The angular momentum of a planet of mass M moving around the sun in
Question:

The angular momentum of a planet of mass $M$ moving around the sun in

an elliptical orbit is $\vec{L}$. The magnitude of the areal velocity of the planet is :

 

  1. (1) $\frac{4 \mathrm{~L}}{\mathrm{M}}$

  2. (2) $\frac{\mathrm{L}}{\mathrm{M}}$

  3. (3) $\frac{2 L}{M}$

  4. (4) $\frac{\mathrm{L}}{2 \mathrm{M}}$


Correct Option: , 4

Solution:

For small displacement ds of the planet its area can be written as

$\mathrm{dA}=\frac{1}{2} \mathrm{rd} \ell$

$=\frac{1}{2} \mathrm{rds} \sin \theta$

A.vel $=\frac{\mathrm{dA}}{\mathrm{dt}}=\frac{1}{2} \mathrm{r} \sin \theta \frac{\mathrm{ds}}{\mathrm{dt}}=\frac{\mathrm{V} \mathrm{r} \sin \theta}{2}$

$\frac{\mathrm{d} \mathrm{A}}{\mathrm{dt}}=\frac{1}{2} \frac{\mathrm{mV}_{\mathrm{r}} \sin \theta}{\mathrm{m}}=\frac{\mathrm{L}}{2 \mathrm{~m}}$

Administrator

Leave a comment

Please enter comment.
Please enter your name.