Circular Motion & Gravitation — Print · Grade 12 Science

Chapter: Forces, Rotation & Equilibrium

Circular Motion & Gravitation

Learning goals

Readings for this unit

Before this unit

Acceleration direction and inverse-square laws.

Use the detailed parent plan: two core 50–60-minute sessions per focused unit, with a third assigned for selected calculation/modeling topics. First study the core explanation and Parts 1–2 with their figures/checks; next study Parts 3–4, the whole-unit application, worked cases, varied practice and exit check. Attempt reasoning before opening guidance. Extend time when actual understanding requires it; investigations and chapter assessments have separate sessions.

Start with something to explain

How can a satellite fall without striking the ground?

Illustrative evidence case. A satellite has tangential motion while gravity turns its velocity inward.

Turning changes velocity even at constant speedTurning changes velocity even at constant speedvelocity tangentacceleration inward“Centripetal” names the inward net-force role.
Original circular-motion geometry; velocity and acceleration arrows have different units and independent scales.

Notice and wonder

Record two observations using the stated quantities or relationships. Keep observations separate from possible causes. Write one question that a new measurement could help answer.

How can a satellite fall without striking the ground?

Your first model

Label the center, tangent and actual inward interaction.

Build the language from the idea

centripetal acceleration
The acceleration of an object moving in a circle, always directed toward the center; a = v²/r.
centripetal force
The net inward force required to keep an object moving in a circle; F = mv²/r. It is supplied by a real force such as tension, friction, or gravity.
period (T)
The time it takes an object to complete one full revolution around a circle.
Newton's law of universal gravitation
Every pair of masses attracts with a force F = Gm₁m₂/r², where r is the distance between their centers.
gravitational constant (G)
The fixed number in the gravitation equation, G = 6.67 × 10⁻¹¹ N·m²/kg², measured first by Henry Cavendish.

An object can move at a perfectly constant speed and still be accelerating — a fact that feels wrong until you remember that acceleration means any change in velocity, and velocity includes direction. A car rounding a curve at a steady 50 km/h is constantly changing direction, so it is constantly accelerating. This is uniform circular motion: constant speed, continuously turning. The acceleration always points toward the center of the circle and is called centripetal acceleration, given by the clean formula a = v²/r, where v is the speed and r is the radius. Notice that tighter turns (smaller r) and higher speeds (larger v) both demand more acceleration — which is exactly why a sharp turn at high speed feels so violent.

Turning changes velocity even at constant speedTurning changes velocity even at constant speedvelocity tangentacceleration inward“Centripetal” names the inward net-force role.
Original circular-motion geometry; velocity and acceleration arrows have different units and independent scales.

By Newton's second law, an acceleration requires a net force in the same direction, so circular motion demands a net inward force called the centripetal force, F = ma = mv²/r. Here is the single most important idea in this topic: *centripetal is not a new kind of force.* The word simply means 'center-seeking' and describes the *direction* of whatever real force is doing the job. For a ball on a string it is tension; for a car on a curve it is friction between tires and road; for the Moon circling Earth it is gravity. Whenever you see circular motion, ask: which real force is pointing toward the center and supplying the mv²/r that the motion requires?

This clears up the most stubborn misconception in the whole unit — the so-called centrifugal 'force.' When a car turns left, you feel flung to the right and swear something is pushing you outward. Nothing is. Your body, obeying the first law, simply wants to continue straight ahead while the car turns underneath you; the door then pushes you inward to make you turn with it. The outward 'force' is an illusion of riding in an accelerating frame. The only real force in play points *inward*, toward the center.

Newton's greatest leap was realizing that the same force pulling an apple to the ground also holds the Moon in its orbit — and that gravity is universal, acting between *every* pair of masses in the cosmos. He captured it in Newton's law of universal gravitation: F = Gm₁m₂/r². The force grows with each mass and shrinks with the square of the distance r between their centers, and G is the gravitational constant, a very small fixed number, 6.67 × 10⁻¹¹ N·m²/kg². Because G is so tiny, the gravitational pull between everyday objects is utterly negligible; it takes a body as massive as a planet for gravity to become the dominant force we feel every second of our lives.

The r² in the denominator makes gravity an inverse-square law, and its consequences are dramatic. Double the distance between two objects and the force drops not to half but to one-quarter; triple the distance and it falls to one-ninth. This same equation, evaluated at Earth's surface, reproduces the familiar weight W = mg — plug in Earth's mass and radius and out comes an acceleration of about 9.8 m/s² for every object, regardless of its own mass. Universal gravitation and everyday weight are not two separate ideas; they are the same law seen at two different scales.

Gravitational separation is measured between centersGravitational separation is measured between centerscenter-to-center rEarthF = GMm/r²; altitude h is not the same as r.For Earth radius R: r = R+h.
Original spherical-source separation model; Earth/satellite sizes and distances are not to scale.

Orbits are the grand payoff, and Newton explained them with a famous thought experiment: imagine a cannon on a mountaintop firing a ball ever faster. Fired gently, the ball arcs down and lands; fired fast enough, the ground curves away beneath it just as quickly as it falls, and the ball 'falls around' the planet forever. That is an orbit — continuous free-fall in which gravity supplies exactly the centripetal force the circular path requires, so Gm₁m₂/r² = mv²/r. This is also why astronauts on the space station float: they are not beyond gravity's reach (Earth's gravity is still nearly as strong up there) but are in constant free-fall together with their station, endlessly falling around the Earth.

An orbit is free fall, not absence of gravityAn orbit is free fall, not absence of gravityTangent velocityInward gravityCurved free fallOrbit: continuously falling while moving sidewaysCircular condition: GM/r² = v²/r.No support force is needed to maintain ideal free fall.
Original causal model for circular orbital free fall, not an orbital scale drawing.

Model check: Centripetal force is the net inward force, not an extra interaction added to gravity or tension. ac=v²/r=8 m/s². Speed can be constant while velocity direction changes.

Build your understanding

Use the detailed parent plan: two core 50–60-minute sessions per focused unit, with a third assigned for selected calculation/modeling topics. First study the core explanation and Parts 1–2 with their figures/checks; next study Parts 3–4, the whole-unit application, worked cases, varied practice and exit check. Attempt reasoning before opening guidance. Extend time when actual understanding requires it; investigations and chapter assessments have separate sessions.

Part 1 · Identify centripetal acceleration without a new force

Uniform circular motion has constant speed but changing velocity direction. Its centripetal acceleration points toward the circle’s center with magnitude v²/r. “Centripetal force” is the inward net-force role, supplied by real interactions such as tension, friction or gravity; it is not an extra force added to those interactions. At each point velocity is tangent while inward acceleration changes its direction. Draw them separately and identify the actual force agent.

Turning changes velocity even at constant speedTurning changes velocity even at constant speedvelocity tangentacceleration inward“Centripetal” names the inward net-force role.
Original circular-motion geometry; velocity and acceleration arrows have different units and independent scales.

Read the visual model

Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.

For a satellite, what supplies the inward force rather than a new “centripetal” interaction?

Visual reasoning — after your attempt

Earth’s gravitational attraction supplies the inward net force in the simple circular-orbit model.

Uniform circular motion has acceleration toward the center because velocity direction changes. The required radial net force is mv²/r; centripetal names the role of the net force, not an additional interaction to draw alongside gravity or tension.

A horizontal ideal turn may obtain radial force from friction. An orbit may obtain it from gravitational attraction.

Pause and explain

Should centripetal force be added to gravity in an orbital diagram?

Check your reasoning after answering

No. Gravity can supply the radial net force; adding a second centripetal interaction double-counts the same requirement.

Part 2 · Check speed and radius dependence

In ac=v²/r, speed is squared but radius is in the denominator. Doubling speed at a fixed radius requires four times the inward acceleration; doubling radius at fixed speed halves it. A larger required acceleration means a larger inward force for unchanged mass. This does not guarantee that friction or tension can supply it: if the available interaction is too small, the assumed circular path cannot continue.

Speed and radius enter the circular model differentlySpeed and radius enter the circular model differentlyInward acceleration for r=1 m (m/s²)048121601234Speed v (m/s)Doubling speed quadruples v²/r at fixed radius.
Calculated values for a fixed 1 m radius; smooth curvature is the square-law model.

Read the visual model

Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.

What happens to ac if both speed and radius double?

Visual reasoning — after your attempt

It changes by 2²/2=2, so it doubles.

Radial acceleration v²/r increases with the square of speed at fixed radius and decreases with radius at fixed speed. Angular speed descriptions give v=ωr and a=ω²r. State which quantity is held constant before making comparisons.

Doubling radius at fixed speed halves acceleration; doubling radius at fixed angular speed doubles it. The conditions determine the comparison.

Pause and explain

Why can radius comparisons give opposite answers?

Check your reasoning after answering

Holding speed and holding angular speed are different experiments because v=ωr links the quantities.

Part 3 · Measure gravitational separation correctly

For ideal point masses or a spherical source outside its surface, Newtonian gravitation gives F=GMm/r² with center-to-center separation r. A satellite’s altitude h above a spherical Earth is not r; use r=R+h. Doubling center distance reduces force and gravitational acceleration to one quarter. The familiar near-surface g≈9.8 m/s² is an approximation over modest height differences, not the same value at every orbital distance.

Gravitational separation is measured between centersGravitational separation is measured between centerscenter-to-center rEarthF = GMm/r²; altitude h is not the same as r.For Earth radius R: r = R+h.
Original spherical-source separation model; Earth/satellite sizes and distances are not to scale.

Read the visual model

Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.

At altitude h=R, what fraction of surface g does this model predict?

Visual reasoning — after your attempt

r=2R, so g is one quarter of its surface value.

Newton’s point-mass gravitational model uses F=Gm₁m₂/r², with r the separation between centers for suitable spherical bodies. Height above a surface must be added to body radius. A doubled height does not mean a doubled center separation.

At a distance twice Earth’s radius from its center, the inverse-square model predicts one-quarter of the near-surface gravitational acceleration.

Pause and explain

Why is altitude alone not the r in the gravity equation?

Check your reasoning after answering

The equation measures center-to-center separation; the surface is already one body radius from the center.

Part 4 · Explain orbital free fall and limitations

In an ideal circular orbit, gravitational acceleration continuously turns a tangential velocity so the object falls around the source. An astronaut can feel weightless because there is little supporting contact force, although gravitational force remains substantial. The circular speed √(GM/r) follows by equating inward acceleration with GM/r². Real orbits can be elliptical and can involve drag or additional bodies, so a circular calculation must state its approximation.

An orbit is free fall, not absence of gravityAn orbit is free fall, not absence of gravityTangent velocityInward gravityCurved free fallOrbit: continuously falling while moving sidewaysCircular condition: GM/r² = v²/r.No support force is needed to maintain ideal free fall.
Original causal model for circular orbital free fall, not an orbital scale drawing.

Read the visual model

Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.

Does apparent weightlessness establish g=0?

Visual reasoning — after your attempt

No. Orbiting objects and astronauts can accelerate together under gravity with little support force.

An ideal circular orbit is continuous free fall with inward acceleration, not absence of gravity. Combining gravity with radial acceleration gives v=√(GM/r). The model neglects drag, nonspherical fields, and other bodies; real trajectories can be elliptical and perturbed.

An orbiting astronaut and spacecraft can share free-fall acceleration, producing little support force even though gravity is present.

Pause and explain

Does weightlessness in orbit imply zero gravitational force?

Check your reasoning after answering

No. It commonly indicates little support interaction while both astronaut and craft accelerate together under gravity.

Apply the whole unit · Calculate a model circular turn

Illustrative 1000 kg vehicle moves at 10.0 m/s around a flat curve of radius 50.0 m. Use g=10.0 m/s² and assume tire static friction supplies the horizontal radial force.

Your task

Find radial acceleration, required force, and minimum friction coefficient in the model. Predict how doubling speed changes the requirement and identify missing real-road evidence.

What your response should show

  • Define the system, quantities, units, and relevant conditions before selecting a model.
  • Show intermediate calculations or a labeled evidence account, including signs or denominators.
  • Explain how the supplied evidence supports the conclusion rather than reporting only an answer.
  • Identify a meaningful uncertainty, model limit, or measurement needed to test the conclusion.
Response guidance — after your attempt
  1. a=v²/r=100/50=2.00 m/s²; required radial force=2000 N.

  2. N≈mg=10000 N; μs must be at least 2000/10000=0.200 in this model.

  3. At 20.0 m/s, required force is 8000 N and modeled minimum μs=0.800.

The squared-speed dependence makes the force requirement rise fourfold, not twofold. This is a supplied mechanics model, not a driving recommendation. Real conditions need evidence about surface, tires, banking, vehicle dynamics, and available friction; the calculation does not certify a safe speed.

Scientific references and source-model foundations

Parent / teacher teaching notes

Require students to name the interaction supplying radial force. Use diagrams and supplied records rather than attempting a circular-motion vehicle test. Ask the learner to explain each symbol, axis, unit and arrow before calculating. Calculated examples, supplied illustrative records and actual historical observations are labeled separately. Selected public OpenSciEd middle-school source figures are foundation examples, not complete high-school editions or an endorsement; all advanced explanations are supplied locally.

planetrvFcGravity supplies the centripetal force: a = v²/r, F = mv²/r
In orbit, velocity (v) points along the path (tangent) while the centripetal force (Fc) points inward toward the planet's center — here supplied entirely by gravity.

The key idea

Circular motion needs a center-seeking net force: a = v²/r and F = mv²/r, supplied by a real force. Gravity is universal, F = Gm₁m₂/r² (inverse-square), and it supplies the centripetal force that holds satellites in orbit.

Centripetal Acceleration

a = v² ÷ r

v is speed in m/s; r is the radius of the circular path in meters; a points toward the center.

Centripetal Force

F = mv² ÷ r

m is mass in kg; this is the net inward force required for circular motion, supplied by a real force like tension, friction, or gravity.

Universal Gravitation

F = G m₁m₂ ÷ r²

m₁ and m₂ are the two masses in kg; r is the distance between their centers in meters; G = 6.67 × 10⁻¹¹ N·m²/kg².

Worked examples

Example 1

A 0.5 kg ball on a string is whirled in a circle of radius 2 m at a speed of 4 m/s. Find the centripetal force on the ball.
  1. Use F = mv²/r.
  2. F = 0.5 × (4²) ÷ 2 = 0.5 × 16 ÷ 2.
  3. = 8 ÷ 2 = 4.
Answer: 4 N (supplied by the string's tension, pointing inward)

Example 2 · Worked application

A 2 kg object moves at 4 m/s in a circle of radius 2 m. Find required net inward force.
  1. a_c=v²/r=16/2=8 m/s².
  2. F_net=ma_c=2×8=16 N.
  3. Identify actual forces supplying that inward sum in the physical setup.
Answer: 16 N inward.
Check the reasoning

Centripetal describes net inward force, not an additional mysterious interaction.

Example 3

Find the gravitational force between Earth (mass 5.97 × 10²⁴ kg) and a 50 kg student standing on its surface, r = 6.37 × 10⁶ m. (G = 6.67 × 10⁻¹¹)
  1. Use F = Gm₁m₂/r² = (6.67 × 10⁻¹¹)(50)(5.97 × 10²⁴) ÷ (6.37 × 10⁶)².
  2. Numerator ≈ 1.99 × 10¹⁶; denominator = 4.06 × 10¹³.
  3. F ≈ 491 N — which is just the student's weight, mg = 50 × 9.8 ≈ 490 N.
Answer: About 491 N (this is the student's weight)

Example 4 · Find and fix the mistake

Context: A 2 kg object moves at 4 m/s in a circle of radius 2 m. Find required net inward force. A student claims: “Uniform circular motion has no acceleration because speed is constant.” Evaluate the claim and write a corrected explanation.
  1. Test the claim against the stated conditions: Centripetal describes net inward force, not an additional mysterious interaction.
  2. Direction changes, so velocity changes and inward acceleration is nonzero.
Answer: Direction changes, so velocity changes and inward acceleration is nonzero.
Check the reasoning

Centripetal describes net inward force, not an additional mysterious interaction.

Example 5 · Reason from the visual model

Gravitational separation is measured between centersGravitational separation is measured between centerscenter-to-center rEarthF = GMm/r²; altitude h is not the same as r.For Earth radius R: r = R+h.
Original spherical-source separation model; Earth/satellite sizes and distances are not to scale.
At altitude h=R, what fraction of surface g does this model predict?
  1. For ideal point masses or a spherical source outside its surface, Newtonian gravitation gives F=GMm/r² with center-to-center separation r. A satellite’s altitude h above a spherical Earth is not r; use r=R+h. Doubling center distance reduces force and gravitational acceleration to one quarter. The familiar near-surface g≈9.8 m/s² is an approximation over modest height differences, not the same value at every orbital distance.
  2. r=2R, so g is one quarter of its surface value.
Answer: r=2R, so g is one quarter of its surface value.
Check the reasoning

Original spherical-source separation model; Earth/satellite sizes and distances are not to scale.

Avoid these mistakes

Check your understanding

v=4 m/s,r=2 m. Find inward acceleration.

Discuss the answer

ac=v²/r=8 m/s². Speed can be constant while velocity direction changes.

Return to the phenomenon

Return to the initial case. How can a satellite fall without striking the ground? Revise your drawing or explanation using two named ideas and one stated limitation.

Compare explanations after your attempt

Tangential velocity and inward acceleration combine into free fall around the source; a circular model has v²/r=GM/r².

What should we investigate next?

Which comparison would test a remaining uncertainty? State what you would change, keep comparable and measure.

Investigate, calculate and revise

Keep a first model and an evidence-supported revision. Attempt each task before opening guidance; all required explanations and supplied-data routes are available locally.

Activity 1 · Return to the opening case

Assigned reading and preparation

Use the opening phenomenon, unit definitions and explained Parts 1–4.

Your task

Return to the initial case. How can a satellite fall without striking the ground? Revise your drawing or explanation using two named ideas and one stated limitation.

Success criteria

  • Use the stated observations and appropriate units.
  • Connect a labeled model with the physical/environmental mechanism.
  • State assumptions and an evidence limitation.
  • Explain a substantive revision or next test.
Hint

Keep object/material identity, material transfer and energy account distinct.

Response guidance — discuss after your attempt

Tangential velocity and inward acceleration combine into free fall around the source; a circular model has v²/r=GM/r².

Activity 2 · Transfer to a changed case

Assigned reading and preparation

Cover the worked answer and use the explained models, relationships and notebook.

Your task

A 0.50 kg mass moves at 4.0 m/s in a 2.0 m horizontal circle. Find inward acceleration and required net force; name what else is needed to identify tension.

Success criteria

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

Name known quantities and conditions, show intermediate steps and check conservation.

Response guidance — discuss after your attempt

ac=16/2=8.0 m/s² and inward net force=0.50×8=4.0 N. Tension equals that net only if the specified geometry and other forces establish it as the sole inward contribution.

Words to know

centripetal acceleration
The acceleration of an object moving in a circle, always directed toward the center; a = v²/r.
centripetal force
The net inward force required to keep an object moving in a circle; F = mv²/r. It is supplied by a real force such as tension, friction, or gravity.
period (T)
The time it takes an object to complete one full revolution around a circle.
Newton's law of universal gravitation
Every pair of masses attracts with a force F = Gm₁m₂/r², where r is the distance between their centers.
gravitational constant (G)
The fixed number in the gravitation equation, G = 6.67 × 10⁻¹¹ N·m²/kg², measured first by Henry Cavendish.

Newton's apple and the unification of the heavens

The famous story of an apple falling in Newton's garden captures a genuine breakthrough: Newton realized that the force pulling the apple down might be the same force holding the Moon in its orbit. This was a stunning claim, because for centuries the heavens and the Earth were thought to obey entirely different rules. His law of universal gravitation, F = Gm₁m₂/r², proposed that every pair of masses in the universe attracts, from apples to moons to distant galaxies. Using it, Newton could explain Kepler's laws of planetary motion, the tides, and the paths of comets with a single equation — one of the greatest unifications in the history of thought.

Cavendish 'weighs the Earth': measuring G

Newton's equation contained the constant G, but he never knew its value. In 1798, more than a century later, the British scientist Henry Cavendish measured it using an exquisitely sensitive torsion balance: two small lead balls on a suspended rod, twisted ever so slightly by the gravitational pull of larger lead spheres brought nearby. From the tiny twist he extracted G ≈ 6.67 × 10⁻¹¹ N·m²/kg². Once G was known, scientists could rearrange the surface-gravity relationship g = GM/r² to calculate Earth's mass for the first time — which is why Cavendish's delicate experiment earned the nickname 'weighing the Earth.'

There is no centrifugal force

When a car turns sharply, you feel thrown against the door and might swear an outward force is pushing you — the so-called centrifugal force. Physically, no such force exists. Your body, obeying Newton's first law, simply tries to continue in a straight line while the car curves underneath you; the door then pushes you inward to make you turn with the car. The outward sensation is an artifact of observing motion from within an accelerating (rotating) frame of reference. In an inertial frame, the only real horizontal force in a turn is the centripetal one, pointing toward the center — supplied here by friction and the car's door.

Orbits, GPS, and Newton's cannonball

Newton explained orbits with a thought experiment: a cannon fired from a tall mountain shoots a ball that arcs to the ground, but fired fast enough, the ground curves away beneath it as quickly as it falls, so the ball circles the planet — an orbit is simply perpetual free-fall. Real satellites embody this: the International Space Station circles about 400 km up at roughly 7.7 km/s, completing an orbit every 90 minutes, while GPS satellites sit much higher and time their signals so precisely that engineers must even correct for effects of relativity. In every case, gravity supplies exactly the centripetal force the circular path demands, setting Gm₁m₂/r² equal to mv²/r.

In the real world

GPS satellites stay in orbit because Earth's gravity supplies precisely the centripetal force their circular paths require — the same physics that keeps the Moon circling Earth.

Exit check

v=4 m/s,r=2 m. Find inward acceleration.

Check your explanation after answering

ac=v²/r=8 m/s². Speed can be constant while velocity direction changes.

Explain your reasoning without notes. If a key idea is missing, revisit the lesson and try again.

Problem 1

A car rounds a curve of radius 50 m at a speed of 10 m/s. Find its centripetal acceleration.
Answer and explanation

2 m/s²

a = v²/r = 10² ÷ 50 = 100 ÷ 50 = 2 m/s², directed toward the center of the curve.

Problem 2

A 2 kg ball moves in a horizontal circle of radius 4 m at 6 m/s. Find the centripetal force on it.
Answer and explanation

18 N

F = mv²/r = 2 × 6² ÷ 4 = 2 × 36 ÷ 4 = 72 ÷ 4 = 18 N.

Problem 3

An object moves in a circle at constant speed. In which direction does the net force on it point?
  1. Along the direction of motion (tangent to the circle)
  2. Outward, away from the center
  3. Toward the center of the circle
  4. There is no net force
Answer and explanation

Toward the center of the circle

Uniform circular motion requires a center-seeking (centripetal) net force. The outward feeling is an illusion of the accelerating frame.

Problem 4

If the distance between two masses is doubled, the gravitational force between them becomes…
  1. one-half as strong
  2. twice as strong
  3. unchanged
  4. one-quarter as strong
Answer and explanation

one-quarter as strong

Gravitation is an inverse-square law (F ∝ 1/r²). Doubling r multiplies the force by 1/2² = 1/4.

Problem 5

A satellite orbits Earth at constant speed. Explain why it is still accelerating, and name the force that provides its centripetal force.
Answer and explanation

Even at constant speed the satellite's direction is always changing, and any change in velocity is acceleration — here it is centripetal acceleration pointing toward Earth's center. The force supplying it is gravity: Earth's gravitational pull provides exactly the mv²/r needed to bend the satellite's path into a circle, so it 'falls around' the planet.

Problem 6

A 1,200 kg car rounds a flat curve of radius 80 m at 20 m/s. What centripetal force must friction supply?
Answer and explanation

6000 N

F = mv²/r = 1,200 × 20² ÷ 80 = 1,200 × 400 ÷ 80 = 1,200 × 5 = 6,000 N, directed toward the center of the curve.

Problem 7

At fixed mass and radius, how does doubling circular speed change required force? Show the reasoning, not just the conclusion.
Hints

Centripetal describes net inward force, not an additional mysterious interaction.

Answer and explanation

It quadruples force because F=mv²/r.

Problem 8

A satellite moves farther from the same spherical planet. How does gravitational field vary outside it?
Hints

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

Answer and explanation

g=GM/r² decreases with distance from the center; doubling r gives one quarter the field.

Problem 9

Gravitational separation is measured between centersGravitational separation is measured between centerscenter-to-center rEarthF = GMm/r²; altitude h is not the same as r.For Earth radius R: r = R+h.
Original spherical-source separation model; Earth/satellite sizes and distances are not to scale.
Interpret the supplied visual. At altitude h=R, what fraction of surface g does this model predict?
Answer and explanation

r=2R, so g is one quarter of its surface value.

For ideal point masses or a spherical source outside its surface, Newtonian gravitation gives F=GMm/r² with center-to-center separation r. A satellite’s altitude h above a spherical Earth is not r; use r=R+h. Doubling center distance reduces force and gravitational acceleration to one quarter. The familiar near-surface g≈9.8 m/s² is an approximation over modest height differences, not the same value at every orbital distance.

r=2R, so g is one quarter of its surface value.

Response criteria

  • Interpret labels, structures, axes, units and directions correctly.
  • Explain the physical or environmental relationship or mechanism.
  • Respect stated assumptions and distinguish observation from inference.

Problem 10

A 0.50 kg mass moves at 4.0 m/s in a 2.0 m horizontal circle. Find inward acceleration and required net force; name what else is needed to identify tension.
Answer and explanation

ac=16/2=8.0 m/s² and inward net force=0.50×8=4.0 N. Tension equals that net only if the specified geometry and other forces establish it as the sole inward contribution.

ac=16/2=8.0 m/s² and inward net force=0.50×8=4.0 N.

Tension equals that net only if the specified geometry and other forces establish it as the sole inward contribution.

Response criteria

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

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