Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Just for reference, I derived t and θ from these two functions:

$\frac{dr}{dt} = \pm\frac{1}{\sqrt{r}}\sqrt{2gR^2 - (2gR-U_r^2)r}$

$\frac{dθ}{dt} = \frac{R^2U_θ}{r^2}$

t & θ:

$t = \frac{1}{\sqrt{(2gR-U_r^2)^3}}[2gR^2[arctan(\sqrt{\frac{(2gR-U_r^2)r}{2gR^2 - (2gR-U_r^2)r}})-arctan(\sqrt{\frac{2gR^2}{RU_r^2} - 1})] + \sqrt{2gR-U_r^2}(RU_r - \sqrt{r}\sqrt{2gR^2 - (2gR-U_r^2)r})]$

$θ = \frac{2U_θ\sqrt{2gR-U_r^2}}{g}[\frac{gR + U_r\sqrt{2gR-U_r^2}}{U_r\sqrt{2gR-U_r^2} - (2gR - U_r^2)} - \frac{gR^2 + \sqrt{2gR-U_r^2}\sqrt{r}\sqrt{2gR^2 - (2gR-U_r^2)r}}{\sqrt{2gR-U_r^2}\sqrt{r}\sqrt{2gR^2 - (2gR-U_r^2)r} - (2gR - U_r^2)r}]$

As you can see, they are both quite complicated and messy functions and as far as I know, there is no simple way to find the inverse functions for either of them. θ seems to be the only function that could possibly have a solution for r, but of course I'd preferably like to avoid going through the tedious calculations if I can.

What do you guys think? Is there any possible solution that you can see in this muddle? At the moment, I really am quite stuck.

share|cite|improve this question

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.