A person in a rotating coordinate system, to do Newtonian mechanics has to use pseudo forces such as Coriolis and centrifugal.

We use such forces for rotating coordinate systems,ones that rotate about a fixed axis.

A person on the surface of earth is not in a rotating but a revolving coordinate system, that revolves around the axis of earth ( neglecting rotation around sun) shown as $E$,that goes around the centre of earth.

How do we describe motion in such a revolving frame? What pseudo forces should I add to analyze motion using newton's laws? When I look at my textbook this difference isn't touched upon and perhaps the motion is same as would've been for a purely rotating one.

I'd be glad if someone pointed out , I don't have a teacher to go to.

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    $\begingroup$ How are you making a distinction between the two cases? You can also consider Coriolis and centrifugal pseudo-forces on earth's surface too. $\endgroup$
    – joseph h
    Commented Feb 20, 2021 at 5:52
  • $\begingroup$ The surface of the Earth is a rotating coordinate system. You just don't happen to be standing at the axis of rotation. $\endgroup$
    – Sandejo
    Commented Feb 20, 2021 at 6:17
  • $\begingroup$ @Joseph h I went through a proof that was about a spinning coordinate system so I was not sure if it'll be the same for a system that revolves. $\endgroup$
    – Kashmiri
    Commented Feb 20, 2021 at 6:27
  • $\begingroup$ @Sandejo, yes I agree but the proof I saw was when the origin was at the axis of rotation, in this case the axis are revolving so im not sure $\endgroup$
    – Kashmiri
    Commented Feb 20, 2021 at 6:34
  • $\begingroup$ Imagine 2 kids on opposite end of a merry-go-round, one of them throws a ball towards the other... their parents observed that the ball is flying in a parabolic but nevertheless straight path! Of course they will never catch the ball unless they cheat by compensating the angle before each throws ;D $\endgroup$
    – user6760
    Commented Feb 20, 2021 at 6:56

4 Answers 4


If the co-ordinates of an object $O$ with mass $m$ are $\vec r(t)$ relative to a rotating co-ordinate system with its origin at the centre of the earth and rotating with the earth, then we know that

$\displaystyle m \frac {d^2 \vec r}{dt^2} = \vec F(\vec r, \dot {\vec r}, t)$

and on the right hand side of these equations of motion we know we need to include fictitious forces such as centrifugal force and Coriolis force, as well as any "real" external forces.

If we now use a revolving co-ordinate system with its origin at a fixed point on the earth's surface, but still rotating with the earth, then the co-ordinates of $O$ in this revolving co-ordinate system are $\vec R((t)$ where

$\vec R(t) = \vec r(t) + \vec c$

Note that there is no time dependence in $\vec c$; we know that $\vec c$ is a constant vector because the origin of the revolving co-ordinate system is fixed on the surface of the earth and its orientaion rotates so that it is always the same as that of the rotating co-ordinate system.

Since $\vec c$ is constant, we have

$\displaystyle m \frac {d^2 \vec R}{dt^2} = m \frac {d^2 \vec r}{dt^2} = \vec F(\vec r, \dot{\vec r},t) = \vec F( \vec R - \vec c, \dot {\vec R}, t)$

so we can see that the equations of motion of $O$ in the revolving co-ordinate system are the same as in the rotating co-ordinate system, as long as we adjust the right hand side to account for the translation of the origin by $\vec c$. One implication of this, for example, is that the magnitude of centrifugal force at the origin of the revolving co-ordinate system is not zero, but is $m|\vec c| \omega^2$.

  • $\begingroup$ Wow. Simple and beautiful. $\endgroup$
    – Kashmiri
    Commented Feb 20, 2021 at 11:01
  • $\begingroup$ As a request, after studying kleppner should I go for Thornton marion cz Goldstein is hard for me. $\endgroup$
    – Kashmiri
    Commented Feb 20, 2021 at 11:03

This should not be too difficult to work out. I wanted to do this a few months ago but never got time to actually write down the equation. The revolving coordinate system does rotate around the inertial coordinate at the same rate as the rotating coordinate system but in addition to that its origin is moving on a circle. My intuition tells me you will still get the Coriolis-like and Centrifugal-like forces plus whatever the result of the rotation of the origin is.

You can think about limiting cases. E.g. if a particle is stationary out there in space along the rotation axis it will look to you as if it is rotating around you, so you will probably need a centrifugal-like force that points towards that axis parallel to the rotation axis but centered on you. On the other hand, if a particle is stationary in the radial direction, it will look like it is making circles with a center at some point away from you. So not as simple as a universal centrifugal force always pointing towards the origin of the rotational frame of reference.

If I manage to work this out on paper I will share.


I looked at a simple example of when the outside object is not moving in the inertial frame (No Coriolis-like forces). I also simplified my life by sticking to 2D. The inertial forces depend on how exactly your revolving system behaves. Is it (a) always facing in the same direction (from the point of view of the inertial frame) or is it (b) always facing away from the rotation axis?

For (a) you will get a Centrifugal force but where it points will depend on where the external object is (as opposed to the rotating system where it always points towards you). The center of attraction will be distance R away from it, away from you when you are at the closest distance to it (R is your distance from you to the revolution axis). For (b) the inertial force actually coincides with that of the rotating system.

For case (b), which is a more natural description of a person on the surface of a rotating planet, it is possible that the Coriolis force is also identical to that of the rotating system. I can not say for certain. And I will probably be too lazy to explicitly compute this tomorrow morning :)

  • $\begingroup$ Waiting for your work dear Dzuku. $\endgroup$
    – Kashmiri
    Commented Feb 20, 2021 at 6:35
  • $\begingroup$ I agree with you. They are not going to be the same. May take me a few days to get the actual equations though. $\endgroup$
    – Bondo
    Commented Feb 20, 2021 at 6:36

The method I'm familiar with is a little strange. It uses the tools of general relativity. I did this awhile ago, but I don't have that notebook anymore. I tried to find it computed somewhere online. (I found it. but ignore the t->t' transformation)

$F^i =\frac{d^{2}{x^i}}{dt^2}+ {\Gamma^i}_{rs} \frac{dx^r}{dt} \frac{dx^r}{dt}$

It's all in this little formula. You just have to mechanically unpack it. By the Einstein Summation Convention, you sum over repeated indices (r and s) and $i=1,2,$ or $3$ where $1$ represents $x$ , $2$ represents y, and $3$ represents z.

For example,

$F^y =\sum_r \sum_s \frac{d^{2}{y}}{dt^2}+ {\Gamma^y}_{rs} \frac{dx^r}{dt} \frac{dx^r}{dt}$


$\Gamma$ is defined in terms of the metric $g_{ij}$.



$g_{ij}$ transforms like a second order tensor. It's easy but tedious to transform from the ordinary flat space metric to the rotating metric. Put rotating metric into the formula for $\Gamma$. The whole problem reduces to finding every $\Gamma$.

The coordinate transformation is

$x'=cos(\omega t) x + sin(\omega t) y$

$y'=-sin(\omega t) x + cos(\omega t) y$


And I recall using the identity $\cos^2 (\theta) + \sin^2 (\theta)=1$.

  • $\begingroup$ Thank you for your effort but I don't see why use GR for a Newtonian geometrical problem $\endgroup$
    – Kashmiri
    Commented Feb 20, 2021 at 8:07
  • $\begingroup$ I agree but the question is from an undergrad mechanics $\endgroup$
    – Kashmiri
    Commented Feb 20, 2021 at 8:18

your position on the sphere surface from the sphere center is:

$$\mathbf R=\left[ \begin {array}{c} r\cos \left( \theta \right) \sin \left( \phi \right) \\ r\sin \left( \theta \right) \sin \left( \phi \right) \\ r\cos \left( \phi \right) \end {array} \right] $$

where $~r~$ is the sphere radius $~\phi~$ the polar coordinate and $~\theta~$ the azimuth coordinate.

the sphere coordinate system if rotating about the z-axes with constant angular velocity $~\omega$, thus the position vector in the rotating system is:

$$\mathbf R_r=\left[ \begin {array}{ccc} \cos \left( \omega\,t \right) &-\sin \left( \omega\,t \right) &0\\ \sin \left( \omega\,t \right) &\cos \left( \omega\,t \right) &0\\ 0&0&1 \end {array} \right] \,\mathbf R$$

from here you can obtain the velocity:

$$\mathbf v=\frac{\partial \mathbf R_r}{\partial \phi}\,\dot{\phi}+ \frac{\partial \mathbf R_r}{\partial \theta}\,\dot{\theta}+\frac{\partial \mathbf R_r}{\partial t}$$

with the kinetic energy $~T=\frac m2 \mathbf v\cdot \mathbf v~$ the potential energy $~U=m\,g\,\mathbf R_r\cdot \mathbf e_z~$ you obtain the equations of motions

$$\left[ \begin {array}{c} \ddot\theta \\ \ddot\phi \end {array} \right] =\left[ \begin {array}{c} 2\,{\frac { \left( \dot\theta +\omega \right) \cos \left( \phi \right) \dot\phi }{\sin \left( \phi \right) }} \\\\ -{\frac {\sin \left( \phi \right) \left( 2\,r\,\dot \theta \,\omega\,\cos \left( \phi \right) +r\,{\dot\theta }^{2}\cos \left( \phi \right) +r{\omega}^{2}\cos \left( \phi \right) +g \right) }{r}} \end {array} \right] $$

The pseudo forces is:

$$\mathbf F_s= m\,\left[ \begin {array}{c} {\omega}^{2}x+2\,\omega\,{\it \dot y} \\ {\omega}^{2}y-2\,\omega\,{\dot x} \\ 0\end {array} \right] $$


$$x=\mathbf R\cdot \mathbf e_x~,\dot x=\frac{\partial x}{\partial \phi}\,\dot \phi+ \frac{\partial x}{\partial \theta}\,\dot \theta$$

$$y=\mathbf R\cdot \mathbf e_y~,\dot y=\frac{\partial y}{\partial \phi}\,\dot \phi+ \frac{\partial y}{\partial \theta}\,\dot \theta$$

you can obtain now the equations of motion with Newton-Euler method

$$m\,\ddot{\mathbf R}=-m\,g\,\mathbf e_z+\mathbf F_s$$


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