Physics: Foundations of Physical Science
Part II — Motion and Forces · Chapter 13
Circular Motion and Gravitation
13.1 Why Circular Motion Is Different from Linear Motion
In linear motion, acceleration changes the magnitude of velocity. In circular motion at constant speed, acceleration changes the direction of velocity while the magnitude stays constant. Both are accelerations — both require a net force by Newton’s second law — but they feel very different and require different analysis.
A key insight: an object moving in a circle at constant speed is always accelerating, even though its speedometer would read constant. The velocity vector is always tangent to the circle, and since the direction of a tangent changes continuously as the object moves around, the velocity is constantly changing. Any change in velocity is acceleration.
The acceleration required to maintain circular motion points toward the center of the circle at every instant. This is called centripetal acceleration. The corresponding force is the centripetal force.
13.2 Velocity in Circular Motion
For an object moving in a circle of radius r with constant speed v, the velocity vector is always tangent to the circle — perpendicular to the radius at every point. The speed v is the magnitude of this velocity vector and remains constant for uniform circular motion.
The period T is the time to complete one full circle. The speed is the circumference divided by the period:
v = 2πr / T
The frequency f is the number of complete circles per second, measured in hertz. Frequency and period are reciprocals:
f = 1 / T
13.3 Centripetal Acceleration
Even at constant speed, an object in circular motion has acceleration because its velocity direction changes. This centripetal acceleration always points toward the center of the circle.
ac = v2 / r
Equivalent expressions using period T or frequency f are:
ac = 4π2r / T2 = 4π2rf2
Centripetal acceleration increases with speed squared and decreases with increasing radius. Double the speed, and the centripetal acceleration becomes four times greater. Double the radius, and the centripetal acceleration becomes half as large.
| Speed | Radius | Centripetal Acceleration |
|---|---|---|
| v | r | v2/r |
| 2v | r | 4v2/r (quadrupled) |
| v | 2r | v2/(2r) (halved) |
| 2v | 2r | 4v2/(2r) = 2v2/r (doubled) |
13.4 Centripetal Force
By Newton’s second law, the centripetal acceleration requires a net force directed toward the center of the circle. This is the centripetal force:
Fc = mv2 / r
This equation is not a new force law; it is Newton’s second law applied to circular motion. Whatever real force provides the centripetal acceleration — gravity, tension, friction, normal force, or any combination — its inward component must equal mv2/r for the object to maintain circular motion.
13.5 Sources of Centripetal Force
In different physical situations, different forces supply the centripetal acceleration. Recognizing the source of centripetal force is the first step in any circular motion problem.
| Situation | Source of Centripetal Force | Direction Toward Center |
|---|---|---|
| Ball on a string, horizontal circle | Tension in string | Along string, toward hand |
| Car rounding a flat curve | Friction from road on tires | Horizontal, toward center of curve |
| Car on banked curve | Component of normal force | Horizontal inward component of N |
| Satellite in orbit | Gravity | Toward Earth’s center |
| Ball in a vertical loop, bottom | Normal force minus weight | Net upward = toward center |
| Ball in a vertical loop, top | Weight plus normal force | Net downward = toward center |
| Planet orbiting Sun | Gravitational attraction | Toward Sun |
13.6 Banking and Circular Motion Applications
A banked curve is tilted at an angle θ inward so that the horizontal component of the normal force provides centripetal acceleration, reducing or eliminating the need for friction. At the ideal banking angle for a given speed, no friction is needed at all.
tanθ = v2 / (rg)
Vertical circular motion introduces an additional complication: the gravitational force changes its contribution to centripetal force depending on position in the loop. At the bottom of a loop, the normal force must exceed weight to provide centripetal acceleration upward. At the top of a loop, weight and normal force both point toward the center; the minimum speed occurs when N = 0.
13.7 Universal Gravitation
Gravity is universal: every object in the universe with mass attracts every other object with mass. This was Newton’s great insight, published in the Principia in 1687: the same force that causes an apple to fall causes the Moon to orbit Earth, and the same force governs planets orbiting the Sun.
Before Newton, gravity and the celestial motions of planets were considered entirely separate phenomena. Newton unified them with a single mathematical law, showed that the inverse-square relationship produced Kepler’s elliptical orbits, and demonstrated that the force causing free fall on Earth was exactly sufficient to keep the Moon in its orbit. This unification was one of the greatest intellectual achievements in scientific history.
13.8 Newton’s Law of Gravitation
The gravitational force between two masses is attractive, acts along the line joining them, and has magnitude:
Fg = Gm1m2 / r2
Key features of the law: the force is always attractive; it is proportional to each mass separately; it is inversely proportional to the square of the separation.
| Change | Effect on Fg |
|---|---|
| Double m1 | Fg × 2 |
| Double both masses | Fg × 4 |
| Double the distance r | Fg ÷ 4 (inverse square) |
| Triple the distance r | Fg ÷ 9 |
| Halve the distance r | Fg × 4 |
13.9 Gravitational Fields
Rather than thinking of gravity as an action-at-a-distance force, physicists often describe it using the concept of a gravitational field. The gravitational field g at any point in space is defined as the gravitational force per unit mass that would act on a small test mass placed there:
g = Fg / m = −GM / r2 (pointing toward M)
Near Earth’s surface, the gravitational field is approximately uniform with magnitude g = 9.81 m/s2, directed downward. This is the g used throughout Chapters 6–12. As altitude increases, g decreases:
g(r) = GMEarth / r2
13.10 Weight vs. Mass Revisited
We can now revisit the weight-mass distinction with greater precision. The weight of an object is the gravitational force on it: W = mg, where g is the local gravitational field strength. Mass m is constant; weight varies with location because g varies.
An astronaut on the ISS, at 400 km altitude, experiences g ≈ 8.69 m/s2. Their weight is W = m × 8.69 N — not zero. The astronaut appears weightless because the station and the astronaut are in the same free-fall orbit. Both accelerate toward Earth at the same rate, so there is no contact force between them. The normal force is zero, giving the sensation of weightlessness.
13.11 Orbital Motion
An orbit is a continuous free fall around a body. The satellite falls toward Earth due to gravity, but moves so fast horizontally that Earth’s surface curves away at the same rate the satellite falls. The result is a circular or elliptical path that never intersects the surface.
For a circular orbit, the gravitational force provides exactly the centripetal force needed:
GMm / r2 = mv2 / r
Solving for the orbital speed:
vorbit = √(GM / r)
All objects at the same orbital radius orbit at the same speed, regardless of mass.
13.12 Satellites and Planetary Motion
The orbital period T is related to orbital radius r by combining v = 2πr/T with v = √(GM/r):
T2 = (4π2 / GM) r3
This relationship, first discovered empirically by Johannes Kepler in 1619 from astronomical observations, was later derived theoretically by Newton from his gravitational law. It is one of the great confirmations that Newton’s mechanics and gravity are correct.
| Object | Orbital Radius (m) | Period | Orbital Speed (m/s) |
|---|---|---|---|
| Moon | 3.84 × 108 m | 27.3 days | 1020 |
| ISS | 6.77 × 106 m | 92.7 min | 7670 |
| GPS satellite | 2.66 × 107 m | 12 h | 3870 |
| Geostationary satellite | 4.22 × 107 m | 24 h | 3070 |
| Earth around Sun | 1.50 × 1011 m | 365.25 days | 29,800 |
13.13 Escape Velocity and Orbital Concepts
Escape velocity is the minimum speed required for an object to escape a planet’s gravitational pull entirely — to reach infinite distance without further propulsion. It is found by setting the object’s kinetic energy equal to the work needed to escape the gravitational well:
vescape = √(2GM / r)
Note that vescape = √2 × vorbit: escape velocity is about 1.41 times the circular orbital speed at the same radius. For Earth’s surface, vescape ≈ 11.2 km/s.
| Body | Surface g (m/s2) | Escape Velocity (km/s) |
|---|---|---|
| Earth | 9.81 | 11.2 |
| Moon | 1.62 | 2.38 |
| Mars | 3.71 | 5.03 |
| Jupiter | 24.8 | 59.5 |
| Sun | 274 | 618 |
| Black hole | Extreme | > c; light cannot escape |
Geosynchronous orbit is the radius at which the orbital period equals Earth’s rotation period, 24 hours. A satellite in geosynchronous orbit above the equator appears stationary from the ground — this is the basis for communication and weather satellites. Geosynchronous radius is about 42,200 km from Earth’s center.
13.14 Common Mistakes in Circular Motion Analysis
Mistake 1: Inventing a “centrifugal force” pointing outward
In an inertial reference frame, there is no outward centrifugal force. The sensation of being pushed outward in a turning car is inertia — your body tends to continue in a straight line while the car turns. Centrifugal force is a fictitious force that appears only in rotating reference frames.
Mistake 2: Setting centripetal force equal to a single force when multiple forces act
The centripetal force is the net inward force. Write Newton’s second law in the radial direction and sum all inward force components. Only then set that sum equal to mv2/r.
Mistake 3: Confusing centripetal acceleration with tangential acceleration
Centripetal acceleration changes the direction of velocity. Tangential acceleration changes the speed. Uniform circular motion has centripetal acceleration only; non-uniform circular motion has both.
Mistake 4: Using weight for centripetal force in orbital problems
In orbital problems, gravity is the centripetal force: set Fg = mv2/r. The satellite’s weight is not zero; it equals Fg, which provides the centripetal acceleration.
Mistake 5: Not converting units before using gravitational formulas
G = 6.674 × 10−11 N·m2/kg2 requires mass in kilograms and distance in meters. Convert all distances and masses before substituting.
Chapter Summary
Circular motion requires a net inward, or centripetal, force. Gravity is the centripetal force for orbits. Newton’s law of universal gravitation governs all gravitational interactions.
- Uniform circular motion: constant speed, continuously changing velocity direction; always accelerating.
- Centripetal acceleration: ac = v2/r, directed toward center of circle.
- Centripetal force: Fc = mv2/r, directed toward center; provided by real forces such as tension, friction, gravity, and normal force.
- Centripetal force is not a new force type — it is the label for whatever net inward force maintains circular motion.
- Banking angle for a curve: tanθ = v2/(rg) for friction-free cornering.
- Newton’s law of gravitation: Fg = Gm1m2/r2; attractive; inverse-square law.
- Gravitational field g = GM/r2; decreases with altitude; near Earth’s surface, about 9.81 m/s2.
- Orbital speed: v = √(GM/r); depends only on M and r, not on satellite mass.
- Kepler’s Third Law: T2 = (4π2/GM)r3.
- Escape velocity: vesc = √(2GM/r) = √2 × vorbit.
- Orbiting astronauts are weightless in sensation but not in gravity — they are in continuous free fall.
Key Terms
- uniform circular motion
- Motion at constant speed along a circular path; velocity is always tangent to the circle; acceleration is always toward the center.
- centripetal acceleration
- The acceleration directed toward the center of a circular path: ac = v2/r; responsible for changing the direction of velocity.
- centripetal force
- The net inward force required to maintain circular motion: Fc = mv2/r; not a new force type but a label for the net inward real force.
- period (T)
- The time for one complete revolution; T = 2πr/v for uniform circular motion.
- frequency (f)
- The number of complete revolutions per second; f = 1/T; units: Hz.
- centrifugal force
- A fictitious outward force that appears in rotating reference frames; does not exist in inertial frames and should not be included in FBDs.
- universal gravitation
- Newton’s law that every mass attracts every other mass: Fg = Gm1m2/r2.
- gravitational constant (G)
- The universal constant in Newton’s gravity law: G = 6.674 × 10−11 N·m2/kg2.
- gravitational field
- The gravitational force per unit mass at a point in space; g = GM/r2 directed toward M; units m/s2.
- orbital speed
- The speed required to maintain a circular orbit at radius r: v = √(GM/r).
- Kepler’s Third Law
- The orbital period squared is proportional to the orbital radius cubed: T2 = (4π2/GM)r3.
- escape velocity
- The minimum speed to escape a gravitational field: vesc = √(2GM/r).
- geosynchronous orbit
- The orbital radius at which the period equals Earth’s rotation period; about 42,200 km from Earth’s center; satellite appears stationary from the ground.
Concept Review Questions
- Explain why an object moving at constant speed in a circle is still accelerating. In what direction is this acceleration?
- What is centripetal force? Is it a new type of force? Give three physical examples of forces that can serve as centripetal force.
- A car moves faster around a curve of the same radius. What happens to the required centripetal force? What happens if the radius doubles at the same speed?
- Why do astronauts on the ISS feel weightless even though gravity is nearly as strong there as at Earth’s surface?
- How does Newton’s law of universal gravitation differ from the simple W = mg formula for weight? When is W = mg a valid approximation?
- Explain why orbital speed decreases with increasing orbital radius. Use the formula v = √(GM/r) in your explanation.
- What is Kepler’s Third Law? Was it discovered before or after Newton’s law of gravitation?
- Explain the concept of escape velocity in your own words. Why is it √2 times the orbital speed?
- Why is the centrifugal force not included in free-body diagrams for circular motion in inertial frames?
Practice Problems
Circular Motion Problems
- A 0.40 kg ball on a 0.75 m string swings in a horizontal circle at 5.0 m/s. Find: (a) centripetal acceleration; (b) tension in the string.
- A car of mass 1100 kg rounds a flat curve (r = 50.0 m) at 15.0 m/s. (a) Find the centripetal force. (b) What is the minimum μs needed?
- A highway curve of radius 200 m is banked at 12°. At what speed can a car round it without relying on friction?
- A roller coaster loop has radius 15.0 m. (a) What minimum speed at the top keeps the car on the track? (b) What normal force does a 70.0 kg rider experience at the bottom if traveling at 22.0 m/s?
- A 1.5 kg mass swings on a 1.2 m rope in a vertical circle. (a) What is the tension at the bottom of the swing when moving at 4.0 m/s? (b) What is the tension at the top if moving at 3.5 m/s?
- A satellite travels in a circular orbit of radius 8.0 × 106 m around Earth. Find: (a) centripetal acceleration; (b) orbital speed; (c) period.
Gravitation Problems
- Find the gravitational force between Earth (m = 5.97 × 1024 kg) and the Moon (m = 7.35 × 1022 kg) separated by 3.84 × 108 m.
- Two identical 5.0 kg balls are 0.50 m apart. Find the gravitational force between them.
- At what height above Earth’s surface does g = 4.905 m/s2, half of surface value? Earth radius = 6.37 × 106 m.
- Compare the gravitational force on a 70 kg person at Earth’s surface to the force at 400 km altitude.
- What is the weight of a 50.0 kg astronaut on the Moon (gMoon = 1.62 m/s2)? On Mars (gMars = 3.71 m/s2)?
Orbital Mechanics Exercises
- Find the orbital speed and period of a satellite at r = 7.0 × 106 m from Earth’s center. MEarth = 5.97 × 1024 kg.
- The Moon orbits Earth in T = 27.3 days at r = 3.84 × 108 m. Verify this is consistent with g = 9.81 m/s2 at Earth’s surface using the inverse-square law.
- Find the radius of geosynchronous orbit (T = 86,400 s). MEarth = 5.97 × 1024 kg.
- Find the escape velocity from Earth’s surface. Find the escape velocity from the Moon’s surface, where MMoon = 7.35 × 1022 kg and RMoon = 1.74 × 106 m.
- Planet X has twice Earth’s mass and twice Earth’s radius. Compare its surface g to Earth’s. Compare escape velocity.
Applied Physics Problems
- A pilot pulls out of a dive at 180 m/s. The radius of curvature at the bottom is 1500 m. Find: (a) centripetal acceleration as a multiple of g; (b) the normal force on a 75.0 kg pilot.
- A 0.30 kg puck slides in a horizontal circle on a frictionless table, connected by a string through a hole to a hanging 0.20 kg mass. If the puck moves at 1.5 m/s, what is the radius of the circle?
- Challenge: A conical pendulum: a 0.50 kg ball on a 1.0 m string swings in a horizontal circle with the string at 30° from vertical. Find: (a) the radius of the circle; (b) the tension; (c) the speed.
- Find the orbital period of a hypothetical satellite orbiting just above Earth’s surface (r = REarth = 6.37 × 106 m). Compare your answer to √(R/g).
Chapter 13 Checkpoint
| I can… | Section |
|---|---|
| Explain why constant-speed circular motion requires acceleration | 13.1 |
| Calculate centripetal acceleration using ac = v2/r | 13.3 |
| Calculate centripetal force using Fc = mv2/r | 13.4 |
| Identify the real force providing centripetal acceleration in various situations | 13.5 |
| Find the ideal banking angle for a given speed and radius | 13.6 |
| Apply Newton’s law of gravitation Fg = Gm1m2/r2 | 13.8 |
| Calculate g at any distance from Earth using g = GM/r2 | 13.9 |
| Find orbital speed using v = √(GM/r) | 13.11 |
| Apply Kepler’s Third Law T2 = (4π2/GM)r3 | 13.12 |
| Calculate escape velocity using v = √(2GM/r) | 13.13 |
| Explain weightlessness in orbit as free fall, not zero gravity | 13.10 |
| Avoid adding fictitious centrifugal force to FBDs | 13.14 |