Question

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

Answer 1

$\huge\boxed{\fcolorbox {pink}{purple}{Answer :-}}$

the distance between sun and aplanet r the angular momentum of planet around sun in circular orbit is proportional to Angular momentum, L\propto r^{1/2}.L∝r

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

Answer 2

Solution :-

So while solving this question we will be using Kepler's law of planetary motion.

It States :- The square of Time period is directly proportional to the cube of distance.

Or T² ∝ R³

$\longrightarrow T^2 \propto R^3$

$\longrightarrow T \propto \sqrt{R^3} \: ...(i)$

Also we have

$\longrightarrow \omega = \dfrac{2\pi}{T}$

From here and from (i) we can say that

$\longrightarrow \omega \propto \dfrac{1}{\sqrt{R^3}}$

• Let k be a constant

$\longrightarrow \omega = k \dfrac{1}{\sqrt{R^3}}$

Now Angular momentum ( L)

$L = mR^2\omega$

$= mR^2 \times k \dfrac{1}{\sqrt{R^3}}$

$= k (m) \times \dfrac{R^2}{R^{3/2}}$

$= k ( m ) \times R^{1/2}$

Or

$L \propto \sqrt{R}$