Let the speed of the planet at the perihelion `P` in figure be `v_(P)` and the Sun planet distance `SP` be `r_(P)`. Relater `r_(P), v_(P)` to the corresponding quantities at the aphelion `(r_(A),v_(A))`. Will the planet take equal times to transverse `BAC` and `CPB`?

Explanation

The magnitude of the angular momentum at P is `L_(p)=m_(P)r_(p)u_(p)`. Since inspection tells us that `r_(p)` and `v_(p)` are mutually perpendicular. Similarly . `L_(A)=m_(p)r_(A).v_(A)`. From angular momentum conservation
`m_(p)r_(p)v_(p)=m_(p)r_(A)v_(A)`
or `(v_(p))/(v_(A))=(r_(A))/(r_(p))`
The area, SBAX bounded by the elipse and the raidus vectors SB and SC is large than SBPC in Fig. 9.1 From kepler's second law, equal areas are swept in equal times. Hence the planat will take a longer time to traverse BAC than CPB.