Deduce Kepler's second laws of planetary motion for a planet.

Explanation

Alternative Method:
In the Fig. 1.35, S is the fixed position of the sun. in a very small interval of time dt, a planet makes an angular displacement `d""theta` moving from A to B in its orbit . If `dtheta` is sufficiently small, SAB is effectively a triangle with `SA~~SB =r`. Also, `AB=rd""theta`= altitude of the triangle.

So, the area described by the planet in time dt is ,
dA=area of the triangle SAB
`=1/2 xx"base" SA xx"altitude" AB`
`=1/2 r*rd""theta=1/2r^2d""theta`
so, the areal velocity of the planet,
`(dA)/(dt)=1/2r^2(d""theta)/(dt)=1/2 r^2omega` [`because omega =(d""theta)/(dt)` [angular velocity]
Now, if m=mass of the planet, then its angular momentum,
`L=Iomega=mr^2omega`
The angular momentum is conserved as there is no external torque on the system.
so, `mr^2omega` = constant
or, `r^2omega` =constant
or, `(dA)/(dt)=1/2r^2omega`= constant