# Question #dea0d

Kepler's second law states that the time an object in an elliptical orbit takes to go from a point $P$ of the orbit to a point $Q$ of the orbit is proportional to the area defined by the line segments joining the points to one of the focci of the ellipse and by the trajectory of that object.
The radius of the circle is a constant $r$ If the distance traveled by Earth in the process of going from point ${P}_{1}$ to the point ${Q}_{1}$ is $s$, and the distance traveled in the process of going from point ${P}_{2}$ to the point ${Q}_{2}$ is also $s$, then the shapes defined by the line segments (radii of the circle) joining those points to the Sun and by the trajectory of the motion is a sector of the circle. Since the area $A$ of the sector of a circle is given by the relation $A = \pi {r}^{2} \theta$, where $\theta$ is the angle (in radians) definied by the line segments, and the angle itself is defined as $\theta = \frac{s}{r}$, we get $A = \pi r s$. Since $s$ is the distance considered for both trajectories, the area "swept" by them is the same, and therefore, by Kepler's second law the time intervales expended in those trajectories are the same.