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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

Option 1) L/m

Option 2) 4L/m

Option 3) L/2m

Option 4) 2L/m

Answers (1)

best_answer

Kepler's 2nd law -

Area of velocity = $\frac{dA}{dt}$

$\frac{dA}{dt}=\frac{1}{2}\frac{\left ( r \right )\left ( Vdt \right )}{dt}=\frac{1}{2}rV$

$\frac{dA}{dt}\rightarrow$ Areal velocity

$dA\rightarrow$ small area traced

- wherein

Simiar to Law of conservation of momentum

$\frac{dA}{dt}= \frac{L}{2m}$

$L\rightarrow$ Angular momentum

Known as Law of Area

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

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

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

L = mr² $\omega$

$\Rightarrow \frac{dA}{dt} = \frac{L}{2m}$

Option 1)

$\frac{L}{m}$

Option 2)

$\frac{4L}{m}$

Option 3)

$\frac{L}{2m}$

Option 4)

$\frac{2L}{m}$

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