Circular velocity vector on the surface of spherical potential (gravity)

Question

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)

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.

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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 $v_{circ} = \sqrt{F(r)r}$

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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.

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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

https://en.wikipedia.org/wiki/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

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.

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*}