Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

In Classical Mechanics, both Goldstein and Taylor (authors of different books with the same title) talk about the centrifugal force term when solving the Euler-Lagrange equation for the two body problem, and I'm a little confused about what it exactly means - is it a real centrifugal force or a mathematical consequence of using polar coordinates for solving the Euler-Lagrange equation.

Their derivations of the Lagrangian $$L=\frac{1}{2}\mu(\dot{r}^{2}+r^{2}\dot{\theta}^{2})-U(r)$$ would lead to one motion of equation (theta) showing that angular momentum is constant and one radial equation of motion shown as $$\mu\ddot{r}=-\frac{dU}{dr}+\mu r\dot{\phi}^{2}=-\frac{dU}{dr}+F_{cf}.$$ They call $\mu r\dot{\phi}^{2}$ the fictitious force or the centrifugal force. I'm quite hazy on my memory of non-inertial frames, but I was under the assumption that fictitious forces only appear in non-inertial frames. The frame of reference in the two body problem was chosen such that the Center of Mass of the two bodies would be the origin so that would be an inertial frame, and I'm assuming that there are no non-inertial frames involved since neither author had talked about it in the previous chapters.

Would calling $\mu r\dot{\phi}^{2}$ an actual centrifugal force be incorrect then? Isn't it a term that describes the velocity perpendicular to the radius? From this two-body problem, it appears as though if I were to use polar coordinates when solving the Euler-Lagrange equations for any other problem, the centrifugal force term will always appear, so it would be a mathematical consequence of the choice of coordinate system rather than it being an actual fictitious force. Is that term being called a centrifugal force because it actually is a centrifugal force or is it because it has a mathematical form similar to it?

share|cite|improve this question
up vote 5 down vote accepted

There are two equivalent descriptions$^1$ of the reduced two-body problem with a central potential $V(r)$:

  1. In an inertial frame with no fictitious forces: Here $\frac{1}{2}\mu r^{2}\dot{\theta}^{2}$ is the angular part of the kinetic energy.

  2. In a rotating frame following the reduced particle with fictitious forces and only 1D radial kinematics: Here $-\frac{1}{2}\mu r^{2}\dot{\theta}^{2}$ is the centrifugal potential (notice the minus sign!).

Each description leads to the same Lagrangian


$^1$ The reduced particle is confined to an orbit plane, so the problem is effectively just two-dimensional described by two coordinates $r$ and $\theta$. In both descriptions, the center-of-mass serves as the origin of the reference frame. The motion of the center-of-mass itself is trivial, as there is no external forces.

share|cite|improve this answer
I see what happened. So there are two interpretations. The first is kinetic energy when deriving using polar coordinates. The second arises when changing to a 1-D problem. For this there are more than one way: in Goldstein one involves integration that is a hassle (Goldstein 3rd ed p.75) and the other is to consider it as a 1-D fictitious problem (p.76-77), which is easier. This fictitious problem changes the interpretation of $\frac{1}{2}\mu r^{2}\dot{\theta}^{2}$ from kinetic energy to potential energy contributed by potential energy. I guess this is mostly reiterating what you've said. – user1604449 Sep 22 '12 at 1:18
@Qmechanic♦ Why did you say that the center of mass is the origin of the reference frame? If you use the reduced mass $\mu$, should't one of the two bodies be the origin of the system? – Sørën Apr 9 at 20:05
$\uparrow$ @Sørën: No. (Note that the formulation of Kepler's first law on Wikipedia treats the ratio $m/M_{\odot}$ as zero, thereby ignoring the motion of the Sun, which is possibly confusing in above context.) – Qmechanic Apr 9 at 20:46
@Qmechanic♦ So actually the center of mass is coincident with $M_{\odot}$ ? – Sørën Apr 9 at 21:00
$\uparrow$ @Sørën: No, only to a very good approximation. – Qmechanic Apr 9 at 21:25

I think you skipped a step...

Once you show that angular momentum $l$ is a constant of the motion you can eliminate $\dot{\theta}$ from the two-dimensional equation of motion and arrive at an equation in 1 independent variable only ($r$): $$ \mu \ddot{r} - \frac{l^2}{\mu r^3} = -\frac{dU}{dr}$$ If you think about it, this new 1-dimensional space is in fact a non-inertial frame (if it were inertial the mass would fall to $r=0$), so it's no surprise to see a new force term in addition to the original potential gradient.

Goldstein then interprets the new term as "the familiar centrifugal force" by re-writing it: $$ \frac{l^2}{\mu r^3} = \mu r \dot{\theta}^2 = \frac{\mu v_\theta^2}{r} $$

And it's true: that new force is indeed one of the non-inertial forces (the centrifugal) that arise if one analyzes the problem in a rotating frame with angular velocity $\omega=\dot{\theta}$.

Aside: The other non-inertial forces in a rotating frame arise from time-varying quantities. It can be shown that the entire non-inertial potential is a velocity-dependent one analogous to that of electromagnetism: $$U_{rot} = \mu \left\{ \left[ - \frac{\left( \boldsymbol{\omega \times r} \right)^2}{2} \right] - \boldsymbol{ \dot{r} \cdot \left( \boldsymbol{\omega \times r} \right) } \right\}$$

The first term gives the centrifugal force, the second the Coriolis and Euler forces (if $\boldsymbol{\omega}$ is non-constant).

share|cite|improve this answer
What a coincidence! I finally replied and it was the same time as you did. I didn't think about the inertial frame and mass falling to $r = 0$. More food for thought. Thanks! – user1604449 Sep 22 '12 at 1:29
@user1604449: yes, it looks like you got to the same place... – Art Brown Sep 22 '12 at 1:33

"They call μrϕ˙2 the fictitious force or the centrifugal force. I'm quite hazy on my memory of non-inertial frames, but I was under the assumption that fictitious forces only appear in non-inertial frames."

all "fictitious" forces still "transfer energy."

Also anything that is an acceleration in math (distance/time^2) is indicative of energy transfer. Also, if it is on one side of the equation, it has to be in both. So if you are dealing with mass in motion around a point, the centrifugal force is real and expressed in the math as acceleration or force in some way.

Weight is the potential energy, velocity x mass is the kinetic energy.

share|cite|improve this answer
"When you look at the values for objects with a mass of 4250kg or more, you will see the centrifugal acceleration begins to exceed the acceleration of gravity." This is absolutely wrong. The mass of the object has nothing to do with the acceleration due to the rotating frame of reference. That acceleration is $v^2/r$. You are confusing force with acceleration. – Bill N Oct 27 '15 at 16:22
@BillN Hey, I see that you are following me and telling everyone I am wrong about everything without showing the math or physics errors. Would you mind explaining yourself with math and physics? I have shown that gravity acting on mass is 143 N and Centrifugal forces of larger masses exceeds gravity's pull. You simply disagree and down vote, thanks for your participation. – d w Oct 28 '15 at 15:09
The contribution to the gravitational field at the surface of the Earth due the Earth's mass is approximately $9.80\, N/kg$. Check it: $GM_E/R_E^2$. The centrifugal acceleration at the equator due to the rotating frame of Earth is $0.034\,m/s^2$. Check it: $\omega^2_ER_E$. The corresponding forces of each would have the mass of the object multiplied by these numbers. The centrifugal contribution will never overcome gravity under present conditions of the Earth and its rotation. I disagree because you are wrong. Go review your Newtonian physics. Thank you. – Bill N Oct 28 '15 at 17:18
@BillN The weight of something on earth is basically constant if the mass is constant. The weight of something under centrifugal force is dependent. With the known constants of our universe, how large does a mass have to be to receive enough centrifugal forces to exceed gravity? – d w Oct 28 '15 at 17:47

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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