0
$\begingroup$

I am trying to solve a problem of a ballistic rocket which is fired from the surface of the Earth. I have not worked with spherical coordinates too much, but the problem necessitates them.

I understand how to specify the initial position of the rocket in spherical coordinates in terms of $r$, $\theta$ and $\phi$ using some picture like this:

enter image description here

However, I struggle to understand how to specify the velocity in terms of spherical coordinates. Let's say my rocket has an initial velocity of 100 m/s, 45 degrees east of north at an angle of 30 degrees relative to the surface. How does that direction "45 degrees east of north at an angle of 30 degrees relative to the surface" translate into some $\vec{v} = [v_r,v_{\theta},v_{\phi}]$?

I am really struggling how to visualize. Any help is appreciated.

$\endgroup$

1 Answer 1

1
$\begingroup$

The best way to understand these different coordinates systems is to picture them exactly as our usual x,y,z system. The only difference is that now our unit vectors $\boldsymbol{\hat{r}},\boldsymbol{\hat{\theta}},\boldsymbol{\hat{\phi}}$ are not constant, they change with time (they change with position, and position changes with time). So lets say that our position is represented as:

$\boldsymbol{r}=r\boldsymbol{\hat{r}}$ (So far so good, we are only saying that the particle is $r$ units in the direction that $r$ increases.

Then the velocity is given by the time derivative of the position vector (same as our normal x,y,z system)

$\boldsymbol{v} = \dot{r} \boldsymbol{\hat{r}} + r \frac{d}{dt}\boldsymbol{\hat{r}}$ (Here is where things change, in our usual x,y,z plane $\frac{d}{dt} \boldsymbol{\hat{i}}$ equals zero)

$\boldsymbol{\hat{r}}=\sin \theta \cos \phi \boldsymbol{\hat{i}} + \sin \theta \sin \phi \boldsymbol{\hat{j}} + \cos \theta \boldsymbol{\hat{k}} \implies \frac{d}{dt} \boldsymbol{\hat{r}} = \dot{\theta} \boldsymbol{\hat{\theta}} + \dot{\phi} \sin \theta \boldsymbol{\hat{\phi}}$

Remember that $\theta$ and $\phi$ they all depend on time.

$\boldsymbol{\hat{v}} = \dot{r} \boldsymbol{\hat{r}} + r\dot{\theta} \boldsymbol{\hat{\theta}}+r \dot{\phi} \sin \theta \boldsymbol{\hat{\phi}} $

Another thing that might help you understand better these coordinate systems is that $\boldsymbol{\hat{a}}$ is a unit vector in the direction where $a$ increases

$\endgroup$
1
  • $\begingroup$ Thanks for the reply. I see your answer and how you derived $\vec{v}$ in terms of components of $\hat{r}$, $\hat{\theta}$ and $\hat{\phi}$. But I guess what I am wondering is how do I solve for $\dot{r}$, $\dot{\theta}$, and $\dot{\phi}$? All I am given is an " initial velocity of 100 m/s, 45 degrees east of north at an angle of 30 degrees relative to the surface". How does that initial velocity relate to the velocity components in spherical coordinates? $\endgroup$
    – Darcy
    Oct 25, 2017 at 20:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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