Earth, Moon & Solar System — Chapter Practice, Grade 9

Model sky patterns, seasons, tides, and predictable orbital motion.

Problem 1

What actually causes the Moon's phases?
  1. Earth's shadow falls on the Moon
  2. Different amounts of the Moon's sunlit half face Earth as it orbits
  3. The Moon changes how much light it produces
  4. Clouds on Earth block our view of the Moon
Answer and explanation

Different amounts of the Moon's sunlit half face Earth as it orbits

Half the Moon is always lit by the Sun; phases come from how much of that lit half faces Earth as the Moon orbits us.

Problem 2

Which phase comes right after a new moon?
  1. Waxing crescent
  2. Full moon
  3. Waning gibbous
  4. Third quarter
Answer and explanation

Waxing crescent

Right after new moon, a thin sliver of light begins to grow — that's the waxing crescent.

Problem 3

About how many days does one full cycle of moon phases take, from new moon back to new moon?
Answer and explanation

29.5 days

One synodic month is about 29.5 days.

Problem 4

Tonight the Moon is waxing gibbous. What does that tell you?
  1. More than half lit, and growing each night
  2. Less than half lit, and shrinking each night
  3. Completely dark
  4. Fully lit and about to shrink
Answer and explanation

More than half lit, and growing each night

Gibbous means more than half lit but not full; waxing means it's still growing toward full moon.

Problem 5

A classmate says, 'There's a lunar eclipse every single month at full moon.' Explain why this isn't correct.
Answer and explanation

A lunar eclipse can only happen at full moon, but it also requires the Sun, Earth, and Moon to line up almost exactly. Since the Moon's orbit is tilted about 5° relative to Earth's orbit around the Sun, that precise lineup only happens a few times a year, not every month.

Problem 6

At full moon, Earth lies roughly between Sun and Moon. Why is the Moon usually not eclipsed? Explain the reasoning in your own words.
Answer and explanation

Orbital tilt usually prevents an exact shadow alignment.

A full moon places the bodies roughly along a line.

The tilted lunar orbit usually places the Moon above or below Earth's shadow.

Response criteria

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

Problem 7

Why can a crescent Moon be visible even when no eclipse occurs? Show the reasoning, not just the conclusion.
Hints

Separate the visible disk from the whole sphere.

Answer and explanation

Only a small part of the sunlit hemisphere faces Earth at that viewing angle.

Problem 8

At full Moon, what approximate Sun–Earth–Moon arrangement applies?
Hints

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

Answer and explanation

The Moon is roughly opposite the Sun in Earth's sky, with Earth between their directions; orbital tilt usually prevents an eclipse.

Problem 9

Phase geometry helps explain approximate viewing timesPhase geometry helps explain approximate viewing timesObserver’s line of sightNear full phase: Moon lies opposite the Sun’s sky directionAs the Sun sets, the full Moon is generally rising.Approximate timing depends on place, date and orbit geometry.
Original ideal alignment for interpreting full-phase viewing. It omits orbital tilt and real horizon/atmospheric effects and does not depict an eclipse. Labels state approximate timing, not an operational observation schedule or encouragement to view the Sun directly.
Interpret the supplied visual model. Why is a full Moon commonly visible at night even though sunlight supplies its illumination?
Answer and explanation

It lies approximately opposite the Sun from Earth, so its illuminated face can be above the night-side observer’s horizon while the Sun is below it. Exact times depend on the stated conditions.

Read the arrangement as an ideal full-phase sky-direction relation. Distances are exaggerated and the Moon’s illuminated side faces back toward the Sun.

It lies approximately opposite the Sun from Earth, so its illuminated face can be above the night-side observer’s horizon while the Sun is below it. Exact times depend on the stated 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 10

A student argues that first quarter must show one quarter of the lunar disk lit and every new Moon causes a solar eclipse. Correct both statements with geometry.
Answer and explanation

First quarter describes about one-quarter of the phase cycle and the visible disk is roughly half lit. New Moon alone does not guarantee a solar eclipse because the tilted lunar orbit usually misses the shadow line; alignment near a node is also needed.

First quarter describes about one-quarter of the phase cycle and the visible disk is roughly half lit. New Moon alone does not guarantee a solar eclipse because the tilted lunar orbit usually misses the shadow line; alignment near a node is also needed.

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 11

What is the main cause of Earth's seasons?
  1. Earth's distance from the Sun changing
  2. The tilt of Earth's axis
  3. The Moon blocking sunlight
  4. The Sun burning hotter and cooler
Answer and explanation

The tilt of Earth's axis

Earth's 23.5° axial tilt changes how directly sunlight hits each hemisphere across the year — that's what drives the seasons.

Problem 12

When is Earth actually closest to the Sun?
  1. Around June 21
  2. Around September 22
  3. Around early January
  4. Around March 20
Answer and explanation

Around early January

Earth reaches perihelion, its closest point to the Sun, in early January — during Northern Hemisphere winter, which shows distance isn't what causes the seasons.

Problem 13

How many equinoxes occur each year, when day and night are nearly equal in length?
Answer and explanation

2

There are two equinoxes each year — the spring (vernal) equinox and the autumn equinox.

Problem 14

What happens during an equinox?
  1. One hemisphere gets far more sunlight than the other
  2. Day and night are roughly equal length nearly everywhere
  3. Earth reaches its closest point to the Sun
  4. Earth's axis points directly at the Sun
Answer and explanation

Day and night are roughly equal length nearly everywhere

At the equinoxes, Earth's tilt is sideways to the Sun, so sunlight is split roughly evenly between hemispheres.

Problem 15

Explain why sunlight hitting the ground at a direct, high angle heats it more than the same sunlight hitting at a low, glancing angle.
Answer and explanation

Direct, high-angle sunlight concentrates its energy over a small area of ground. Low-angle, glancing sunlight spreads that same amount of energy over a much larger area, so each patch of ground receives and absorbs less energy and heats up less.

Problem 16

Why can Australia have summer when New York has winter? Explain the reasoning in your own words.
Answer and explanation

Opposite seasonal sunlight patterns arise from axial tilt, not separate Sun distances.

The Southern Hemisphere tilts toward the Sun while the Northern Hemisphere tilts away.

This changes illumination angle and day length in opposite ways.

Response criteria

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

Problem 17

Why does low-angle sunlight warm a given surface less strongly under similar conditions? Show the reasoning, not just the conclusion.
Hints

Seasons oppose across hemispheres despite sharing Earth's Sun distance.

Answer and explanation

The same incoming beam spreads over a larger surface area and passes through more atmosphere.

Problem 18

Does Earth's axis reverse its tilt every six months?
Hints

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

Answer and explanation

No. It maintains approximately the same orientation as Earth orbits, changing which hemisphere tilts toward the Sun.

Problem 19

Opposite hemispheres challenge a distance-only explanationOpposite hemispheres challenge a distance-only explanationIllustrative matched-latitude daylight (h)Northern June daylight14Southern June daylight10Northern Decemberdaylight10Southern Decemberdaylight14Opposite hemispheric geometry occurs at the same Earth–Sun distance.
Original illustrative paired-hemisphere daylight durations for conceptual comparison. Exact values depend on latitude, date and horizon/refraction conventions. The chosen matched-latitude example isolates seasonal geometry rather than supplying a local almanac.
Interpret the supplied visual model. Why would “Earth is nearer the Sun, so both hemispheres have summer” fail to explain these pairs?
Answer and explanation

Both hemispheres share the same instantaneous distance but have opposite daylight/tilt effects. Axis orientation and latitude explain that contrast; distance alone does not.

Compare June and December pairs under one shared planet-to-Sun distance at each date. The bars measure daylight duration rather than actual temperature.

Both hemispheres share the same instantaneous distance but have opposite daylight/tilt effects. Axis orientation and latitude explain that contrast; distance alone does not.

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 20

Under a simple comparison, two surfaces receive equal absorbed power per square metre, but one receives it for 9 hours and the other 15. Compare energy received and explain why this still does not uniquely fix their temperatures.
Answer and explanation

The second receives 15/9 = 5/3, about 1.67 times the energy per area under the assumptions. Temperature also depends on heat capacity, mass, losses, prior conditions and varying real radiation.

The second receives 15/9 = 5/3, about 1.67 times the energy per area under the assumptions. Temperature also depends on heat capacity, mass, losses, prior conditions and varying real radiation.

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 21

Which planet is closest to the Sun?
  1. Mercury
  2. Venus
  3. Earth
  4. Mars
Answer and explanation

Mercury

Mercury is the innermost planet, orbiting closest to the Sun.

Problem 22

About what percent of the solar system's total mass is contained in the Sun?
Answer and explanation

99.8 %

The Sun holds about 99.8% of all the mass in the solar system.

Problem 23

Jupiter and Saturn are best classified as…
  1. Terrestrial planets
  2. Gas giants
  3. Ice giants
  4. Dwarf planets
Answer and explanation

Gas giants

Jupiter and Saturn are huge, made mostly of hydrogen and helium gas, with rings and many moons — that makes them gas giants.

Problem 24

Uranus and Neptune are made mostly of icy materials (water, ammonia, methane) plus hydrogen and helium gas. What group are they in?
  1. Terrestrial planets
  2. Gas giants
  3. Ice giants
  4. Asteroids
Answer and explanation

Ice giants

Uranus and Neptune's icy composition places them in their own group: the ice giants.

Problem 25

What is the correct order as a small rocky object from space enters the atmosphere and lands?
  1. Meteoroid → meteor → meteorite
  2. Meteor → meteoroid → meteorite
  3. Meteorite → meteor → meteoroid
  4. Meteoroid → meteorite → meteor
Answer and explanation

Meteoroid → meteor → meteorite

It's a meteoroid while in space, becomes a meteor (a streak of light) while burning up in the atmosphere, and any piece that survives to hit the ground is a meteorite.

Problem 26

Explain why astronomers use astronomical units (AU) instead of miles or kilometers to describe distances in the solar system.
Answer and explanation

AU keeps the numbers small and easy to compare. Instead of writing hundreds of millions or billions of miles or kilometers, astronomers can just say a planet is, for example, 5 AU from the Sun, using Earth's own distance from the Sun as a simple, familiar yardstick.

Problem 27

A model places Earth 1 m from the Sun. At the same distance scale, a planet at 5 AU is how far away? Explain the reasoning in your own words.
Answer and explanation

5 m from the model Sun; planet-size scale may still differ.

One AU corresponds to 1 m in this model.

Multiply 5 AU by 1 m/AU.

Response criteria

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

Problem 28

Why is an astronomical unit useful for planetary distances? Show the reasoning, not just the conclusion.
Hints

Larger radius alone does not determine greater mean density.

Answer and explanation

It provides a convenient scale based on Earth's average Sun distance, avoiding very large kilometer values.

Problem 29

Must a moon be smaller than every planet?
Hints

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

Answer and explanation

No. Categories concern orbital and physical criteria; some large moons exceed some planets in diameter.

Problem 30

Object class and observed activity are different questionsObject class and observed activity are different questionsSunPlanet and moonA comet tail responds to solar effects, not only the travel directionSelected motionTail away from Sun
Original qualitative comet/planet/moon scene with exaggerated body sizes and one selected ion-tail direction. It is not a real encounter image or simultaneous orbital-position map. The caption distinguishes the ion-tail example from possible curved dust-tail geometry.
Interpret the supplied visual model. Why can a comet tail fail to point behind its direction of travel?
Answer and explanation

Solar wind/radiation affect released material, so tail direction depends on the Sun and material type as well as the orbit. An ion tail points broadly away from the Sun.

The blue motion arrow and green away-from-Sun tail arrow point in different directions. The tail is one selected ion-tail direction, not a universal dust-tail shape.

Solar wind/radiation affect released material, so tail direction depends on the Sun and material type as well as the orbit. An ion tail points broadly away from the Sun.

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 31

A Solar System distance model places a body at a = 10 AU. Convert that scale to kilometres using 1 AU ≈ 150 million km. Explain why the distance and an image of a water-related feature do not alone establish its size or life.
Answer and explanation

The scale is about 1.5 billion km. It is an orbital distance, not a diameter. A water-related surface interpretation concerns geological conditions and does not alone prove organisms or all habitability requirements.

The scale is about 1.5 billion km. It is an orbital distance, not a diameter. A water-related surface interpretation concerns geological conditions and does not alone prove organisms or all habitability requirements.

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 32

The strength of the gravitational pull between two objects depends on
  1. their masses and the distance between them
  2. their colors and temperatures
  3. only the time of day
  4. whether they are on Earth
Answer and explanation

their masses and the distance between them

Gravitational attraction grows with greater mass and weakens as the distance between the objects increases.

Problem 33

Why do planets stay in orbit instead of flying off in a straight line or falling into the Sun?
  1. Gravity alone holds them perfectly still
  2. Gravity continually changes the direction of their motion
  3. There are no forces acting on planets
  4. Planets have engines that keep firing
Answer and explanation

Gravity continually changes the direction of their motion

An orbit is the curved path created when an object's forward motion (inertia) is continuously bent by gravity's inward pull.

Problem 34

Which planet orbits the Sun faster: one close to the Sun, or one far away?
  1. The closer planet, because it feels stronger gravity
  2. The farther planet, because it has more room to speed up
  3. Both orbit at exactly the same speed
  4. Distance doesn't affect orbital speed
Answer and explanation

The closer planet, because it feels stronger gravity

Gravity is stronger at shorter distances, and that stronger pull corresponds to a faster orbital speed and a shorter 'year.'

Problem 35

What mainly causes Earth's ocean tides?
  1. Wind patterns blowing across the ocean
  2. Differences in lunar and solar gravitational effects across Earth
  3. Earthquakes on the ocean floor
  4. The Sun heating the water unevenly
Answer and explanation

Differences in lunar and solar gravitational effects across Earth

The Moon produces the larger tidal effect because the gravitational difference across Earth matters; this is not the same as comparing total force.

Problem 36

Explain why spring tides happen at BOTH new moon and full moon, even though the Moon is on opposite sides of Earth in those two cases.
Answer and explanation

At new and full phase, the Sun, Earth and Moon lie roughly along one line, so the lunar and solar tidal effects have aligned axes and can produce larger ranges. This concerns differential gravitational effects across Earth; the total Sun/Moon forces need not point in the same direction. Local ocean dynamics modify actual ranges.

Problem 37

A mean lunar day—the time for Earth to turn once relative to the Moon—is about 24 hours 50 minutes. Approximately how many hours is that as a decimal?
Answer and explanation

24.8 hours

50 minutes is about 0.83 of an hour, so 24 hours 50 minutes ≈ 24.8 hours.

Problem 38

An orbiting satellite moves at constant speed but continually changes direction. Is it accelerating? Explain the reasoning in your own words.
Answer and explanation

Yes; gravitational acceleration changes its direction of motion.

Velocity includes direction as well as speed.

A changing velocity direction means acceleration, directed inward in the circular model.

Response criteria

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

Problem 39

Why can both near and far sides of Earth experience tidal bulges in a simplified model? Show the reasoning, not just the conclusion.
Hints

Being in orbit does not mean gravity is absent.

Answer and explanation

Tidal effects arise from differences in lunar gravitational acceleration across Earth, not just the total pull on one side.

Problem 40

If a satellite's forward velocity vanished, could the same circular orbit continue?
Hints

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

Answer and explanation

No. Without tangential motion the given circular trajectory would not persist.

Problem 41

Tidal effects depend on differences across EarthTidal effects depend on differences across EarthMoonCompare lunar acceleration with Earth-center accelerationNear side: stronger lunar acceleration than Earth’s center.Far side: weaker lunar acceleration than Earth’s center.Relative effects produce two bulges in the simplified equilibrium model.
Original exaggerated equilibrium-tide geometry. Red arrows represent relative sizes of Moonward lunar acceleration at selected Earth positions, not actual force values. The elongated outline summarizes an ideal two-bulge model; real coastal tides depend on rotation, basins, coastlines and dynamics.
Interpret the supplied visual model. How can the far-side relative effect be outward from Earth’s center while lunar gravity there still points toward the Moon?
Answer and explanation

The far side accelerates toward the Moon less than Earth’s center does. In the Earth-following comparison, that difference is away from the center, while the actual lunar attraction remains Moonward.

All red lunar-attraction arrows point toward the Moon, with different magnitudes. The far-side bulge is not drawn as lunar gravity pulling away from the Moon.

The far side accelerates toward the Moon less than Earth’s center does. In the Earth-following comparison, that difference is away from the center, while the actual lunar attraction remains Moonward.

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 42

At fixed central mass, a satellite’s center separation doubles while its mass triples. Find the gravitational-force factor. Explain why this does not uniquely give the new orbit or period.
Answer and explanation

The force factor is 3/2² = 3/4. A trajectory also needs velocity/initial conditions and an appropriate orbital model; force ratio alone does not define an entire new orbit.

Apply the mass factor 3 and separation-square factor 1/4.

Multiply to obtain 3/4.

Distinguish force ratio from full motion and request the needed initial/model conditions.

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 43

For a solar orbit with a = 9 AU, use P² = a³ to find P.
Answer and explanation

27 Earth years

P² = 729; P = 27.

Problem 44

A circular orbit at constant speed has…
  1. zero acceleration
  2. inward acceleration
  3. an outward inertia force balancing gravity
  4. no gravity
Answer and explanation

inward acceleration

Direction changes continuously.

Problem 45

Why cannot P² = a³ in Earth years and AU be applied unchanged to any orbit around any object?
Answer and explanation

The relation’s constant depends on the central mass and chosen units.

Problem 46

For the same star, orbit B has semimajor axis 4 times A's. What is its period ratio? Explain the reasoning in your own words.
Answer and explanation

Orbit B's period is 8 times orbit A's under this model.

T² scales with a³, so the squared period ratio is 4³=64.

Take the square root: T_B/T_A=8.

Response criteria

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

Problem 47

Using P²=a³, find P when a=9 AU. Show the reasoning, not just the conclusion.
Hints

The relation requires the stated units and central mass.

Answer and explanation

P=√729=27 years.

Problem 48

Why cannot the solar-unit formula be applied unchanged to a moon orbiting a planet?
Hints

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

Answer and explanation

The central mass differs; the numerical relationship and units must be adjusted.

Problem 49

Period grows faster than semimajor axis in the solar relationPeriod grows faster than semimajor axis in the solar relationOrbital period P (years)016324864014916Semimajor axis a (AU)Same dominant solar mass: P² = a³, so P = √(a³).
Original graph of the chapter’s supplied ideal solar-system records. The points follow P² = a³; straight connecting guides do not replace the smooth underlying curve. It is not a dataset of four named planets and is not valid with the same constant for arbitrary central masses.
Interpret the supplied visual model. Find P when a = 9 AU in the supplied solar model.
Answer and explanation

P = √(9³) = √729 = 27 years. The relation assumes the same dominant solar mass and appropriate years/AU units.

Read a on the horizontal axis and P vertically. Neither variable is instantaneous speed or planet diameter.

P = √(9³) = √729 = 27 years. The relation assumes the same dominant solar mass and appropriate years/AU units.

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 50

In the ideal same-Sun model, a body has a = 6.25 AU. Find P and the transformed pair (a³, P²). Explain why the same numerical constant need not apply around another star.
Answer and explanation

a³ = 244.140625, so P = 15.625 years and the transformed pair is (244.140625 AU³, 244.140625 years²). The proportional constant depends on the dominant central mass; the convenient solar years/AU form is not universal.

Compute 6.25³ = 244.140625.

Take its square root: 15.625 years.

Keep transformed units and identify the central-mass assumption.

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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