Module 6 · circular motion and gravitation
Nothing in this module travels in a straight line. Yet every single thing in it obeys the same law as a ball rolling off a table. Along the way: a stone on a string, a car on a bend, a third law that finally makes sense, and the moment Newton realised the Moon is falling.
Tie a stone to a string and whirl it round your head at a steady speed. Its speed never changes. Ask most people whether it is accelerating and they say no. They are wrong. Why they are wrong is the most important idea in this module, and it is the next paragraph.
Acceleration is the rate of change of velocity, and velocity is an arrow: it has a size and a direction. Change either one and you have accelerated. The stone's arrow is constantly swinging round, so the stone is constantly accelerating, even though a speedometer glued to it would never move.
You are whirling the stone in a flat circle above your head. The string snaps at the instant the stone is at the far side of the circle, straight out in front of you. Looking down from above, which way does the stone fly off?
Ignore gravity for a moment — imagine looking straight down on it.
So the string was never holding the stone out. It was holding the stone in. The circle exists only for as long as something pulls inward, and the moment that pull stops, the stone's own straight-line motion is all that is left. Everything remaining in this module is a hunt for the thing doing the pulling.
This is the one derivation in the module worth doing slowly, because it is pure geometry — no forces, no masses, nothing but arrows.
Two things this formula does not say. It does not say what is doing the pushing — that is the next section, and it is a different question entirely. And it does not apply only to circles: at every point of any curved path there is a v²/r component of acceleration pointing at the centre of the circle that best hugs the curve there, where r is that circle's radius. If the speed is also changing, there is a second component along the path, but that one never bends anything. A gentle bend has a huge r and needs almost nothing; a hairpin has a small r and needs a lot.
Half-fill a bottle with water, tie it to a strong string, and swing it in a vertical circle — fast, in the garden, not in the kitchen. The water stays in even at the top, when the bottle is upside down. Nothing is holding it up. It is falling, exactly as fast as the bottle is falling, and the bottom of the bottle keeps arriving underneath it. Slow below a critical speed and you get wet. That speed comes from the top of the swing: gravity alone must supply the mv²/r there, so the slowest speed the water still holds is v = √(gr) — Bench 3 of Module 5 works it through with an energy ledger.
Module 3 covered this with a horse and a cart, and if that section left you unsure, it was moving too fast. Here it is again, at walking pace, because everything after this section depends on being able to answer one question without hesitating: what force is pushing this thing toward the centre?
A force is an interaction between two objects. It is never a property of one. So there is no such thing as "the force of the ball" — only "the force of the bat on the ball" or "the force of the ball on the bat". Get into the habit of never writing down a force without naming both objects and which way round it goes. Half of all mechanics confusion in the world dies right here.
Newton's third law is then almost obvious: if A pushes B, then B pushes A, equally hard, in exactly the opposite direction. You cannot touch without being touched. A pair of third-law forces is always:
When you want to know how one object moves, you draw a box around that object and list only the forces acting on it. Forces on other things do not enter, ever. A third-law pair therefore never appears in the same list, so a third-law pair can never cancel each other out. They cannot cancel because they are not even in the same conversation.
Now the bit that was too quick last time. The horse is harnessed to a cart. Draw the box around the horse alone and list what touches it horizontally.
Here is where P comes from, step by step. The horse's muscles drive its hooves backward against the road — the hoof is trying to scrape the road backwards, the way your foot tries to scrape the floor backwards when you start walking. Friction is the force that opposes that scraping, so friction acts on the hoof in the forward direction. That forward friction is P: the road pushing the horse forward.
And its third-law partner? The hoof pushing the road backward, equally hard. The Earth does accelerate backward in response — by an amount so absurdly small (its mass is 6 × 10²⁴ kg) that no instrument will ever see it, but the force is there and is exactly as large as the one that moves the horse.
So the horse accelerates because P is bigger than T. The cart accelerates because the rope pulls it forward with T and its own friction drags it back with f, and T is bigger than f. Both statements are true at once. The rope pulls each of them with the same T, and that contradicts neither one. The tension is a hindrance to the horse and a help to the cart, which is exactly what a rope between two objects should be.
The test that proves it. Put the horse on wet ice. Its muscles are as strong as ever. The rope tension is unchanged. But now the hooves slip: the road cannot push forward, P drops to nearly zero, and nothing moves. Nothing about the third law changed — what changed was the one force that was actually driving the system, and it was never the horse's pull on the cart. The horse does not move by pulling the cart. The horse moves because the ground pushes it.
| The force | acts on | Its third-law partner | acts on |
|---|---|---|---|
| rope pulls horse backward | horse | horse pulls rope forward | the rope |
| rope pulls cart forward | cart | cart pulls rope backward | the rope |
| road pushes horse forward | horse | hooves push road backward | the Earth |
| Earth pulls the book down (its weight) | book | book pulls the Earth up | the Earth |
| table pushes book up | book | book pushes table down | table |
| you push the water backward | water | water pushes you forward | you, swimming |
| rocket pushes exhaust backward | exhaust gas | exhaust pushes rocket forward | rocket |
The pull on the horse and the pull on the cart are equal because the rope is light enough to have no net force on it — that is the second law applied to the rope, not the third law.
Read the fourth row and the fifth row together, because that pair is the classic trap. The book on the table has two forces on it: the Earth pulling down, the table pushing up. They are equal — but not because of the third law. They are equal only because the book is not accelerating. Put the whole table in a lift and accelerate it upward: the table's push would instantly become larger than the book's weight. Every genuine third-law pair in the room would stay exactly equal throughout. The true partner of the book's weight is the book's gravitational pull on the Earth, which acts on the planet, 6,400 km away at its centre, and never appears in the book's list at all.
Take any two objects that interact and nothing else. A pushes B with force F, so B's momentum changes at rate +F. B pushes A with force −F, so A's momentum changes at rate −F. Add them: the total momentum of the pair changes at a rate of exactly zero. That is conservation of momentum, and it is not so much a consequence of the third law as the same statement in different clothes. This is why physicists trust momentum conservation even more than they trust Newton. Twentieth-century physics overturned two of Newton's ideas: that F = ma always holds, and that a particle always has one definite path. Momentum conservation survived both, untouched.
Here is the mistake almost every student makes on a free-body diagram: they draw the weight, the normal force, the tension — and then add an extra arrow labelled "centripetal force" pointing at the centre. That arrow is always wrong, because it is double-counting.
"Centripetal force" is not a new kind of force alongside gravity, tension and friction. It is a job description. Something in your existing list has to do the job of pointing the object at the centre with size mv²/r. Your task is always to identify which one. Two warnings about that word "which". First, the job can be shared: on the banked bend below, the normal force and friction each supply part of it, and it is only their combined inward pull that has to come to mv²/r. Second, the job is only about the inward direction. If the object is also speeding up or slowing down, the forces on it have a forward or backward part as well, and only the part aimed at the centre is doing the centripetal job.
| Situation | Who actually does the centripetal job |
|---|---|
| Stone on a string | the string's tension |
| Car going round a flat bend | sideways friction between tyres and road — nothing else, which is why rain is dangerous |
| The Moon orbiting Earth | gravity |
| An electron in an atom (roughly) | the electric pull between the negative electron and the positive nucleus |
| A motorcyclist riding round the inside of a vertical cylinder — the fairground stunt called the wall of death | the horizontal push of the wall (normal force) |
| A plane in a banked turn | the horizontal component of the wings' lift |
| Clothes in a spin dryer | the drum pushes the clothes inward; the water, with nothing to push it inward, goes straight and leaves through the holes |
That last one is worth pausing on, because it inverts the usual story. Nothing flings the water out. The water is simply not being pushed inward any more, so it does what Newton's first law says and travels in a straight line — which, from inside a spinning drum, looks like flying outward. A spin dryer does not throw water out. It removes the reason the water was going round.
You are in a car taking a fast left bend, and you feel thrown against the right-hand door. What is actually happening?
Everything a passenger feels on a bend is the seat and the door doing the centripetal job on their body. "Centrifugal force" is the name for what that feels like from inside the turning car. The sensation is real; the force is not. Later, in class 11, you will meet a legitimate trick in which you pretend the turning car is standing still and add an outward "centrifugal force" to keep the sums balanced — it works, and physics uses it — but it is bookkeeping, not a push from any object. Nothing outside the car is pushing you outward. There is no third-law partner for it, which is the giveaway: every real force has one.
Module 1's Bench 4 built this bend end-on and found tan β = v²/rg; here it is again with the friction budget shown explicitly, so you can watch grip run out. Module 1 also flagged the limit of the model — as µ tan β approaches 1 the predicted maximum speed runs away to infinity, and long before that the car tips over instead of sliding. This bench never tips.
Start flat (bank 0°) and raise the speed until it slides. Then add bank and watch the speed it can take climb. Then drop µ, the coefficient of friction, to 0.1 — a sheet of ice — and find the one speed at which the car still holds the bend perfectly, needing no friction at all. That speed is the design speed of the bank, where β is the bank angle:
A velodrome — the steeply banked oval track used for indoor cycle racing — is banked at about 42° on the tightest part of its bends, where the radius is about 22 m: at that geometry the frictionless design speed is roughly 50 km/h, and a rider cruising at that pace is barely using grip sideways at all. An aircraft has no road to grip, and so no sideways friction at all — the only thing touching it is air — so it must bank until the horizontal component of lift does the whole job, which is why the world tilts in the window every time you turn. And a fighter pilot pulling 9g is being pushed along his spine by his seat with nine times his own weight; mg of that holds him up, and the remaining √(9²−1) = 8.9 mg does the centripetal job. Either way the blood drains from his head, and that is why g-suits exist.
The apple story is probably true, though Newton only told it when he was old and famous, and by then he had every reason to make it sound better than it was. What is certainly true is the calculation. It is one of the greatest single pieces of reasoning in the history of science, and it is nothing harder than arithmetic. So let us do it ourselves.
Newton's leap was this. Everybody knew apples fall. Everybody knew the Moon goes round. Nobody had suggested these were the same phenomenon — that the Moon is simply an apple that has been thrown sideways so fast that it keeps missing the Earth. The picture is exact in one way and loose in another. It is exact in that the Moon really is in free fall, with nothing but gravity acting on it, precisely like the apple. It is loose in two ways: a thrown apple is slowed by the air and the Moon is not, because there is no air out there; and the Moon is heavy enough to swing the Earth slightly toward itself in return, which an apple never does. If that same gravity that gives the apple 9.8 m/s² is giving the Moon its v²/r, then gravity must get weaker with distance, or the Moon would be falling as hard as the apple. Newton's guess for how it weakens was the inverse square: at twice the distance, a quarter of the strength. There was a reason to guess it. Whatever spreads out from the Earth is spread over the surface of a sphere, and a sphere's area grows as the square of its radius, so the strength should be diluted as 1/r². Kepler's measurements of the planets pointed the same way. Now test it on the Moon.
The Moon is 60 times further from the Earth's centre than the apple is. If gravity weakens as the square of the distance, the Moon's acceleration should be:
Now check it against the Moon's actual motion, using only two numbers a seventeenth-century astronomer already had: the Moon's distance (3.84 × 10⁸ m) and the time it takes to go round (27.3 days).
Sit with that for a second. Two numbers measured by astronomers who had never heard of gravity, combined with the acceleration of a dropped stone in an English garden, agree to three significant figures with a guess about how a force falls off with distance. Newton said he "compared the force requisite to keep the Moon in her Orb with the force of gravity at the surface of the earth, and found them answer pretty nearly." In today's English: he compared the force needed to hold the Moon in its orbit with ordinary gravity at the ground, and found the two matched closely. That is one of the great understatements. It is the moment the heavens and the Earth became one subject.
And notice what falls out of it. The force on a falling object is GMm/r², where M is the planet's mass, and Newton's second law says a = F/m. The m cancels. Every object falls with the same acceleration, whatever its mass — the thing Galileo had discovered by experiment a century earlier, now appearing as a two-line consequence. That cancellation hides something strange. The m on top measures how strongly gravity grabs the object. The m underneath measures how stubbornly it resists being accelerated. Those are two completely different properties, and there is no obvious reason at all that they should be the same number. Einstein spent ten years being bothered by that, and general relativity was the result.
Newton drew a mountain so tall it poked above the atmosphere, and a cannon on top firing horizontally. Fire slowly and the ball lands nearby. Fire faster and it lands further round the curve of the Earth. Fire fast enough and the ground curves away exactly as fast as the ball falls — and the ball never lands. It is in orbit. Nothing has changed about the physics; it is still just falling.
Three speeds matter, and the bench will let you find all three.
For a circular orbit, gravity does the centripetal job: GMm/r² = mv²/r. Cancel m and one r: v² = GM/r. Now the period is T = 2πr/v, so T² = 4π²r²/v² = 4π²r³/GM. That is T² ∝ r³ — Kepler's third law, which took him a decade of staring at Tycho Brahe's tables to extract from data, derived here in four lines from a force law.
Worth doing yourself: put T = 24 hours into that equation with GM = 3.986 × 10¹⁴ and solve for r. You get 4.22 × 10⁷ m, which is 35,870 km above the surface. (The commonly quoted figure for geostationary orbit is 35,786 km, because the satellite must keep pace with the stars, not the Sun: the Earth turns once in 23 h 56 min, not 24 h. Put T = 86,164 s in and you get it.) That is where every television satellite in the sky sits, because at that one radius the orbit takes exactly one day and the satellite hangs motionless over one spot on the equator. The number was calculated long before anyone could get there.
The commonest misconception in all of school physics: astronauts float because there is no gravity in space. At the height of the space station, gravity is about 89% as strong as it is in your bedroom. They float because they are falling — the station, the astronaut and the pen floating beside her are all on the same orbital path, all accelerating toward the Earth at the same 8.7 m/s², and so they do not press on each other at all.
You can have exactly the same experience for half a second by jumping off a wall. During the fall nothing supports you, and that unsupported sensation is precisely what an astronaut feels for six months. There are two different things people call weight, and mixing them up is what makes this confusing. Your weight in the sense your textbook uses is mg, the Earth's gravitational pull on you, and that does not switch off in orbit. What a bathroom scale measures is something else: the force between you and the floor, which physicists call your apparent weight. Take the floor away and the apparent weight reads zero — in orbit, or in a falling lift — while mg carries on unchanged.
1. A car goes round a flat, unbanked bend at a steady 30 km/h without slipping. The driver doubles the speed to 60 km/h on the same bend. The sideways friction force the tyres must supply is now:
2. A satellite in a circular orbit is moved to an orbit twice as far from the Earth's centre. Its orbital speed is:
3. A lorry collides head-on with a motorbike. Which statement about the forces during the collision is true?