A planet of mass `m` is the elliptical orbit about the sun `(mlt ltM_("sun"))` with an orbital period `T`. If `A` be the area of orbit, then its angular momentum would be:

Options

A. `(2mA)/T`
B. mAT
C. `(mA)/(2T)`
D. 2mAT

Answer

A

Explanation

Area of the triangle shown is `dA=1/2rxxdr`
`rArr dA=1/2 xx r xx rd theta = r^2/2 d theta`
Dividing both side by dt ,
`(dA)/(dt)=r^2/2 (d theta)/(dt)=r^2/2 omega =(m r^2 omega)/(2m) = (Iomega)/(2m) =L/(2m) rArr dA=L/(2m)dt`
Integrating both sides, we get `rArr A=L/(2m)xxT rArr L=(2mA)/T`