Gravity, Orbits & Tides — Practice · Grade 9 Science

Chapter: Earth, Moon & Solar System

What this practice covers

By the end of Gravity, Orbits & Tides you should be able to:

11 questions, multiple choice, written answer and numeric, each with a worked explanation.

Vocabulary used: gravity, orbit, tide, spring tide & neap tide.

Problem 1

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 2

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 3

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 4

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 5

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 6

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 7

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 8

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 9

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 10

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 11

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.

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