Ordinary Differential Equations - Teschl, Problem 1.1

Question

Consider the case of a stone dropped from the height $h$. Denote by $r$ the distance of the ston from the surface. The initial condition reads $r(0)=h, \dot{r}(0)=0$. The equations of motion reads $$ \ddot{r} = \frac{-\gamma M }{(r+R)^2} (exact \ \ model)$$ respectively $$\ddot{r} = -g \ \ (approximate \ \ model),$$ where $g=\gamma M/R^2$ and $R, M$ are the radius, and mass of the earth, respectively.

Transform both equations into a first-order system.

I have transformed the approximate equation by integrating boths sides, and I got that $\dot{r} = -gt +C$ but I do not know how to convert the exact model, I think the biggest problem I am facing is that I do not think we can just simply integrate both sides since the $r$ in the denominator is a function and not just a variable.

Answer 1

Multiplying by $\dot{r}$ on both sides, we get:

$$\dot{r}\ddot{r} = -\frac{\gamma M\dot{r}}{(r+R)^2}$$

Note that

$$\dot{r}\ddot{r} = \frac{d}{dt}\frac{(\dot{r})^2}{2}$$

And

$$-\frac{\gamma M\dot{r}}{(r+R)^2} = \frac{d}{dt} (\frac{\gamma M}{r+R})$$

Thus, the equality can be rewritten as:

$$\frac{d}{dt}\frac{(\dot{r})^2}{2} = \frac{d}{dt} (\frac{\gamma M}{r+R})$$

Integrating both sides with respect to $t$ and substituting the appropriate limits at $t = 0$, we get:

$$\frac{\dot{r}^2}{2} - 0 = \gamma M\cdot (\frac{1}{r+R} - \frac{1}{h+R})$$

Or,

$$\dot{r}^2 = 2\gamma M\cdot (\frac{1}{r+R} - \frac{1}{h+R})$$

Which is a first order differential equation, as required.

You can also verify this by using the conservation of energy. You will receive the same result.

Answer 2

Another approach but highly precise for $r \ll R$:

$$\ddot{r} = \frac{-\gamma M}{(r+R)^2} \approx -\frac{\gamma M}{R^2}+\frac{2 \gamma M r}{R^3}=-g+2g\frac{r}{R}$$

With initial conditions $r(0)=h, \dot{r}(0)=0$ we can find the solution:

$$r(t)=\frac{1}{2} ((2 h-R)\cdot \cosh (\sqrt{\frac{2g}{R}}\cdot t)+R)$$