Test Index

Gravitation Part 1

Section: Physics

© examsnet.com

Question : 19 of 100

Marks: +1, -0

The angular momentum of a planet of mass

M

moving around the Sun in an elliptical orbit is

L

. The magnitude of the areal velocity of the planet is

[18 Mar 2021 Shift 2]

$\;\frac{4L}{M}$
$\;\frac{L}{M}$
$\;\frac{2L}{M}$
$\;\frac{L}{2M}$

Solution:

According to the Kepler's second law,

\frac{dA}{dt}=

constant

Area covered from

P

to

P'

is

dA

, then

dA = \frac{1}{2} r^{2} d\theta \Rightarrow \frac{dA}{dt} = \frac{1}{2} r^{2} \frac{d\theta}{dt}

\Rightarrow \frac{dA}{dt} = \frac{1}{2} r^{2} \omega

...(i)

Angular momentum,

L = M v r = M (\omega r) r = M \omega r^{2}

\Rightarrow \omega = \frac{L}{m r^{2}}

\therefore

From Eqs. (i) and (ii), we get

\;\frac{dA}{dt} = \frac{1}{2} r^{2} \left( \frac{L}{M r^{2}} \right)

\Rightarrow \;\frac{dA}{dt} = \frac{L}{2M}

© examsnet.com

Go to Question:

Prev Question

Next Question