The angular momentum of a planet of mass M mo | vedclass

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(D) Consider a small displacement $ds$ of the planet along its elliptical orbit in a small time interval $dt$.The area $dA$ swept by the position vect

Vedclass · Accepted Answer

$\frac{L}{2M}$

Vedclass · Answer

(D) Consider a small displacement $ds$ of the planet along its elliptical orbit in a small time interval $dt$.The area $dA$ swept by the position vector $\vec{r}$ in this time is given by the area of a triangle with sides $r$ and $ds$:$dA = \frac{1}{2} |\vec{r} 	imes d\vec{s}| = \frac{1}{2} r ds \sin 	heta$where $	heta$ is the angle between the position vector $\vec{r}$ and the displacement vector $d\vec{s}$.The areal velocity is the rate of change of area with respect to time:$	ext{Areal velocity} = \frac{dA}{dt} = \frac{1}{2} r \sin 	heta \frac{ds}{dt}$Since the velocity of the planet is $v = \frac{ds}{dt}$,we have:$\frac{dA}{dt} = \frac{1}{2} r v \sin 	heta$Multiplying and dividing by the mass $M$ of the planet:$\frac{dA}{dt} = \frac{1}{2M} (M v r \sin 	heta)$Since the magnitude of angular momentum is $L = Mvr \sin 	heta$,we get:$\frac{dA}{dt} = \frac{L}{2M}$

The angular momentum of a planet of mass $M$ moving around the sun in an elliptical orbit is $\overrightarrow{L}$. The magnitude of the areal velocity of the planet is:

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