Solving equation of motion for two body problem

[Robin04]

Homework Statement

$\ddot{r} = c \frac{1}{r^2}$, where $c$ is a constant, and $r$ is the position of one object with respect to the other. I need to find the function $\dot{r}(r)$

We are in one dimension.

Homework Equations

The Attempt at a Solution

I don't really have any idea how to start.
The acceleration is a function of the position, but the position is also a function of the acceleration. To get $\dot{r}$ I should integrate with respect to time but $r$ is dependent on time which makes it problematic as I won't need that in my solution. There has to be some magic trick here.

 

[George Jones]

There has to be some magic trick here.

Let $v=dr/dt$. Then,
$$\frac{d^2 r}{dt^2} = \frac{dv}{dt} = \frac{dv}{dr} \frac{dr}{dt} = \frac{dv}{dr} v.$$

 

[Robin04]

Let $v=dr/dt$. Then,
$$\frac{d^2 r}{dt^2} = \frac{dv}{dt} = \frac{dv}{dr} \frac{dr}{dt} = \frac{dv}{dr} v.$$

So $c\frac{1}{r^2}= \frac{dv}{dr}v$, then $\frac{c}{vr^2}=\frac{dv}{dr}$, and $v = \int \frac{c}{vr^2} dr$, whis means $v^2=c\int \frac{1}{r^2}dr$.
$v= \sqrt{C-\frac{c}{r}}$ is this correct? ($C$ is the constant of integration)

 

[gneill]

So $c\frac{1}{r^2}= \frac{dv}{dr}v$, then $\frac{c}{vr^2}=\frac{dv}{dr}$, and $v = \int \frac{c}{vr^2} dr$, whis means $v^2=c\int \frac{1}{r^2}dr$.
$v= \sqrt{C-\frac{c}{r}}$ is this correct? ($C$ is the constant of integration)

No, you need to separate the r and v.

$c\frac{1}{r^2}~dr = v~dv$

Now both sides can be integrated.

 

[Robin04]

No, you need to separate the r and v.

$c\frac{1}{r^2}~dr = v~dv$

Now both sides can be integrated.

$C-\frac{c}{r} = \frac{v^2}{2}$
$v=\sqrt{K-\frac{2c}{r}}$, where $K=2C$

 

[George Jones]

$C-\frac{c}{r} = \frac{v^2}{2}$
$v=\sqrt{K-\frac{2c}{r}}$, where $K=2C$

Instead of doing indefinite integrals and having a constant of integration, you might want to do definite integrals with limits from $r_0$ to $r$ and $v_0$ to $v$. This automatically incorporates initial conditions.

 

[Robin04]

Instead of doing indefinite integrals and having a constant of integration, you might want to do definite integrals with limits from $r_0$ to $r$ and $v_0$ to $v$. This automatically incorporates initial conditions.

$\int_{r_0}^r \frac{c}{r^2}dr= \int_{v_0}^v v dv$
$-\frac{c}{r}+\frac{c}{r_0}=\frac{v^2}{2}-\frac{v_0^2}{2}$
$v=\sqrt{v_0^2+2c(\frac{1}{r_0}-\frac{1}{r})}$

 

[Robin04]

@gneill liked my last comment, so I guess it's correct. Thank you very much for your help! :)