How to solve this nonlinear diff eq of celestial mechanics?

Solution 1:

Rewrite the integral as

$$\frac{1}{\sqrt{2}}\int \sqrt{\frac{r}{\mu+hr}}\:dr$$

and use the substitution $r = \frac{\mu}{h}\sinh^2\rho$

$$\sqrt{\frac{\mu^2}{2h^3}}\int2\sinh^2\rho\:d\rho = \sqrt{\frac{\mu^2}{2h^3}}\int\cosh2\rho-1\:d\rho = \sqrt{\frac{\mu^2}{2h^3}}\left(\sinh\rho\cosh\rho-\rho\right)$$

Then undo the substitution with

$$\sinh \rho = \sqrt{\frac{hr}{\mu}} \implies \cosh\rho = \sqrt{1+\frac{hr}{\mu}}$$

giving an equation

$$\sqrt{\frac{hr}{\mu}}\sqrt{1+\frac{hr}{\mu}}-\sinh^{-1}\left(\sqrt{\frac{hr}{\mu}}\right) = \frac{\sqrt{2h^3}}{\mu}(t+C)$$

This is of course only the case where an object keeps moving further and further away from a gravitational body because of your choice for $\dot{r}$ to be strictly positive.