why would that make the turnoff harder to observe?

  4. Only the lowest-mass true stars would still remain on the main sequence - the coolest, faintest red dwarfs just above the hydrogen-burning limit.

  5. The turnoff would lie at the very faint, cool end of the main sequence, near the lower-right part of the H-R diagram.

  6. That would make the turnoff harder to observe because the relevant stars would be extremely dim. It is much harder to identify a faint red turnoff than a bright blue one in a younger cluster.

Challenge 3: Reasoning From Populations

Restatement: Suppose you observed 10,000 red giants across the sky.

  1. Would that mean the red-giant phase lasts forever?

  2. What would the large number of red giants actually tell you about stellar populations and evolutionary timescales?

  3. Why are population snapshots so important in astronomy?

  4. No. A large number of red giants does not mean the red-giant phase lasts forever.

  5. It tells you that many stars are passing through that phase across the Galaxy and that the Galaxy contains a large, mixed-age stellar population. Population counts help astronomers infer which phases are common, rare, short-lived, or long-lived.

  6. Population snapshots are important because astronomy usually cannot watch one star evolve from birth to death directly. Instead, astronomers compare many stars at different stages and use those snapshots to reconstruct the life story.