Central orbit - Find eccentricity of the orbit A particle moving in an ellipse under the action of a force towards the focus O, moves from greatest distance from O to an extremity of the minor axis in time t, and then to the least distance from O in time $\frac{t}{k}$. Show that the eccentricity of the orbit is
$$\frac{k-1}{k+1}\frac{\pi}{2}.$$ 
Approach:
Let the equation of an ellipse with focus as a pole is $$\frac{l}{r}=1+e \cos{\theta}$$
We know that law of force with focus as center of force is $$F=\frac{\mu}{r^2}$$ 
Let ACA' and BCB' are the major and minor axes of the ellipse, C being center. Given the particle moves from A' (as it is at the greatest distance from O) to B in time t and particle takes time t/k to move from B to A (as it is the least distance from O). 
Now how to find the eccentricity? 
 A: Well first let's define some properties of the ellipse to be used in this problem:
$a$ defines the semi major axis.
$b$ defines the semi minor axis.
$e$ is the eccentricity.
$2c$ is the distance between the two fociis.
$S$ is the total surface of the ellipse
We have the following relations:
$$
S=\pi ab\\
e=\frac{c}{a}
$$
Now I you know Kepler second law that states that the surface swept at in different time intervals are equal or:
$$
\frac{dS}{dt}=Constant
$$
Let $S_1$ the surface swept from $A'$ to $B$ in the interval $t$. Ans $S_2$ be the surface swept from $B$ to $A$ in the interval $\frac{t}{k}$, so using kepler second law:
Here is a (picture)[http://i72.fastpic.ru/big/2015/0905/ae/eab21b69f3d76bbdeaab189f4fa6f2ae.jpg]
$$
\frac{S_1}{t}=\frac{S_2}{\frac{t}{k}}\\
S_1=kS_2
$$
Let's calculate $S_1$ :
I don't have a figure now but I hope you understand. We note that $S_1$ is made of a quarter ellipse plus the right triangle $COB$.
So $S_1$ is the sum of the area of a quarter ellipse plus the are of the triangle hence:
$$
S_1=\frac{\pi ab}{4}+\frac{CB\times OC}{2}
$$
But $BC=b$ and $OC=c$
So :
$$
S_1=\frac{\pi ab}{4}+\frac{bc}{2}\\
S_1=\frac{\pi ab}{4}+\frac{abe}{2}\\
S_1=ab(\frac{\pi}{4}+\frac{e}{2})
$$
Now for $S_2$ note that it is made of a quarter ellipse minus the same latter triangle $COB$. So it would be the same as $S_1$ except that we use a minus instead of a plus:
$$
S_2=ab(\frac{\pi}{4}-\frac{e}{2})
$$
Now returning to the formula obtained by kepler law:
$$
S_1=kS_2\\
ab(\frac{\pi}{4}+\frac{e}{2})=kab(\frac{\pi}{4}-\frac{e}{2})\\
\frac{\pi}{4}+\frac{e}{2}=k\frac{\pi}{4}-k\frac{e}{2}\\
\frac{\pi}{2}+e=k\frac{\pi}{2}-ke\\
e(k+1)=\frac{\pi}{2}(k-1)\\
e=\frac{\pi}{2}\frac{k-1}{k+1}
$$
