Stars & the Universe — Chapter Practice, Grade 9

Use stellar models, diagrams, and spectra to explain energy, elements, and cosmic history.

Problem 1

The Sun’s main energy source is…
  1. coal combustion
  2. nuclear fusion
  3. reflected moonlight
  4. friction with air
Answer and explanation

nuclear fusion

Hydrogen fusion supplies energy in its core.

Problem 2

How does solar energy cross the near-vacuum to Earth?
Answer and explanation

As electromagnetic radiation.

Problem 3

Why does more stellar fuel not necessarily mean a longer lifetime?
Answer and explanation

A higher-mass star can consume fuel much faster.

Problem 4

Why can a massive star have a shorter main-sequence lifetime than a smaller star? Explain the reasoning in your own words.
Answer and explanation

The larger fuel supply can be consumed disproportionately faster.

Its energy output and fuel-consumption rate can be much greater.

Compare available fuel with the rate of use rather than fuel amount alone.

Response criteria

  • Use the stated conditions.
  • Show the evidence or calculation.
  • Explain the conclusion and any stated limitation.

Problem 5

What distinguishes fusion from chemical burning in a star? Show the reasoning, not just the conclusion.
Hints

Compare supply with consumption rate.

Answer and explanation

Fusion changes nuclei and releases nuclear energy; ordinary chemical reactions rearrange electrons and atoms without that nuclear transformation.

Problem 6

Why does a star's color provide temperature evidence?
Hints

State what the supplied observations or conditions establish, and separate that from any assumption.

Answer and explanation

Its emitted spectrum depends on surface temperature; bluer stars generally have hotter surfaces than redder ones.

Problem 7

More fuel does not guarantee a longer lifetimeMore fuel does not guarantee a longer lifetimeSupplied relative fuel / rate indicesStar A: available fuel1Star B: available fuel5Star A: use rate1Star B: use rate100Lifetime scale B/A = 5/100 = 0.05 in this model.
Original relative fuel/use-rate model. A and B are invented examples and the index values are not a calibrated stellar mass relation. Full-scale bars make the much greater rate visible; the lifetime inference uses the explicitly supplied normalization and assumptions.
Interpret the supplied visual model. If A’s model lifetime is 10 billion years, what lifetime follows for B under the supplied ratios?
Answer and explanation

0.05×10 billion = 0.5 billion years, or 500 million years. This is a teaching comparison, not a measured lifetime for a named star or all stars of a given mass.

Fuel amount and use rate are unlike kinds of quantity, with separate normalization. Calculate their ratio rather than adding bar heights.

0.05×10 billion = 0.5 billion years, or 500 million years. This is a teaching comparison, not a measured lifetime for a named star or all stars of a given mass.

Response criteria

  • Read the labeled quantities, reference frame or objects correctly.
  • Explain the relation between the visual evidence and the mechanism.
  • Keep the drawing’s stated assumptions and limits in the conclusion.

Problem 8

Model star B has four times A’s available fuel but eighty times its fuel-use rate. Estimate its lifetime relative to A and explain why this is a ratio model rather than a complete stellar-evolution forecast.
Answer and explanation

The ratio is 4/80 = 0.05, so B’s lifetime scale is one twentieth of A’s. Actual evolution also depends on initial mass, composition, core structure and changing conditions; these supplied ratios are not a universal stellar formula.

The ratio is 4/80 = 0.05, so B’s lifetime scale is one twentieth of A’s. Actual evolution also depends on initial mass, composition, core structure and changing conditions; these supplied ratios are not a universal stellar formula.

Response criteria

  • Use the stated measurements or conditions, with correct units.
  • Connect the result to the physical mechanism or evidence relationship.
  • Identify the assumption, model limit or further evidence requested.

Problem 9

A cool but very luminous star may be a…
  1. red giant
  2. small white dwarf only
  3. dark planet only
  4. star with no surface area
Answer and explanation

red giant

A large radiating area can produce high luminosity at lower temperature.

Problem 10

Why can apparent brightness not determine luminosity by itself?
Answer and explanation

Distance and intervening material affect the observed brightness.

Problem 11

On many H–R diagrams, temperature increases toward the…
  1. left
  2. right always
  3. top only
  4. bottom only
Answer and explanation

left

The conventional horizontal temperature axis is often reversed; verify labels.

Problem 12

A star is cool but has high luminosity. What does this suggest about its size compared with a hot dim star? Explain the reasoning in your own words.
Answer and explanation

It likely has a much larger radiating surface, consistent with a giant-type star.

Cool surfaces emit less energy per unit area than hot ones.

High total luminosity at low temperature requires a large emitting area.

Response criteria

  • Use the stated conditions.
  • Show the evidence or calculation.
  • Explain the conclusion and any stated limitation.

Problem 13

On an H–R diagram, a star is cool but very luminous. What broad size inference follows? Show the reasoning, not just the conclusion.
Hints

Luminosity is intrinsic output, distinct from apparent brightness.

Answer and explanation

It likely has a large radiating surface, consistent with a giant or supergiant region rather than a small main-sequence star.

Problem 14

Why must the temperature-axis direction be checked on an H–R diagram?
Hints

State what the supplied observations or conditions establish, and separate that from any assumption.

Answer and explanation

Many diagrams place hot temperatures on the left, opposite the usual increasing numerical axis.

Problem 15

At equal surface temperature, greater emitting area raises luminosityAt equal surface temperature, greater emitting area raises luminosityEqual surface temperature, different radiusRadius RRadius 3RArea factor = (3R)² / R² = 9; luminosity factor is 9 in this model.
Original matched-temperature radius comparison. Disk colors are a shared temperature label, not exact observed stellar colors. The drawn circle area represents the radius-squared relation; the physical emitting surface is spherical and the inference depends on matched conditions.
Interpret the supplied visual model. If two equal-temperature stars differ in luminosity by a factor of 16, what radius factor follows in the basic model?
Answer and explanation

A factor of four, because area and luminosity scale with radius squared at equal temperature/emissive conditions.

Compare radius rather than diameter of the illustrated disks, and use the written three-to-one ratio for the spherical model.

A factor of four, because area and luminosity scale with radius squared at equal temperature/emissive conditions.

Response criteria

  • Read the labeled quantities, reference frame or objects correctly.
  • Explain the relation between the visual evidence and the mechanism.
  • Keep the drawing’s stated assumptions and limits in the conclusion.

Problem 16

Two ideal stars have equal surface temperature; star B has radius twice A’s. Find B/A luminosity. At the same observer distance and unobscured path, compare received brightness. Name one condition that would break the simple comparison.
Answer and explanation

Luminosity factor is 2² = 4 and received brightness is also four at equal distance/path conditions. Different temperatures, extinction or emission/measurement conditions would require additional factors.

Luminosity factor is 2² = 4 and received brightness is also four at equal distance/path conditions. Different temperatures, extinction or emission/measurement conditions would require additional factors.

Response criteria

  • Use the stated measurements or conditions, with correct units.
  • Connect the result to the physical mechanism or evidence relationship.
  • Identify the assumption, model limit or further evidence requested.

Problem 17

A 400 nm rest-wavelength line is observed at 440 nm. Find z.
Answer and explanation

0.1

(440 − 400) ÷ 400 = 0.10.

Problem 18

A line at longer wavelength than its rest value is…
  1. redshifted
  2. blueshifted
  3. proof of a new element by itself
  4. unmeasurable
Answer and explanation

redshifted

Redshift refers to increased wavelength.

Problem 19

Why compare multiple spectral lines?
Answer and explanation

A pattern makes identification more reliable and helps distinguish composition from shifts or measurement errors.

Problem 20

A known line at 500 nm appears at 510 nm. Calculate z=(observed−rest)/rest. Explain the reasoning in your own words.
Answer and explanation

Redshift z=0.02; interpretation needs the full pattern and relevant model.

Wavelength difference=510−500=10 nm.

z=10/500=0.02; the wavelength units cancel.

Response criteria

  • Use the stated conditions.
  • Show the evidence or calculation.
  • Explain the conclusion and any stated limitation.

Problem 21

A line is at 510 nm instead of 500 nm. Calculate fractional wavelength shift. Show the reasoning, not just the conclusion.
Hints

Not every redshift alone determines a unique distance or cause.

Answer and explanation

(510−500)/500=0.02.

Problem 22

Why does a galaxy redshift–distance pattern support expansion more strongly than one spectrum?
Hints

State what the supplied observations or conditions establish, and separate that from any assumption.

Answer and explanation

A systematic relationship across many galaxies tests a global model rather than one object's local motion.

Problem 23

A broad distance–redshift relation supports expansion with qualificationsA broad distance–redshift relation supports expansion with qualificationsNearby-model recession speed (km/s)0700140021002800010203040Illustrative distance (Mpc)A supplied slope 70 km/s per Mpc; real estimates have uncertainty.
Original deliberately supplied linear relation with slope 70 km/s/Mpc for model reading. It is not a current observed galaxy dataset or an exact cosmological conversion for all redshifts. The graph makes its distance unit and nearby-approximation limit explicit.
Interpret the supplied visual model. At 30 Mpc in the supplied relation, what speed is predicted, and could a local motion cause a deviation?
Answer and explanation

2100 km/s. Local gravitational motion and measurement/calibration differences can cause deviations from the broad relation; a nearby exception does not automatically refute the expansion model.

Read distance and speed as different axes. The straight line is an illustrative nearby approximation, not a route traveled by one galaxy over time.

2100 km/s. Local gravitational motion and measurement/calibration differences can cause deviations from the broad relation; a nearby exception does not automatically refute the expansion model.

Response criteria

  • Read the labeled quantities, reference frame or objects correctly.
  • Explain the relation between the visual evidence and the mechanism.
  • Keep the drawing’s stated assumptions and limits in the conclusion.

Problem 24

A 500 nm laboratory line is observed at 575 nm, and a 600 nm line at 690 nm. Calculate z for each. Explain what the agreement supports and what it does not uniquely establish.
Answer and explanation

Both give z = 0.15: 75/500 and 90/600. Agreement supports a common fractional shift of that identified pattern, while the physical cause, distance and full cosmological interpretation need context, calibration and additional evidence.

Compute (575−500)/500 = 0.15.

Compute (690−600)/600 = 0.15.

Interpret the common fractional shift without assigning a unique cause from arithmetic alone.

Response criteria

  • Use the stated measurements or conditions, with correct units.
  • Connect the result to the physical mechanism or evidence relationship.
  • Identify the assumption, model limit or further evidence requested.

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