QJEEM19Phc462: JEE Main 2019 Physics - Class 11 JEEM19Phc462 Solution: If the angular momentum of a planet of mass m, moving around…

Question

Verified Solution • 2025–26 Session moderate

If the angular momentum of a planet of mass m, moving around the Sun in a circular orbit is L, about the center of the Sun, its areal velocity is :

Accepted Answer

dA =

{1 \over 2}

r²d

\theta

\therefore

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

\Rightarrow

{{dA} \over {dt}} = {1 \over 2}{r^2}\omega

. . . . . (1)

We know,

angular momentum,

L =

mvr

=

m\left( {\omega r} \right)r

= mr²

\omega

\therefore

\omega

=

{L \over {m{r^2}}}

. . . . . (2)

Put value of

\omega

in equation(1),

{{dA} \over {dt}}

=

{1 \over 2}{r^2}

(

{L \over {m{r^2}}}

)

=

{L \over {2m}}

Step-by-Step Verified Solution