# What provides the centrifugal force for planets orbiting a star?

this is a question I had when I was a kid. I'm a bit ashamed because I think I am missing out on something very obvious since I have the same question despite almost being an engineer now!

From Newtonian physics, I understand how although the gravitational force pulls the planet towards a star, and the planet 'falls' towards the star, due the angular momentum, it also moves laterally. In case of a planetary orbit, this is just enough to keep it moving around it in an elliptical orbit. (With speed and radius being such that angular momentum and energy is conserved)

But let's consider a single planet and star system where the planet moves around the star in a circular path for simplicity. If gravitational force provides centripetal force, what can account for the centrifugal force?

I have a feeling it is related to some inaccurate view I have, of centrifugal force. The xkcd comic and wikipedia article talk about two concepts: fictitious and reactive centrifugal force. But it would be safe to assume the star(EDIT: sorry, I meant to choose a frame of reference such that the centripetal force on the planet has to be cancelled by a centrifugal force on the planet in the direction: star to planet) as a stationary frame of reference for the purpose of this question, right? Which means that it is not a fictitious force, right? Or is it not considered a centrifugal force if you take the star as the reference frame?

• Perhaps my answer to another question here might help, since the satellite-planet system from my example there has the exact same physics as the planet-star example you're thinking about. (The question is related but certainly not the same.) – Wouter Jul 12 '13 at 13:57
• Thanks, that helped too! The planet is a non inertial reference frame. That was the critical thing I ignored! – mehfoos Jul 12 '13 at 15:45
• You know star gravity will not just act on the planet, but on any observer on the planet also, keeping them in orbit. – ja72 Jul 12 '13 at 18:18
• See my comment on the truly correct answer. – randomstring Jul 15 '13 at 22:14

Perhaps you are thinking of the centrifugal force as something that prevents the planet from falling into the star? There is only centrifugal force in the orbiting frame of the planet. In this frame, the planet is not accelerating, so you a need centrifugal force to balance the centripetal force. It is perfectly valid to consider the star as your reference frame, but then there is no centrifugal force at all, and the planet is accelerating towards the star.

• Yes, you are right about the fact that I am thinking about centrifugal force as something that prevents the planet from falling into the star. Sorry, I should have chosen the reference frame in such a way that a centrifugal force is required to balanced the gravitational centripetal force on the planet. In that case, what provides the centrifugal force? – mehfoos Jul 12 '13 at 14:48
• In that case, it is a "fictitious" force that is simply a result of changing to an accelerating coordinate frame. This is much easier to understand in the inertial frame of the star. Then, the planet is accelerating, but it does so such that it's speed does not change, and it maintains its distance. – ZachMcDargh Jul 12 '13 at 15:40
• Makes sense. I should have know that the planet is a non-inertial reference frame! Today has been a productive day haha – mehfoos Jul 12 '13 at 15:44

Newton's third law tells us that action and reaction are equal and opposite, so the gravitational force between e.g. the Sun and the Earth pulls equally on both bodies:

The Sun moves a lot less, but that's because it's a lot heavier than the Earth. In fact both the Sun and the Earth are accelerating towards their mutual centre of mass so there aren't inwards and outwards forces - both forces are inwards towards the centre of mass. Anyhow, we call the (apparently) outward force on the Sun the centrifugal force from the Latin for fleeing the centre. The (apparently) inwards force on the Earth we call the centripetal force from the Latin for seeking the centre.

• Sorry, I need 10 reps for posting an image than can illustrate what I mean. But basically, I had intended to upload an image similar to yours but with the centrifugal arrow starting from the centre of the planet. Would that be right? – mehfoos Jul 12 '13 at 14:17
• No, there is no force pulling the planet outwards. The planet is just attempting to move in a straight line, and the only force acting on it is the inward centripetal force that is bending it's trajectory into a curve. – John Rennie Jul 12 '13 at 14:49
• I see, but then in your diagram, the centrifugal force is directed away from the centre of revolution (whether we consider the planet or sun as the reference frame), isn't it? The Wikipedia article I cited says that Centrifugal force is the apparent force that draws a rotating body away from the centre of rotation. – mehfoos Jul 12 '13 at 14:58
• I think the OP is thinking entirely in the co-rotating frame. – user10851 Jul 12 '13 at 15:47
• I do not understand the abbreviation "OP" used by Chris White, but the illustration above shows two centripetal forces, towards the common center of mass (think of two stars of equal size), and no centrifugal force. @user1218748 is correct that the centrifugal force is balancing the centripetal one. It is produced by inertia, in the accelerated (here rotating) frame. If one considers a centrifugal force on the sun, it goes the opposite way. – babou Jul 16 '13 at 17:05

I would think that centrifugal force is provided by inertia. Inertia is usually what will resist acceleration, when you apply a force to a body that can move freely (but my knowledge of physics is pretty old, and for all I know, it may have changed :)

• This is the right answer. There is no such thing as centrifugal force. It is just inertia that makes you want to go in a straight line. The straight line takes you away from the object so it feels like an outward push. It takes something unusual to prevent you from going off on a tangent and move in a circle. Here this unusual thing is gravity. In other cases it could be electromagnetic force, or a rope attached to you. It is solely for convenience that we give a single name, centripetal force, to all these different forces. But the outward "force" results from just one single cause: inertia. – randomstring Jul 15 '13 at 22:14
• @randomstring Thanks for the comment. I am a bit surprised people call it a fictitious force. Jet pilots seem to consider it very real, and car racers too. I do too when I fall off my skidding bicycle. – babou Jul 15 '13 at 23:50
• @babou and randomsting: I would be more than willing to accept this answer, if you can clarify some problem I have with this answer: 1) Inertia is orthogonal to gravitational centripetal force. So how can it be the centrifugal force? 2)Centripetal & centrifugal forces should be equal in magnitude and opposite in direction. My first issue talks about the direction. What about the magnitude? How is it equal to magnitude of gravitational attraction? 3)What reference frame are you taking for the scenario where inertia is the centrifugal force? – mehfoos Jul 16 '13 at 10:56
• Inertia will take you on a tangent if no force is applied. It does not because an orthogonal force is applied, creating an orthogonal acceleration. The inertial force (many names for it) is the force that resist acceleration. You can even feel it when accelerating you car (fictitious as it may be), though it is parallel to motion, or when turning, when it is orthogonal. It is the exact opposite to the force causing the acceleration. I guess it is related to the local frame, since it is the only force a local experimenter in a closed box can observe (for a small satellite, not on a planet). – babou Jul 16 '13 at 13:42
• I agree with your description about fictitious force. I also get how inertia is the cause of fictitious force in the case of a car. But I don't get how it applies here. Sorry if I'm beginning to sound annoying, but can you address each of those 3 points separately? It would make it much easier for me to comprehend your explanation. I read on wiki that centrifugal force is cause by inertia; I was not convinced for the reasons mentioned above, which is why I asked this question. I then though that Zach's explanation that the force exists because the frame in itself is accelerated makes sense. – mehfoos Jul 16 '13 at 17:24

The previous answers are well and good. Except I find it a bit strange, nonetheless, saying that the moon does not fall to Earth because its inward centripetal acceleration is balanced by its outward centrifugal acceleration. More correct might be to say that the inward gravitational force is balanced by the outward inertial or centrifugal force. Or that the centripetal force caused by gravity is balanced by the centrifugal force. Thus gravity plays the same role as the tension in the string holding a stone as we swing it round our heads. It is interesting to think of the centrifugal force as real one. Astronauts in a centrifuge feel it pressing them outward. And the fact that these devices are called centrifuges implies that in certain cases it is is indeed more convenient to consider this fictitious force as in some sense real.