Banked curves, vertical circles and universal gravitation AP Physics 1 · Lesson 6 of 7
Circular motion in the real world: cars on flat and banked curves, roller coaster loops and hills, and the gravity that keeps the Moon and satellites in orbit. You will use Newton's law of universal gravitation, think about gravitational fields and see where Kepler's laws come from.
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This lesson covers banked vs unbanked curves, vertical circles and loops, universal gravitation, gravitational forces and fields, and Kepler's laws:
- Banked vs unbanked curves
- Vertical circles and loops
- Universal gravitation and fields
- Orbits and Kepler's laws
All numbers use g = 9.8 m/s², as on the AP Physics 1 equation sheet. Every formula used here is on that sheet: AP Physics 1 equation sheet.
Banked vs unbanked curves
On a flat (unbanked) curve, the only horizontal force on a car is static friction from the road, pointing toward the centre. As for the coin on a turntable (Lesson 5), the fastest safe speed is when friction reaches its maximum:
μsmg = mv²/r ⇒ vmax = √(μsgr)
A banked curve tilts the road toward the centre. Now the normal force is tilted too, so part of it points inward. At the right speed, the design speed, no friction is needed at all.
Vertical: FN cosθ = mg Inward: FN sinθ = mv²/r ⇒ tanθ = v²/(rg)
| Car speed on a banked curve | What friction does |
|---|---|
| Exactly the design speed | No friction needed |
| Faster than the design speed | Friction acts down the slope (toward the centre) to add to the inward force |
| Slower than the design speed | Friction acts up the slope, to stop the car sliding down the bank |
Worked example 1: flat curve
A car rounds a flat curve of radius 50 m. μs between the tyres and road is 0.70. What is the maximum speed?
- vmax = √(μsgr) = √(0.70 × 9.8 × 50) = 18.5 m/s (about 41 mph).
Worked example 2: banking angle
Engineers want cars to round a 120 m radius curve at 25 m/s with no need for friction. What banking angle is needed?
- tanθ = v²/(rg) = 625/(120 × 9.8) = 0.531.
- θ = 28°. The mass of the car does not matter.
Common trap
On a banked curve, FN is bigger than mg (FN = mg/cosθ), unlike on a slope where a block slides down and FN = mg cosθ. The difference is the acceleration: on the curve it is horizontal, so the vertical forces must balance.
Vertical circles and loops
In a vertical circle, gravity sometimes points toward the centre (at the top) and sometimes away from it (at the bottom). The speed usually changes around the loop, but at the very top and bottom the acceleration toward the centre is still v²/r. Take toward the centre as positive in each position.
| Position | Toward the centre is | Newton's second law | Normal force (or tension) |
|---|---|---|---|
| Bottom of a loop or dip | Up | FN − mg = mv²/r | FN = mg + mv²/r (bigger than mg) |
| Top, inside a loop | Down | FN + mg = mv²/r | FN = mv²/r − mg |
| Top of a hill (car on the outside) | Down | mg − FN = mv²/r | FN = mg − mv²/r (less than mg) |
Minimum speed at the top of a loop. The cart stays on the track only while FN ≥ 0. At the slowest possible speed, the track pushes with zero force and gravity alone provides the inward force: mg = mv²/r, so
vmin, top = √(gr)
The same speed is the maximum for a car going over the top of a hill: faster than √(gr) and gravity cannot pull it round the curve, so the car leaves the road. A ball whirled on a string in a vertical circle has the same minimum speed at the top, with tension in place of the normal force.
Worked example 1: bottom of a dip
A 70 kg rider on a roller coaster goes through the bottom of a dip of radius 12 m at 15 m/s. What is the normal force from the seat?
- Up (toward the centre) is positive: FN − mg = mv²/r.
- FN = 70(9.8) + 70(225)/12 = 686 + 1313 = 2,000 N, about 2.9 times the rider's weight.
Worked example 2: top of a loop
The same rider goes over the top of a loop of radius 8.0 m, upside down, at 12 m/s. What is the seat's force on the rider? What is the slowest safe speed?
- Down is toward the centre: FN + mg = mv²/r, so FN = 70(144)/8.0 − 686 = 1260 − 686 = 574 N (downward, from the seat above the rider).
- Slowest speed: vmin = √(gr) = √(9.8 × 8.0) = 8.9 m/s.
Worked example 3: hill
A car drives over a hilltop shaped like an arc of radius 40 m. How fast can it go before its wheels leave the road?
- At the limit FN = 0, so mg = mv²/r.
- v = √(gr) = √(9.8 × 40) = 19.8 m/s.
Universal gravitation and gravitational fields
Every pair of masses attracts each other with a gravitational force along the line between their centres. Newton's law of universal gravitation:
Fg = Gm1m2/r² G = 6.67 × 10−11 N·m²/kg²
r is the distance between the centres. For a satellite or a planet's surface, that is the planet's radius plus the height. The force on each mass is the same size (Newton's third law), even if one mass is far bigger.
| Change | Effect on Fg |
|---|---|
| Double one mass | ×2 |
| Double both masses | ×4 |
| Double the distance | ×1/4 |
| Triple the distance | ×1/9 |
| Halve the distance | ×4 |
The gravitational field g is the gravitational force per kilogram at a point. Near a planet of mass M, at distance r from its centre:
g = Fg/m = GM/r²
Worked example 1: g at Earth's surface
Use G = 6.67 × 10−11 N·m²/kg², Earth's mass 5.97 × 1024 kg and radius 6.37 × 106 m to find g at the surface.
- g = GM/R² = (6.67 × 10−11)(5.97 × 1024)/(6.37 × 106)² = 9.81 m/s².
Worked example 2: ratio reasoning
The International Space Station orbits about 400 km above Earth's surface. What is g there, as a fraction of g at the surface?
- r = 6370 + 400 = 6770 km from Earth's centre.
- g/gsurface = (6370/6770)² = 0.885, so g ≈ 8.7 m/s².
- Gravity is still almost 90% as strong up there. Astronauts float because they are in free fall around Earth, not because there is no gravity.
Key idea
AP questions on gravitation are mostly ratio questions. Write the formula, change one variable at a time, and track the factor. If a planet has twice Earth's mass and twice its radius, surface g = G(2M)/(2R)² = (1/2) Earth's g.
Orbits and Kepler's laws
For a satellite or planet in a circular orbit, gravity is the only force, and it provides the net inward force:
GMm/r² = mv²/r ⇒ v = √(GM/r) T² = (4π²/GM)r³
The satellite's mass m cancels, so every object at the same orbital radius has the same speed and period. A closer orbit is faster. The second equation comes from putting v = 2πr/T into the first. It is Kepler's third law for circular orbits.
| Kepler's law | What it says | Why (in AP Physics 1 terms) |
|---|---|---|
| First | Planets move in ellipses with the Sun at one focus | A result of the inverse-square gravitational force (circles are a special case) |
| Second | The line from the Sun to a planet sweeps out equal areas in equal times | The planet speeds up as it falls closer and slows as it moves away (angular momentum is conserved, Unit 6) |
| Third | T² ∝ r³ (r = average distance, or the radius of a circular orbit) | From GMm/r² = m(2πr/T)²/r |
Worked example 1: orbital speed and period
A satellite orbits 600 km above Earth's surface. Find its speed and period.
- r = 6.37 × 106 + 0.60 × 106 = 6.97 × 106 m.
- v = √(GM/r) = √(6.67 × 10−11 × 5.97 × 1024/6.97 × 106) = 7,560 m/s.
- T = 2πr/v = 2π(6.97 × 106)/7560 = 5,790 s ≈ 97 minutes.
Worked example 2: Kepler's third law as a ratio
Moon A orbits a planet at radius R with period 4.0 days. Moon B orbits the same planet at radius 4R. What is B's period?
- T² ∝ r³, so TB/TA = (4)3/2 = 8.
- TB = 8 × 4.0 = 32 days.
Key equations
| Flat curve, maximum speed | vmax = √(μsgr) |
|---|---|
| Banked curve, no friction | tanθ = v²/(rg) |
| Top of a loop, minimum speed | vmin = √(gr) |
| Universal gravitation | Fg = Gm1m2/r², G = 6.67 × 10−11 N·m²/kg² |
| Gravitational field | g = GM/r² |
| Circular orbit | v = √(GM/r), T² = (4π²/GM)r³ |
Practice problems
18 problems, from warm-up to exam level. Answer each one before you look: use the hint if you are stuck, then reveal the worked solution one step at a time. Number answers are marked within 2.5%, so g = 10 m/s² still counts.
Hint
Maximum static friction provides mv²/r.
Worked solution
- v = √(μsgr) = √(0.60 × 9.8 × 40) = 15.3 m/s.
Hint
tanθ = v²/(rg).
Worked solution
- tanθ = 400/(80 × 9.8) = 0.510.
- θ = 27.0°.
Hint
Rearrange tanθ = v²/(rg).
Worked solution
- v = √(rg tanθ) = √(150 × 9.8 × 0.268) = 19.8 m/s.
Hint
Faster means a bigger inward force is needed than the normal force alone can give.
Worked solution
- The car tends to slide up and out of the curve.
- Friction opposes that, so it acts down the bank, adding to the inward force. The answer is B.
Hint
Only the vertical part of FN balances the weight.
Worked solution
- FN cosθ = mg, so FN = mg/cosθ.
- cosθ < 1, so FN > mg. The answer is C.
Hint
At the bottom, the centre is upward: FN − mg = mv²/r.
Worked solution
- FN = 60(9.8) + 60(196)/10 = 588 + 1176 = 1760 N.
Hint
At the minimum speed the track pushes with zero force.
Worked solution
- mg = mv²/r, so v = √(gr) = √(9.8 × 6.0) = 7.67 m/s.
Hint
At the top, both tension and gravity point down, toward the centre.
Worked solution
- FT + mg = mv²/r.
- FT = 0.40(25)/0.90 − 3.92 = 11.1 − 3.92 = 7.19 N.
Hint
Compare FN = mg − mv²/r and FN = mg + mv²/r.
Worked solution
- At the bottom, the seat must push up enough to cancel gravity and supply the inward force: FN > mg.
- At the top, FN < mg. The answer is B.
Hint
F = Gm1m2/r².
Worked solution
- F = 6.67 × 10−11 × 70 × 70/1.0² = 3.27 × 10−7 N, far too small to notice.
Hint
Change one variable at a time.
Worked solution
- Doubling a mass multiplies F by 2.
- Tripling the distance multiplies F by 1/9. Overall 2/9. The answer is A.
Hint
Newton's third law.
Worked solution
- The two forces are a third-law pair, equal in size.
- Their effects differ (Earth's acceleration is tiny), but the forces are equal. The answer is B.
Hint
g = GM/R². Scale each factor.
Worked solution
- gX = 9.8 × 3/2² = 7.35 N/kg.
Hint
One quarter means double the distance from Earth's centre.
Worked solution
- g ∝ 1/r², so g/4 at r = 2RE.
- That is a height of 2RE − RE = RE above the surface. The answer is C.
Hint
Gravity provides mv²/r, with r measured from Earth's centre.
Worked solution
- r = 8.37 × 106 m.
- v = √(GM/r) = 6900 m/s.
Hint
v = √(GM/r) does not depend on the satellite's mass.
Worked solution
- v ∝ 1/√r.
- 4 times the radius gives 1/2 the speed. B's mass does not matter. The answer is C.
Hint
Kepler's second law: equal areas in equal times.
Worked solution
- Close to the Sun, the line from the Sun is short, so the comet must cover more of its path to sweep the same area.
- So it moves fastest at closest approach. The answer is B.
Hint
Kepler's third law: T² ∝ r³.
Worked solution
- T2/T1 = 93/2 = 27.
- T2 = 27.0 days.
Free-response practice
Original questions in the style of the AP exam's free-response section. Write a full answer on paper, with words as well as equations, then open the scoring guide and tick each point you earned.
Qualitative/Quantitative Translation (QQT) style · 8 points · about 15 minutes
A small cart of mass m rolls around the inside of a vertical circular loop of radius R. Friction and air resistance are negligible. At the top of the loop the cart moves at speed vtop, and at the bottom it moves faster, at speed vbot.
- (a) Draw free-body diagrams for the cart at the top and at the bottom of the loop. Show relative sizes within each diagram.
- (b) A student says the normal force at the top must be larger than at the bottom, “because at the top the cart has to be held up against gravity”. Without equations, explain why the student is wrong.
- (c) Derive expressions for the normal force at the top and at the bottom, in terms of m, R, vtop, vbot and g.
- (d) Explain how your expressions in (c) support your reasoning in (b). Then find the smallest vtop for which the cart stays on the track.
Write your full answer on paper first, then open the scoring guide and tick each point you earned.
Scoring guide and model answer
Part (a)
Top: normal force down and weight down (any relative lengths, consistent with the speed). Bottom: normal force up, longer than the weight down, because the net force must point up toward the centre.
Part (b)
At the top the track is above the cart and pushes it down, not up; gravity helps provide the inward force, so the track needs to push less. At the bottom the track must cancel gravity and also provide the whole inward force, and the cart is also moving faster there. So the normal force is larger at the bottom.
Part (c)
Top (down positive): FN,top + mg = mvtop²/R, so FN,top = mvtop²/R − mg. Bottom (up positive): FN,bot − mg = mvbot²/R, so FN,bot = mvbot²/R + mg.
Part (d)
FN,bot has mg added and uses the larger speed; FN,top has mg subtracted and the smaller speed, so FN,bot > FN,top. The cart stays on while FN,top ≥ 0: mvtop²/R ≥ mg, so vtop ≥ √(gR).
Your mark: 0 of 8
Mathematical Routines (MR) style · 7 points · about 12 minutes
A newly discovered planet has a moon in a circular orbit. Astronomers measure the moon's orbital radius r and period T. A second, much smaller moon orbits the same planet.
- (a) Derive an expression for the mass M of the planet in terms of r, T and physical constants.
- (b) The first moon has r = 4.2 × 108 m and T = 2.6 × 106 s. Calculate M.
- (c) The second moon has 1/100 of the first moon's mass and orbits at radius 2r. A student says its period will be shorter, because a lighter moon is easier to move. Explain what is wrong, and find its period as a multiple of T.
- (d) Explain why astronauts in a spacecraft orbiting this planet feel weightless, even though the planet's gravity acts on them.
Write your full answer on paper first, then open the scoring guide and tick each point you earned.
Scoring guide and model answer
Part (a)
GMm/r² = mv²/r with v = 2πr/T gives GM/r² = 4π²r/T², so M = 4π²r³/(GT²).
Part (b)
M = 4π²(4.2 × 108)³/[(6.67 × 10−11)(2.6 × 106)²] = 6.5 × 1024 kg.
Part (c)
The moon's mass cancels in GMm/r² = mv²/r, so the period does not depend on it. The period depends only on the radius: T² ∝ r³, so at 2r the period is 23/2T ≈ 2.8T, longer, not shorter.
Part (d)
The astronauts and the spacecraft are both in free fall: gravity is the only force on each, and it gives them the same acceleration toward the planet. So the floor does not need to push on the astronauts (FN = 0), and there is no normal force for them to feel.
Your mark: 0 of 7
Common mistakes
- Measuring the orbital radius from the planet's surface instead of its centre.
- Forgetting to square r in F = Gm1m2/r², so that doubling the distance only halves the force.
- Writing FN = mg cosθ on a banked curve. With a horizontal acceleration, FN = mg/cosθ.
- At the top of a loop, drawing the normal force upward. The track is above the cart and pushes down.
- Saying astronauts are weightless because there is no gravity in space. Gravity is about 90% as strong at the ISS; they are in free fall.
- Thinking a more massive satellite must orbit faster or slower. Orbital speed at a given radius does not depend on the satellite's mass.
Frequently asked questions
Why are highway curves banked?
So that part of the normal force points toward the centre of the curve. At the design speed, cars need no friction at all to turn, which makes the curve safe even in rain or ice. Racing tracks are banked steeply for the same reason.
Why do you feel heavy at the bottom of a roller coaster dip?
The seat must push up with enough force to cancel gravity and also provide the upward net force for the circular motion: FN = mg + mv²/r. You feel that larger normal force as extra weight.
Why doesn't the water fall out of a bucket swung in a vertical circle?
At the top, as long as the bucket moves at least √(gr), gravity alone is not enough to provide the inward force needed, so the bucket bottom must push down on the water. The water is accelerating downward at least as fast as it would in free fall, so it stays in the bucket.
Is G the same as g?
No. G = 6.67 × 10−11 N·m²/kg² is the universal gravitational constant, the same everywhere in the universe. g is the gravitational field strength at a particular place, about 9.8 N/kg at Earth's surface, and depends on the planet and the distance from it.
Do I need to know elliptical orbit calculations for AP Physics 1?
No calculations. You should know Kepler's laws qualitatively, especially that a planet moves faster when closer to the Sun, and be able to use T² ∝ r³ for circular orbits, which you can derive from Newton's laws.
Is gravitational field strength the same as acceleration due to gravity?
Yes, numerically. g in N/kg (force per kilogram) equals g in m/s² (the acceleration of a freely falling object) because 1 N/kg = 1 m/s².
Why does a satellite not fall to Earth?
It is falling all the time. It moves sideways so fast that, as it falls, Earth's surface curves away beneath it at the same rate. Gravity changes its direction, not its speed, so it keeps circling.
Keep going
Related guides and practice:
- Centripetal force explained
- AP Physics 1 forces summary
- AP Physics 1 equation sheet
- AP Physics 1 forces practice problems
All lessons in the AP Physics 1 forces and energy course · AP Physics 1 past papers