Centripetal Force Calculator

centripetal force = mass × velocity² ÷ radius centripetal acceleration = velocity² ÷ radius g-force = acceleration ÷ 9.80665

m

Mass in kilograms — the field also takes grams, tonnes and pounds. It sets the force but not the acceleration: a heavier car needs more grip for the same corner, and goes round it in exactly the same way

v

Speed in metres per second, and this is the squared term — the field also takes km/h, mph and ft/s. Going 10 percent faster needs 21 percent more force, which is why speed kills on corners more efficiently than anything else

r

Radius of the turn in metres, from the centre of the circle to the object, not the width of the road. It is in the denominator: a wider curve needs less force, in inverse proportion

a

Centripetal acceleration, v²/r, in m/s². It points at the centre even when the speed is constant, because the direction of the velocity is changing every instant

g

The same acceleration expressed in standard gravities, 9.80665 m/s² to the g. On a road this number is also the friction coefficient the tyres need; in orbit it is the gravity that is doing the pulling

Use this page whenever something has to be made to follow a curve and you need to know what that costs: a car or a bicycle in a corner, a train on a bend, a ball swung on a string, a satellite or a space station in orbit, a rider on a roller coaster loop, a centrifuge sample, a charged particle bent by a magnetic field. The question is always the same — how much inward force, and does the available supplier have it. Two habits help. First, read the g figure rather than the newtons, because it is the one that carries across situations: 0.637 g on a road is a friction coefficient the tyres can meet, 4 g in a loop needs a seat belt and a strong neck, and 0.884 g in orbit is simply the gravity that was already there. Second, check the supplier before believing the answer. If the page says a corner needs a friction coefficient of 1.2 in the wet, the honest reading is not that the force is large but that the corner is not takeable at that speed, because no wet tyre will give you 1.2.

1. A 3,300 lb car at 56 mph on a 328 ft curve

   1. 56 mph = 25 m/s; 328 ft = 100 m (the fields take SI units and convert)
   2. Acceleration: 25² ÷ 100 = 625 ÷ 100 = 6.25 m/s²
   3. Force: 1500 × 6.25 = 9375 N, which is 2107.6 lbf
   4. In gravities: 6.25 ÷ 9.80665 = 0.637 g

   The g figure is the one to carry away. On a road it is also the coefficient of friction the tyres have to deliver, and 0.637 is well inside the 0.7 to 0.9 that a dry tyre on good asphalt can manage — so this corner is fine, with a modest margin. In the wet the same tyre gives about 0.4 to 0.5, the number goes above it, and the honest answer changes from how much force is needed to how fast this corner can be taken at all: about 20 m/s, or 45 mph. The force itself is a different quantity — 9375 N is roughly the weight of a small car, pulling sideways on the tyres for the two or three seconds of the corner.

2. The same car on a 164 ft radius, half the size

   1. Only the radius changes: 164 ft = 50 m, half of the previous example
   2. Acceleration: 625 ÷ 50 = 12.5 m/s²
   3. Force: 1500 × 12.5 = 18750 N, or 4215.2 lbf
   4. In gravities: 12.5 ÷ 9.80665 = 1.275 g

   Halving the radius doubles the demand, and 1.275 g is past what any road tyre can do — which is the point of the example. The force needed does not fail gradually as a corner tightens; it crosses the available friction at a specific radius, and past that radius the car simply does not turn. It is also why the same speed on a motorway slip road feels entirely different from the same speed on a roundabout: nothing about the car changed, and the radius went from 100 m to 50 m. The mass cancels out of the g figure, so a lighter car is not safer here — it needs proportionally less force from tyres that have proportionally less grip.

3. The International Space Station: 420 t at 17,100 mph

   1. 7660 m/s is 17,100 mph; the orbital radius is 6771 km, measured from the centre of the Earth (about 400 km above the surface)
   2. Acceleration: 7660² ÷ 6771000 = 58675600 ÷ 6771000 = 8.666 m/s²
   3. Force: 420000 × 8.666 = 3639603 N, or 818215.3 lbf
   4. In gravities: 8.666 ÷ 9.80665 = 0.884 g

   The station is 420 tonnes and the inward force on it is 3.6 meganewtons — and nothing is pulling it, in the sense that no rope or tyre is involved. Gravity is the whole supplier, and 0.884 g is exactly the strength of gravity at that altitude. That is the general pattern for orbits: the centripetal force needed always equals the gravity available, so the body goes round instead of falling in or flying off. It is also why the crew float: they are falling at 0.884 g and so is the station around them, so nothing presses them against a floor. Note also that the acceleration is not small — the station is accelerating towards the Earth harder than a car cornering at its limit.

The formula describes motion in a circle at constant speed, and the acceleration it gives points at the centre only because the speed is constant — a car that is braking or accelerating through the corner also has a tangential acceleration, and the total is the vector sum of the two, which is more than this page reports. It assumes the turn is a circle; a real road curve is a clothoid or a spiral, so the radius changes along it and the peak demand is at one point rather than throughout. It says nothing about where the force comes from, which is the assumption most likely to be wrong in practice: a tyre can only supply what its friction allows, a rope has a breaking load, and a banked curve supplies part of the force through the normal force rather than through friction, which this page does not model. It treats the object as a point mass, so it ignores the fact that a real car rolls, that its tyres share the load unevenly, and that its centre of mass is above the road — a tall vehicle can tip before it slides. It assumes the object stays at a fixed radius, so a satellite spiralling down, a car understeering wide or a rider leaning in are all outside it. Nothing here is relativistic: at orbital speeds the correction is in the eighth decimal place, but the field will accept speeds where it is not, and a charged particle in a magnetic field is a separate problem with its own formula.

What is the centripetal force formula?

F = mv²/r, and the acceleration alone is a = v²/r. A 1500 kg car at 25 m/s on a 100 m radius needs 1500 × 625 ÷ 100 = 9375 N, which is an acceleration of 6.25 m/s², or 0.637 g. Note where the radius sits: underneath. Tightening the corner from 100 m to 50 m doubles the force, which is why the same speed feels completely different on a slip road and a roundabout.

What is the difference between centripetal and centrifugal force?

Centripetal is the real, inward force in the frame of the ground, and it always has a supplier — the friction of the tyres, the tension in a string, gravity on a satellite. Centrifugal is the outward apparent force that appears if you do the analysis in the rotating frame, and nothing is actually pushing outward. The magnitudes are the same, so both pages give the same number; the difference is which question you are asking. If you are the car, you want to know whether the tyres can deliver it. If you are the machine holding the load, you want to know what your bearings must take.

What provides the centripetal force when a car goes round a corner?

Friction between the tyres and the road, and on a banked curve a share of it comes from the normal force instead. That is why the g figure on this page does double duty on a road: it is numerically equal to the coefficient of friction the tyres must supply, with no mass involved. A dry tyre on good asphalt manages about 0.7 to 0.9; in the wet, about 0.4 to 0.5; on ice, under 0.1. If the page says the corner needs more than the surface can give, the honest answer is not a bigger force but a lower speed.

Why does a wider curve need less force?

Because the radius is in the denominator. Going from a 100 m radius to a 200 m one halves the required force at the same speed, and going from 100 m to 50 m doubles it. That inverse relationship is the reason motorway junctions are built with long sweeping curves, and also why a roundabout taken at the same speed as an open bend feels violent. It is also the cheapest way to reduce the demand, since reducing the speed has to overcome the square while widening the radius works directly.

Does a heavier car need more grip in a corner?

It needs more force in proportion to its mass, and it also has more available, because the friction a tyre can supply is proportional to the load on it — so to a first approximation the cornering speed is the same. That is why the g figure on this page does not contain the mass at all: 0.637 g is the demand on the tyres whether the car weighs 1500 kg or 3000 kg. The mass matters for everything else — the force in newtons, the heat in the tyres, the load on the suspension — and the tidy cancellation breaks down when the tyre is near its limit, which is why heavy vehicles are slower through corners in practice.

Why does the space station need a centripetal force if nothing is pulling it?

Something is: gravity. At 400 km up the Earth's gravity is still about 0.884 g, and the station's 420 tonnes need 3.6 meganewtons of inward force to follow its orbit at 7660 m/s. Gravity supplies exactly that, which is what being in orbit means — the station is falling towards the Earth continuously, and moving sideways fast enough that the ground curves away at the same rate. Nothing is holding it up, and that is also why the crew float: they and the station are falling together.

• Centripetal Acceleration (College Physics 2e, §6.2) — a = v²/r, why the acceleration points inward at constant speed, and the worked car and orbit examples — OpenStax
• Centripetal force — the force that always has a supplier, the banked-curve and orbiting-satellite cases, and the friction coefficient a corner demands — Wikipedia
• NIST Guide to the SI — standard gravity as 9.80665 m/s², and the exact mile, foot and pound factors used in the examples above — National Institute of Standards and Technology