Question

the distance between sun and a planet is r. the angular momentum of planet around the sun in circular orbit is proportional to ?​

Answer 1

Answer:

Angular momentum, $L\propto r^{1/2}.$

Explanation:

Given that, $r$ is the radial distance between the sun and the planet.

The angular momentum of the planet around the sun is given by

$L=mvr$.

where,

$m$ = mass of the planet.

$v$ = speed of the planet.

The speed of the planet also varies with $r$.

The centripetal force with which the planet is revolving around the sun must be balanced with the gravitational force on the planet due to sun.

The centripetal force is given by

$F_c = \dfrac{mv^2}{r}$.

The Gravitational force on the planet due to Sun is given by

$F_g =\dfrac{GMm}{r^2}$.

where,

$G$ = Universal Gravitational constant.

$M$ = mass of the Sun.

On putting these two equal,

$F_e=F_g\\\dfrac{mv^2}{r}=\dfrac{GMm}{r^2}\\v^2=\dfrac{GM}{r}\\v=\sqrt{\dfrac{GM}{r}}$

Putting this value of speed in the expression of angular momentum, we get,

$L=mvr=m\left ( \sqrt{\dfrac{GM}{r}} \right )r=\left(m\sqrt{GM \right) r^{1/2}\\\Rightarrow L\propto r^{1/2}.$

Answer 2

see the attachment..

hope it helps!