You'd think it would have an opposing force due to Newton's Third Law, but it is usually portrayed alone, without any other force there cancelling it out. I know there exists a concept of Cetrigufal Force (https://en.wikipedia.org/wiki/Centrifugal_force) which seems to do the work, but most agree that it is also pretty much not a real force (at least when you look at all this from non-inertial frame of reference); besides, there is also a Reactive Centrifugal Force (https://en.wikipedia.org/wiki/Reactive_centrifugal_force) that now actually compensates that, but some real-life examples don't utilise it: I don't think a car going in circular motion really has it, yet its velocity certainly changes direction.

Now, my confusion is really related to all the cases: why does Centripetal Force seem to be balanced here but not there, is it actually secretly balanced in all cases or is there some magic mumbo-jumbo that makes it an exception to the rule? Perhaps I was just missing context all along and it's actually quite simple and basic.


9 Answers 9


You'd think it would have an opposing force due to Newton's Third Law,

There is an opposing force, which is the Newton's Third Law force partner.

but it is usually portrayed alone,

Most forces acting on a given body are "usually portrayed alone." The Newton's Third Law equal-and-opposite force acts on the other body.

For example, if you are spinning a ball around on a string, the force on the ball is directed down the string towards you. The force on you is up the string towards the ball. Those forces are equal in magnitude and opposite in direction. They don't both act on the ball. One acts on the ball. One acts on you. If you only draw the force on the ball, you only draw one force.

Same thing, for example, with a moon orbiting a planet. There's a force on the moon and an equal-magnitude opposite-direction force on the planet.

The Newton's Third Law force pairs act on different bodies. If they acted on the same body they would just cancel each other and cause no acceleration for that one body.

  • $\begingroup$ But what about, let's say, a car in circular motion? It isn't attached anything, and there doesn't seem to be any Reactive Centrifugal Forces acting on it, creating a force pair. I'm primarily pondering the cases where it doesn't exist, where it is only an ordinary Centrifugal Force, the pseudoforce version of it. There, it seems both it and the Centripetal force act on the same body, yet this does harm to Newton's Third Law. Perhaps it is like this due to us viewing the whole ordeal from Non-inertial Frame of Refercence. I'd like to know what you can tell me about this. $\endgroup$
    – Ainis
    Commented Aug 20, 2022 at 6:54
  • $\begingroup$ The car is attached to the road via its tires. The car feels a force and the road feels and equal and opposite force. This is why we need to have people out there in yellow vests with orange cones and bulldozers rebuilding our roads every few years--they get beaten apart by the force of tires (and weather, etc). $\endgroup$
    – hft
    Commented Aug 20, 2022 at 16:55

I am afraid that the first concept you should clarify is related to Newton's Third Law, even without reference to centripetal/centrifugal forces.

The words used in the question ("unbalanced centripetal force," "an opposing force due to Newton's Third Law" ) seem to indicate that you think the action-rection pair should balance or equilibrate.

That is not the case. Action and reaction in Newton's Third Law are always forces on different bodies. As such, they do not compensate in the sense that in the equation of motion of each body, there is only one force acting, not the sum of the two. Unless bodies are constrained in some specific way, both accelerate and no one of them "opposes" the other.

Only after understanding such a starting point can we move to examine the centripetal force in the case of a circular motion. The circular motion is a particular case, one can have a centripetal force even for a more general motion. But it is better to start with the simplest case.

The Uniform circular motion of a mass requires one constant radial force pointing towards the center of the circle. It should not be balanced by other forces on the same body. However, since such a force would be due to the interaction with another body, there will be an equal and opposite force on such a second body.

  • $\begingroup$ Could you give an example of the last paragraph in work? Let's say I'm riding a car in circular fashion, does the road produce an opposing force? What does it to with that force, for what other reasons does it exists apart from satisfying Newton's Third Law? $\endgroup$
    – Ainis
    Commented Aug 20, 2022 at 7:04
  • 1
    $\begingroup$ @Ainis Forces do not exist to satisfy Newton's Third Law. It is the other way around. The properties of the interaction between most of the systems are such that Newton's Third Law is satisfied. In order to have a body moving along a circle, a force is required. Otherwise, the body would go straight (in the absence of friction or dissipation). That's the behavior of a body on ice. In order to bend the trajectory, a force pointing in the direction of bending is required. A body has to provide such a force, but interactions are symmetric: an equal and opposite force will act on the body. $\endgroup$ Commented Aug 20, 2022 at 14:43

Tie a rock to a string and swing it in a circle around your head.

The rock goes in a circle because of the centripetal force you apply. You pull on the string.

The equal and opposite balancing force is the force the rock exerts on you. You feel the rock pulling on the string.


You'd think it would have an opposing force due to Newton's Third Law, but it is usually portrayed alone, without any other force there cancelling it out.

There is an equal and opposite force to the centripetal force per Newton's 3rd law, but they don't "cancel" each other out. In my opinion it's not a good idea to think in terms of action-reaction pairs of forces as "balancing" or "cancelling" out as it reinforces the erroneous notion that things can't accelerate because the action-reaction forces are equal and opposite to each other. You need to apply Newton's 2nd law on each object of a system individually accounting for all external forces acting on each object.

Consider a ball of mass $m$ at the end of a string undergoing uniform circular motion in a horizontal plane. The other end of the string is tied to a fixed post at the center of rotation. This is the example shown in the second Wikipedia article. Now let's consider the forces and action-reaction pairs of forces involved.

Take the ball first. Neglecting gravity (which doesn't create the circular motion), there is only one external horizontal force acting on the ball, and that's the tension in the string that acts towards the center of the circular motion. That is the centripetal force given by


Which causes a centripetal acceleration of


Now let's consider the string. The ball exerts an equal and opposite force of $F_c$ on the string per Newton's 3rd law. That is the "reactive centrifugal force" referred to in the Wikipedia article. The force the ball exerts on the string is an external force on the string. But it is not the only external force on the string. The other is the force that the post exerts on the string which is equal an opposite to the force the string exerts on the post, i.e. $F_c$, again an action reaction pair per Newton's 3rd law. So there are two action reaction force pairs on each end of the string but no net force on the string.

Finally consider the post. There are two external forces acting on the post. One is the force exerted by the string (tension force) on the post. It is equal and opposite to the force the post exerts on the string per Newton's 3rd law. The other external force acting on the post is the ground. It is equal and opposite to the force the string exerts on the post, for a net external force of zero on the post. The post does not move.

Bottom line: There is only one net external force acting on the ball, string, and post components. It is the centripetal force acting on the ball. It is responsible for the centripetal acceleration of the ball.

Hope this helps.


Let's say I'm riding a car in circular fashion, does the road produce an opposing force?

Let's make it a box truck, driving clockwise around a circular track, and you are in the windowless box holding a heavy ball stationary in your hands. There are two interesting points of view from which the situation can be described:

From your own point of view, you feel a mysterious force that tries to pull the ball out of your hands toward the left side of the truck, but you apply an equal and opposite force with your hands to keep the ball from moving. Then, you let go, and you see the ball accelerate toward the left side. You conclude that the acceleration is caused by the mystery force, which now is acting unopposed on the ball.

The other point of view belongs to an observer looking down on the scene from a tower in the center of the track. This observer happens to be Superman, so he is able to see through the walls of the box with his X-ray vision.

What Superman sees before you let go of the ball is that you and the truck and the ball all are accelerating toward the center of the circle. Uniform circular motion is accelerated motion, and the acceleration always is directed toward the center of the circle. The force that accelerates the truck is the static friction between its wheels and the track, the force that accelerates you is the static friction between the floor and your body, and the force that accelerates the ball is from your hands.

Both you and superman are aware of the force that your hands apply to the ball, but you think that your hands balance a mystery force acting on the ball, and that "balance" allows the ball to remain stationary (unaccelerated.) Superman is aware of the same force between your hands and the ball, but he perceives no mystery force. He thinks that the force between your hands and the ball is unbalanced, and he sees the ball accelerating for as long as you hold it.

When you let go of the ball, you see it accelerate toward the side of the truck. You think that the mystery force accelerates it. But, Superman sees the ball stop accelerating when you let go. He sees that there now are no forces acting on the ball, and he sees it roll in a perfectly straight line at constant speed until it hits the wall of the truck.

Superman sees unaccelerated, linear motion where you see accelerated circular motion. He sees no force where you see the mystery force.

The name of the mystery force is "centrifugal," and it only exists when we are trying to describe the laws of motion in a rotating coordinate system (e.g., in a coordinate system that is rigidly attached to the truck.) In Superman's description of the same scene, in a non-rotating, unaccelerated coordinate system (a.k.a., an inertial coordinate system) he uses the word "centripetal" to describe the forces and accelerations that are directed toward the center of the circular motion. The centripetal forces accelerate the truck, yourself, and the ball in uniform circular motion.


Centripetal force, being a real force, is always balanced by Newton's 3rd Law. When we talk about centripetal force, we are in a reference frame where some object is rotating. This is what the "reactive centrifugal force" is supposed to balance. In the case of a car going in circles, the reactive force is the friction force of the car's tires on the ground.

On the other hand, centrifugal force, being a fictitious force arising from a non-inertial reference frame that is rotating along with the object, is never balanced by Newton's 3rd Law.

  • $\begingroup$ To finish the final sentence: because the object/reference frame is accelerating. $\endgroup$
    – user121330
    Commented Aug 20, 2022 at 5:24

In an inertial frame, when a body ( B ) with mass moves in a curved path, some ( centripetal ) force, from some other body ( A ), is causing the curve.

The centripetal force may be gravity, a string, train tracks, Electric field, or any medium that can pass force.

But body B also applies an equal and opposite force to body A, backwards through the same medium, which may be labeled "reactive centrifugal force".

Sometimes physics problems are presented as "a particle travelling in a ( electric, gravity ) field", in which case we are simply ignoring the effect of B on A, in order to focus on the effect on B.

  • $\begingroup$ What about the cases where it is the ordinary Centrifugal Force, the pseudoforce version? Let's say I'm riding a car in circular motion, does the road on which I'm riding create an opposing force to my Centripetal one? I know the Centripetal Force in this case is the force of static friction, but what force then can be considered its opposition? $\endgroup$
    – Ainis
    Commented Aug 20, 2022 at 6:56
  • $\begingroup$ @Ainis the road provides centripetal force to the car tires, accelerating the car sideways. The car tires provide reactive force to the road. The earth receives the impulse of momentum, but due to the earth's great mass, the change in its velocity is unmeasurably small. The car seats provide centripetal force to your body, and your body provides reactive force to the seats. You can see the cushions bulge due to this force pair. If your seat was not bolted firmly to the car, both you and the seat would slide sideways ( actually, the car would move sideways underneath you ). $\endgroup$ Commented Aug 20, 2022 at 15:17

Suppose a car starting a race in an circular circuit. It is accelerating from the rest, that is: its speed is increasing. But the direction of its velocity is also changing due to the shape of the circuit.

The increase of the speed is the tangential acceleration, and it is possible due to the friction force of the rolling tires with the road. The tires make a backward force on the ground. The reaction force from the ground on the car causes the tangential acceleration of the car.

On the other hand, the car is making a circle, and the tires are forcing the surface of the road radially outward. Again, its is only possible due to the friction with the road. As it is radially outwards, it is called centrifugal force (acting on the road). The reaction of the ground on the tires is a centripetal force (acting on the car).

It is interesting to note that the net force from the car on the road, and the reaction force from the road on the car are not radial, so they are not centripetal or centrifugal. It is only when we focus on the radial and tangential components of the vector that centrifugal or centripetal forces and accelerations arises.

That forces are not balanced on the car (because in that case it would not be accelerating). The (apparent) balance of forces happens however for an observer inside the car. The driver feels a forward force from the seat back, and he intuitively postulates that there is a backward force on him. The 'balance' of that forces explains being at rest inside the car.

The same with the side force from the seat belt on the driver, a centripetal force. He intuitively postulates that there is a centrifugal force on him that makes the balance, keeping him at rest.


"Centripetal force" does not exist, so there's no reason for it to obey Newton's 3rd law (or any other law, for that matter).

Ok, I will admit that this was not a good answer. I'm still confident it's correct, but I was too terse, and I didn't actually try to address OP's concern. (I am also pretty sure the downvoters agree with me, but I didn't state my point clearly.)

When people talk about the centripetal force on an object in circular motion$^1$, usually what they mean is the component of the net force on the object in the direction toward the center of the circle. Let's call this component $F_{\text{net},c}$. An object moving in a circle has to have a component $a_c$ of its acceleration directed toward the center of the circle, and we can work out from the kinematics of circular motion that $a_c = v^2/r$. Newton's 2nd law then forces us to conclude that $F_{\text{net},c} = ma_c$.

When we apply Newton's 2nd law, we add up all the forces acting on an object to get the net force, \begin{equation} \vec{\mathbf{F}}_{\text{net}} = \vec{\mathbf{F}}_1 + \vec{\mathbf{F}}_2 + \ldots \tag1 \end{equation} The forces $\vec{\mathbf{F}}_1$, $\vec{\mathbf{F}}_2$, etc. may be all sorts of forces exerted by other, different objects on the object in question, like gravity, electric forces, magnetic forces, tension, normal/contact forces, friction, and whatever else.

When I say that centripetal force does not exist, what I mean is that we will never include it in the sum of forces on the right hand side of Eq. 1. The centripetal force is (a component of) the net force get when we add up all the actual forces acting on our particle, but it is not itself a force.

In some cases, there is only a single force $\vec{\mathbf{F}}_1$ on the right hand side of Eq. 1. For example, if you have a satellite orbiting a planet in circular motion, the only force on the satellite is the gravitational force from the planet, and as a shorthand you might want to refer to this gravitational force as the centripetal force.$^2$

In such cases, what we are calling "the centripetal force" does obey Newton's 3rd law$^3$, in the sense that, whatever body B is exerting the force $\vec{\mathbf{F}}_1$ on our object A, A will exert a force $-\vec{\mathbf{F}}_1$ back on B. Just like every other example of Newton's 3rd law, the two forces $\vec{\mathbf{F}}_{\text{B on A}} = \vec{\mathbf{F}}_1$ and $\vec{\mathbf{F}}_{\text{A on B}} = -\vec{\mathbf{F}}_1$ don't "cancel" because they act on different objects. Going back to our concrete example, A is our orbiting satellite, B is the planet it's orbiting, and the two bodies are exerting gravitational forces on each other with equal magnitude and opposite direction.

However, I find it highly problematic even in this simple case to say that the centripetal force obeys Newton's 3rd law. We now have two different meanings of centripetal force that we are liable to mix up. The first is $F_{\text{net},c}$, and the second is $\vec{\mathbf{F}}_1$, and it is not necessarily true that Newton's 3rd law applies to the first meaning. For example, even if the gravitational force from the planet is the only force on the satellite, maybe there is some other force acting on the planet that is keeping it in equilibrium. Then the net force on the planet is zero, and it is surely not the case that $(F_{\text{net},c})_{\text{planet}} = -(F_{\text{net},c})_{\text{satellite}}$.

In general, once we have to include more than one force on the right hand side of Eq. 1, talking about "the centripetal force" makes less sense. Consider the case of a marble rounding a vertical, circular loop. Both a gravitational force $\vec{\mathbf{F}}_g$ exerted by the Earth and a normal force $\vec{\mathbf{N}}$ exerted by the circular track act on the marble. Each of these forces individually obey Newton's 3rd law: the marble exerts a gravitational force back on the Earth and it also exerts a normal force back on the track.

Now, which of these forces is "the centripetal force"? Of course, it's neither of them, because what we really mean by centripetal force is the radially-inward component of the net force \begin{align} F_{\text{net},c} = N + F_{g,c} \end{align} and this is not any single force. It's a sum of components of two different forces exerted by different objects, and it is not the kind of entity that Newton's 3rd law has anything to say about.

I could go on, but hopefully I've cleared myself up and offered a better answer to the question. I would just like to offer a final remark. When teaching physics, we work hard to impress on our students that the words we use, like work, power, acceleration, and so on, have specific technical meanings. It seems ridiculous to me to then just ignore that and refer to a "centripetal force" which isn't actually a force at all. I am sure you can sympathize with me if you've ever had to field a question like "How come we don't have to draw the centripetal force on our free-body diagram?"

[1] As @GiorgioP points out in a comment, we can define centripetal force for more general motion by talking about the direction toward the center of the instantaneous osculating circle of our particle's path, but let's keep this simple and focus on purely circular motion.

[2] Of course, this is an idealization, and in real life there will be (at least) gravitational forces from every other mass in the universe that we need to account for.

[3] We have to be careful here because there are cases — for example, involving magnetic forces — in which Newton's 3rd law is violated, but the more general principle of momentum conservation is obeyed.

  • $\begingroup$ Don't you mean centrifugal? $\endgroup$ Commented Aug 19, 2022 at 21:27
  • $\begingroup$ @Not_Einstein No. Centrifugal force does exist: it's an inertial force detected by observers in a rotating reference frame. It's no less real than gravity is. But there is no such thing as "centripetal force." $\endgroup$
    – d_b
    Commented Aug 19, 2022 at 21:37
  • 2
    $\begingroup$ Centripetal force is any force or component of a force directed towards the instantaneous curvature center of the trajectory. Maybe you do not like them, but I can reassure you: they do exist. The force Sun is exercising on you just now is mostly a centripetal force. $\endgroup$ Commented Aug 19, 2022 at 22:05
  • $\begingroup$ @GiorgioP I don't think that definition entirely captures what you want it to. For example, it doesn't account for cases when the "centripetal force" is the sum of components of multiple different forces. That aside, I don't disagree that you can define "centripetal force" to be something, but that something will surely not be a force, so it's misleading to call it one. It's the same reason we now refer to the emf instead of the "electromotive force" — it's not a force, so it's confusing to everybody involved to call it one. $\endgroup$
    – d_b
    Commented Aug 19, 2022 at 23:04
  • $\begingroup$ Centripetal forces are real. What are you thinking? $\endgroup$
    – Bob D
    Commented Aug 19, 2022 at 23:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.