I am pretty sure this is a really basic question, but after the summer, not using integrals / derivatives at all, I just can not remember how to do math anymore.

I have a function for acceleration that I need to integrate with respect to time $t$ in order to obtain speed.

My acceleration function is (for angle $\phi$) $$\ddot \phi = -2 \frac{\dot r}{r} \dot \phi - \dot \phi \dot \theta \frac{cos(\theta)}{sin(\theta)}$$

where $$\ddot \phi = \frac{d^2}{dt^2}\phi $$ and $$\dot \theta = \frac{d}{dt}\theta $$ and $$\dot \phi = \frac{d}{dt}\phi$$ and so on.

so I would need to obtain the $\dot \phi$ by integrating the acceleration once.

$$\dot \phi = \int \ddot \phi dt= \int \Big(-2 \frac{\dot r}{r} \dot \phi - \dot \phi \dot \theta \frac{cos(\theta)}{sin(\theta)} \Big) dt$$

I just do not remember at all how to start to process this problem. I know it's almost a bit shameful but indeed could use a helping hand here.

(I have 2 similar equations to be solved for $\ddot \theta$ and $\ddot r$ but I am pretty sure I get these done when I once remember how to calculate things)

Thank you so much if you could help me out.

Those 2 other equations are following : (if someone is interested)

$$\ddot{r} = r \dot \theta ^2 + r\dot\phi^2 sin^2(\theta) -\frac{GM}{r^2}$$ $$\ddot \theta = \dot \phi^2 sin(\theta)cos(\theta) - 2 \frac{\dot r}{r} \theta$$

  • $\begingroup$ I'm assuming $r$, $\theta$, and $\phi$ functions of $t$, right? And when you integrate acceleration, you get velocity as a result, not necessarily speed because speed is just how fast something's going regardless of its direction. So if I'm correct, you need to find the velocity function. $\endgroup$ Aug 25, 2022 at 17:54
  • $\begingroup$ Yes you are correct. I have solved other parts of my problem (so far), and got to this problem where I needed some help. I will post the whole problem/task tomorrow and tag you so you can see it. Shortly, I want to solve initial theta and phi velocoties so that the particle stays on the 'surface of sphere' while orbiting center/mass under potential (i.e. Newtonian). This also means that the 'r' (radial) acceleration and velocities are zero as the distance r stays constant (radius of sphere). $\endgroup$
    – mamark
    Aug 25, 2022 at 19:29

1 Answer 1


You can divide the equation by $\dot\phi$ to obtain $$\frac{\ddot\phi}{\dot\phi}=-2\frac{\dot r}{r}-\dot\theta\frac{\cos\theta}{\sin\theta}.$$ Integrating over time yields $$\ln \dot\phi=-2\ln r-\ln(\sin\theta)+C_1,$$ hence $$\dot\phi=\frac{C_0}{r^2\sin\theta}.$$

  • $\begingroup$ Just a side note: In the solution above we obviously "lose" the trivial solution $\dot\phi=0$ when taking the logarithm. Also, the constant $C_0=e^{C_1}$ is always positive. Symmetrically, $\dot\phi$ can also be negative; then, the second equation would be modified to $\ln(-\dot\phi)=...$ leading to $\dot\phi=\frac{-C_0}{r^2\sin\theta}$. Putting all this together, we see that $\phi\in C^2((a,b))$ is a solution if and only if there is $C\in\mathbb{R}$ such that $\dot\phi=\frac{C}{r^2\sin\theta}$ in $(a,b)$. $\endgroup$
    – junjios
    Aug 25, 2022 at 10:59
  • $\begingroup$ Thank you so much. Im on train rn, but after reading your solution it makes completely sense and things are coming back to my mind. I will dig deep to this question in coming days and ask if there was something unclear (dont think so tho). Thank you! $\endgroup$
    – mamark
    Aug 25, 2022 at 17:47

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