Analyzing Central Force

Everyday we often meet various kind of force. For example: gravitational force, frictional force, tidal force, etc. Perhaps some of us only know forces on linear trajectory like pull or push forces but they don't know how the forces happening on non-linear trajectory like round trajectory. How forces work on round trajectory?

img scr: http://www.autobild.es/sites/default/files/1-marquez-cruzado.jpg

Let we see image above. How can marquez cruzado almost touching the ground when he turns with high velocity? the answer is he is affected by centripetal force which the direction to center of trajectory (turn trajectory which means the trajectory is round).

Force which works to marquez is centripetal force.

Centripetal force which works on marquez is central force that affected by tangential velocity (velocity which is proportional to r) and the direction to the center.

Let we define a central force :

$\vec{F}(r)=F(r)\hat{r}$

with the equation of motion:

$m\frac{\mathrm{d}^{2}\vec{r} }{\mathrm{d} t^{2}}=F(r)\hat{r}$

Central force has angular momentum which the direction proportional to the direct of motion. (see image below).

$L=mr^{2}\frac{\mathrm{d} \theta}{\mathrm{d} t}$

$\vec{L}=\vec{r}\times \vec{p}=\vec{r}\times m\frac{\mathrm{d} \vec{r}}{\mathrm{d} t}$

because $\vec{r}\perp \vec{v}$ then $\left \| \vec{L} \right \|= mr\frac{\mathrm{d} r}{\mathrm{d} t}$

we define that:

$\frac{\mathrm{d} r}{\mathrm{d} t}=r\omega =r\frac{\mathrm{d} \theta }{\mathrm{d} t}$

then the angular momentum is:

$L=mr^{2}\frac{\mathrm{d} \theta}{\mathrm{d} t}$

Okay, Back to the equation of motion $m\frac{\mathrm{d}^{2}\vec{r} }{\mathrm{d} t^{2}}=F(r)\hat{r}$

In analyzing the central force we don't use cartesian coordinate but we use polar coordinate.

$\vec{r}=r\hat{r} \rightarrow \frac{\mathrm{d}\vec{r} }{\mathrm{d} t}=\hat{r}\frac{\mathrm{d}r }{\mathrm{d} t}+r\frac{\mathrm{d}\hat{r} }{\mathrm{d} x}$

use law $\frac{\mathrm{d} \hat{r}}{\mathrm{d} t}=\frac{\mathrm{d}\hat{r} }{\mathrm{d} \theta}\cdot \frac{\mathrm{d} \theta}{\mathrm{d} t}$

at the image we can know that $\hat{r}=cos\theta\, \hat{i}+sin\theta\, \hat{j}$ and $\hat{\theta}=-sin\theta\,\hat{i}+cos\theta\,\hat{j}$

$\frac{\mathrm{d} \hat{r}}{\mathrm{d} \theta}=-sin\theta\,\hat{i}+cos\theta\,\hat{j}=\hat{\theta}$

then $\frac{\mathrm{d} \vec{r}}{\mathrm{d} t}=\frac{\mathrm{d} r}{\mathrm{d} t}\hat{r}+r\frac{\mathrm{d}\theta }{\mathrm{d} t}\hat{\theta}$

second derivative of $\hat{r}$ :

$\frac{\mathrm{d^2} \vec{r}}{\mathrm{d} t^2}=\frac{\mathrm{d^2} \hat{r}}{\mathrm{d} t^2}\hat{r}+\frac{\mathrm{d} r}{\mathrm{d} t}\frac{\mathrm{d} \theta}{\mathrm{d} t}\hat{\theta}+\frac{\mathrm{d} r}{\mathrm{d} t}\frac{\mathrm{d} \theta}{\mathrm{d} t}\hat{\theta}+r\frac{\mathrm{d^2} \theta}{\mathrm{d} t^2}\hat{\theta}+r\frac{\mathrm{d} \theta}{\mathrm{d} t}\frac{\mathrm{d} \theta}{\mathrm{d} t}(-\hat{r})$

$=\left (\frac{\mathrm{d^2} r}{\mathrm{d} t^2}-r\left (\frac{\mathrm{d} \theta}{\mathrm{d} t} \right )^2 \right )\hat{r}+\left ( 2\frac{\mathrm{d} r}{\mathrm{d} t} \frac{\mathrm{d} \theta}{\mathrm{d} t}+r\frac{\mathrm{d^2} \theta}{\mathrm{d} t^2} \right )\hat{\theta}$

the equation above multiplied by m

$m\frac{\mathrm{d^2} \vec{r}}{\mathrm{d} t^2}=m\left (\frac{\mathrm{d^2} r}{\mathrm{d} t^2}-r\left (\frac{\mathrm{d} \theta}{\mathrm{d} t} \right )^2 \right )\hat{r}+m\left ( 2\frac{\mathrm{d} r}{\mathrm{d} t} \frac{\mathrm{d} \theta}{\mathrm{d} t}+r\frac{\mathrm{d^2} \theta}{\mathrm{d} t^2} \right )\hat{\theta}$

we get that:

$m\left ( \frac{\mathrm{d^2} r}{\mathrm{d} t^2} -r\left ( \frac{\mathrm{d} \theta}{\mathrm{d} t} \right )^2\right ) = F(r)$ and

$m\left ( 2\frac{\mathrm{d} r}{\mathrm{d} t}\frac{\mathrm{d} \theta}{\mathrm{d} t}+r\frac{\mathrm{d^2} \theta}{\mathrm{d} t^2}\right ) = 0$

Look at $L=mr^{2}\frac{\mathrm{d} \theta}{\mathrm{d} t}$ then we derive it:

$\frac{\mathrm{d} L}{\mathrm{d} t}=2mr\frac{\mathrm{d} r}{\mathrm{d} t}\frac{\mathrm{d} \theta}{\mathrm{d} t}+mr^2\frac{\mathrm{d^2}\theta }{\mathrm{d} t^2}$ then devided by mr (math manipulation)

then we get $\left ( 2\frac{\mathrm{d} r}{\mathrm{d} t}\frac{\mathrm{d} \theta}{\mathrm{d} t}+r\frac{\mathrm{d^2} \theta}{\mathrm{d} t^2}\right ) = 0$ → L is constant

we take a force that the direction is opposite to the position $F(r)=-f(r)$

because L is constant then $m\frac{\mathrm{d} }{\mathrm{d} t}\left ( r^2\frac{\mathrm{d} \theta}{\mathrm{d} t} \right )=0$ we symbolize $r^2\frac{\mathrm{d} \theta}{\mathrm{d} t} = h$

$\left ( \frac{\mathrm{d^2}r }{\mathrm{d} t^2}-r\left ( \frac{\mathrm{d} \theta}{\mathrm{d} t} \right )^2 \right )=-f(r)$

$\left ( \frac{\mathrm{d^2}r }{\mathrm{d} t^2}-\frac{h^2}{r^3} \right )=-\frac{f(r)}{m}$

Let $u=\frac{1}{r} \rightarrow r=\frac{1}{u}$

$\frac{\mathrm{d}r }{\mathrm{d} t}=-\frac{1}{u^2}\frac{\mathrm{d} u}{\mathrm{d} t}$

$r^2\frac{\mathrm{d} \theta}{\mathrm{d} t}=h \rightarrow \frac{\mathrm{d} \theta}{\mathrm{d} t}=\frac{h}{r^2}=hu^2$

$=-\frac{1}{u^2}\frac{\mathrm{d} u}{\mathrm{d} \theta}\frac{\mathrm{d} \theta}{\mathrm{d} t}$

$\frac{\mathrm{d} r}{\mathrm{d} t}=-\frac{1}{u^2}\cdot hu^2\frac{\mathrm{d}u }{\mathrm{d} \theta}=-h\frac{\mathrm{d}u }{\mathrm{d} \theta}$

$\frac{\mathrm{d^2} r}{\mathrm{d} t^2}=-h\frac{\mathrm{d^2}u }{\mathrm{d} \theta^2}\frac{\mathrm{d} \theta}{\mathrm{d} t}=-h\frac{\mathrm{d^2}u }{\mathrm{d} \theta^2}(hu^2)=-h^2u^2\frac{\mathrm{d^2}u }{\mathrm{d} \theta^2}$

$-h^2u^2\frac{\mathrm{d^2} u}{\mathrm{d} \theta^2}-h^2u^3=-\frac{f(u)}{m}$

Then we get the non-homogenous differential equation $\frac{\mathrm{d^2} u}{\mathrm{d} \theta^2}+u=\frac{f(u)}{mh^2u^2}$

The general solution of homogen differential equation: $u=Acos\theta+Bsin\theta$

using inverse square force: $f(u)=\propto u^2$

$\frac{\mathrm{d^2} u}{\mathrm{d} \theta^2}+u=\frac{\propto u^2}{mh^2u^2}=\frac{\propto }{mh^2}=constant$

Then the general solution: $u=Acos\theta+Bsin\theta+\frac{\propto }{mh^2}$

with trigonometry formula then:

$u=C\,cos(\theta-\theta_{0})+\frac{\propto }{mh^2}$

$\frac{1}{r}=C\,cos(\theta-\theta_{0})+\frac{\propto }{mh^2}$

defined that $l=\frac{mh^2}{\propto }$ then $\frac{1}{r}=C\,cos(\theta-\theta_{0})+\frac{1 }{l}$

Then derived by l and defined C constant as e, so we ge

$\frac{l}{r}=1+e\,cos(\theta-\theta_{0})$ which is the equation of motion on elips trajector

Now we know the solution of Central force's equation of motion is elips equation. All body have energy. How we analyze the energy of the motion which affected by central force?

First of all we know that the total energy or mechanic energy E=K+V(r) with V(r) depending on position.

We look back at a force that the direction is opposite to vector of position $\vec{F}=-f(r)\hat{r}$
Potential energy is obtained from negative integration of force:

$V(r)=-\int \vec{F}(r)\,\mathrm{d}\vec{r}$
   $=\int f(r)\mathrm{d}r$
velocity can we define with a review of the kinetic energy :

$v^2=\left ( \frac{\mathrm{d} \vec{r}}{\mathrm{d} t} \right )^2=\left ( \frac{\mathrm{d}r }{\mathrm{d} t}\hat{r}+r\frac{\mathrm{d} \theta}{\mathrm{d} t}\hat{\theta} \right )^2=\left ( \frac{\mathrm{d} r}{\mathrm{d} t} \right )^2+r^2\left ( \frac{\mathrm{d} \theta}{\mathrm{d} t} \right )^2$

$E=K+V(r)$
$E=\frac{1}{2}mv^2+V(r)$
$E=\frac{1}{2}m\left ( \left ( \frac{\mathrm{d} r}{\mathrm{d} t} \right )^2+r^2\left ( \frac{\mathrm{d} \theta}{\mathrm{d} t} \right )^2 \right )+V(r)$
$E=\frac{1}{2}m \left ( \frac{\mathrm{d} r}{\mathrm{d} t} \right )^2+\frac{mr^2\left ( \frac{\mathrm{d} \theta}{\mathrm{d} t} \right )^2}{2}+V(r)$

Before we have known that $L=mr^{2}\frac{\mathrm{d} \theta}{\mathrm{d} t}$    then $\frac{\mathrm{d} \theta}{\mathrm{d}t }=\frac{L}{mr^2}$
the equation above we can input it to the equation E

So we can obtain $E=\frac{1}{2}m \left ( \frac{\mathrm{d} r}{\mathrm{d} t} \right )^2+\frac{L^2}{2mr^2}+V(r)$ with    $\frac{L^2}{2mr^2}+V(r)$ is the effective potential.
Let $V(r)=-\frac{GMm}{r}$ then $V_{eff}=\frac{L^2}{2mr^2}-\frac{GMm}{r}$
If the equation above we plot it to graph then:
Image result for effective potential