So, I would like to integrate paths of particles on circular velocity on the surface of a sphere (due to some potential, i.e. gravity).

The problem is to fix the two angular velocities $\dot \theta $ and $\dot \phi $ (explained more later)

The situation at the start/beginning :

1) I have a set of x y z positions located on the surface of the sphere (see image) Particles on the surface of sphere

2) From those locations, I can solve $\theta$ and $\phi$ angles (using some trigonometry, have not yet tried).

3) Then I would need to solve velocity components ($\dot \theta , \dot \phi$) so that the particle stays on circular velocity on the surface of the sphere (generated by potential, i.e. gravitational potential).

4) I am aware that there would be an infinite amount of different orientations for such orbit. Thus one could select a random value for one of the angular velocities, and based on that the other angular velocity would be fixed/known.

I think my method is kinda close, I just do not figure out some set of equations that I could use to fix the angles.

Some background equations:

Potential (in this example Newtonian for simplicity) $\phi(r) = -\frac{GMm}{r}$

Force $F(r)=-\frac{\partial \Phi(r)}{\partial(r)} = \frac{GMm}{r^2}$

and in vector form this would be $F(r)\frac{\bar r}{r}$

where $\bar r$ is position vector and $r = ||\bar r ||$ is the length of the vector. $G=1$ (grav. constant) , $M$ is the mass of potential, $m$ is the mass of the test particle (can be ignored).

Circular velocity is case $F_c = F_g$ (centrifugal force equals grav. force) and thus yields $ma = m \frac{v^2}{r} = \frac{GMm}{r^2}$ and solving for $v = v_{circle} = \sqrt{ \frac{GM}{r} }$

This can also be written in terms of potential function as $v_{circ} = \sqrt{\Phi(r)}$ or with force $v_{circ} = \sqrt{F(r)r}$

First, a 2D example of circular velocities, so here I'm creating a disk around the central point. For this, I am using polar coordinates (which later would be changed to Cartesian coordinates).

method 1

Cartesian coordinates

$$x = r \cos(\theta)$$

$$y = r \sin(\theta)$$

$$\dot x = \dot r \cos(\theta) - r \sin (\theta) \dot \theta$$

$$\dot y = \dot r \sin(\theta) - r \cos (\theta) \dot \theta$$

and note $\dot r = 0$ because we are staying in a circular orbit.

One can now link the $\dot \theta$ to circular velocity as follows :

$$\dot \theta = \frac{2\pi}{T} ; v = \frac{2\pi r}{T} $$

$$ \dot \theta = \frac{v}{r} $$

where now v = $v_{circ}$

Thus this now gives the velocity components as follows

$$v_x = -\sin (\theta) r \frac{v}{r} = -\sin (\theta) v_{circle}$$

$$v_y = -\sin (\theta) r \frac{v}{r} = \cos (\theta) v_{circle}$$

Method 2

Now this uses polar coordinates

First, define unit vectors as follows

$$\hat e_r = \cos (\theta) \hat i + \sin (\theta) \hat j $$

$$\hat e_{\theta} = -\sin (\theta) \hat i + \cos (\theta) \hat j $$

and so one can write the position vector $$\bar r = r \hat e_r = r [\cos(\theta)\hat i + \sin (\theta) \hat j ]$$

Lets now take the derivative with respect to time (so we get the velocity)

$$\bar v = \dot {\bar r} = \dot r \hat e_r + r \dot \theta \hat e_{\theta}$$

and again we have that $\dot r = 0$ and $\dot \theta = v/r$

giving solution $$\bar v = v \hat e_{\theta} = v_{circ} \Big(-\sin(\theta) \hat i + \cos(\theta)\hat j \Big) $$

which is identical to the method 1 .

And indeed, when integrated over time, the orbits are circular as they should be. Here in the image is some quick test for disk orbits. Example of circular orbits on disk

Thus, I decided to try Spherical coordinates for solving the velocity components for given point $\theta$ and $\phi$ . This is how far I got:

Spherical coordinate system

Wikipedia link to spherical coordinate system


First, define unit vectors

$$\hat e_r = \sin (\theta) \cos (\phi) \hat i +\sin (\theta) \sin (\phi) \hat j + \cos(\theta) \hat k $$

$$\hat e_{\theta} = \cos (\theta) \cos (\phi) \hat i +\cos (\theta) \sin (\phi) \hat j - \sin(\theta) \hat k $$

$$\hat e_{\phi} = -\sin (\phi) \hat i + \cos (\theta) \hat j $$

and position vector $$\bar r = r \hat e_r = r \Big( \sin (\theta) \cos (\phi) \hat i +\sin (\theta) \sin (\phi) \hat j + \cos(\theta) \hat k \Big)$$

from where one can get the velocity vector

$$\dot {\bar r} = \bar v = \dot r \hat e_r + r \dot \theta \hat e_{\theta} + r \dot \phi \sin(\theta) \hat e_{\phi}$$

So now I have a problem that I have 2 unknown variables $\dot \theta$ and $\dot \phi$ so I need 2 independent equations to solve the set of these variables. Any idea what those two equations could be ?

I have tried to think one equation from the angular velocity perspective, such that $\sqrt{\dot \phi ^2 + \dot \theta ^2} = v/r$ (similarly as in the case of 2D , polar coordinate).

Another equation could have been from Angular momentum $\bar L = \bar r \times m\bar v $ (cross product)

and one need to remember that since we are in a circular orbit, $\bar r$ and $\bar v$ are orthogonal to each other, thus $\bar L = \bar r m \bar v $ and since we are interested on test particles, the mass $m$ can be neglected.

Or similarly maybe kinetic energy $E_{kin} = 1/2 m v ^2 = 0.5 \Big( (r\dot \theta)^2 + (r\dot \phi \sin(\theta))^2\Big)$ . Note, $v^2$ comes from above where velocity for spherical coordinate system has been solved.

However, I do not get anything sensible out for $\dot \theta$ or $\dot \phi$ .

I chose two of the above equations into Mathematica to be solved as pair of equations. The result was horrible, so I think I have made somewhere a stupid mistake. Here is an example of what the Mathematica gave Mathematica solution for set of equations

So in any case, I for some reason do not notice a mistake I have made in solving the velocity components for a specific point.

  • $\begingroup$ I think from the equations of motion and the first integral you can solve your problem $\endgroup$
    – Eli
    Commented Jul 26, 2022 at 19:22
  • $\begingroup$ I do not notice how I could get the angular velocities $\dot \theta$ and $\dot \phi$ from doing so @Eli . I am not sure if I managed to explain everything as well as I could have (I got super tired from trying to think this problem). But yeah, I would need those two angles in order to get the direction of the particles' path so that it would stay on the circular orbit (with circular velocity all the time). I hope the question makes even somewhat sense, if not, I can try to explain it in more details. $\endgroup$
    – mamark
    Commented Jul 27, 2022 at 11:19

1 Answer 1


\begin{align*} &\text{position vector} \\ &\mathbf{R}=\left[ \begin {array}{c} r\cos \left( \phi \right) \sin \left( \theta \right) \\ r\sin \left( \theta \right) \sin \left( \phi \right) \\ r\cos \left( \theta \right) \end {array} \right]\\\\ &\text{from here the velocity vector}\\\\ &\mathbf{v}=\left[ \begin {array}{c} \cos \left( \phi \right) \sin \left( \theta \right) {\dot{r}}+r\cos \left( \phi \right) \cos \left( \theta \right) \dot{\theta} -r\sin \left( \theta \right) \sin \left( \phi \right) \dot{\phi} \\ \sin \left( \theta \right) \sin \left( \phi \right) {\dot{r}}+r\cos \left( \theta \right) \sin \left( \phi \right) \dot{\theta} +r\cos \left( \phi \right) \sin \left( \theta \right) \dot{\phi} \\ \cos \left( \theta \right) {\dot{r} }-r\sin \left( \theta \right) \dot{\theta} \end {array} \right] \\ &\Rightarrow\\ &v=\sqrt {{{\dot{r}}}^{2}+{r}^{2}{\dot{\theta} }^{2}+{r}^{2}{\dot{\phi} }^{2} \left( \sin \left( \theta \right) \right) ^{2}} \end{align*}

\begin{align*} &\text{with}\\ &T=\frac{m}{2}\mathbf{v}\cdot\mathbf{v}\\ &U=-\frac{G\,m\,M}{r}\\ &\text{the EOM's}\\ &\ddot{r}=r{\dot{\phi} }^{2} \left( \sin \left( \theta \right) \right) ^{2}+ r{\dot \theta }^{2}-{\frac {GM}{{r}^{2}}} \\ &\ddot{\theta}=\ldots \\ &\ddot{\phi}=\ldots \\ &\text{and the first integral}\\ &r^2\dot{\phi}\,\sin^2(\theta)=\text{constant} \end{align*} substitute from the first integral $~\dot{\phi}~$ to the equation $~\ddot{r}=\ldots~$ and with $~\ddot{r}=0~$ (r=constant) you obtain $~\dot{\theta}$

\begin{align*} &\Rightarrow\\ &v=\sqrt{\frac{G\,M}{r}} \end{align*}

  • $\begingroup$ I know its been a while since your answer, sorry about that. I tried to follow your method, but I just do not see where you got to your EOMs, as I got lost at some point. Thus I left this problem till another day, and tried again but yet still did not get the same result as you did. $\endgroup$
    – mamark
    Commented Aug 23, 2022 at 14:47
  • $\begingroup$ @mamark with Euler -Lagrange L=T-U you obtain the EOM's $\endgroup$
    – Eli
    Commented Aug 23, 2022 at 14:56
  • $\begingroup$ okay I try that aswell.. $\endgroup$
    – mamark
    Commented Aug 23, 2022 at 14:59

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