Question

A planet of mass M is revolving round the sun
in an elliptical orbit. If its angular momentum is
I then the area swept per second by the line joining
planet to sun will be:-
​

Answer 1

${ \huge \bf { \mid{ \overline{ \underline{Question}}} \mid}}$

A planet of mass M is revolving round the sun in an elliptical orbit. If its angular momentum is J then the area swept per second by the line joining planet to sun will be?

$\huge{\bold{\underline{\underline{Answer}}}}$

$\huge{\bold{\underline{Given:-}}}$

• Angular momentum = J
• Let the Radius be "r"
• And, Let it swipe an angle "dθ" at centre.

$\huge{\bold{\underline{Explanation:-}}}$

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Applying Kepler's Second law:-

Let dA be the Area swept under a time dt.

So, dA = Area swept.

$\large{\boxed{\tt dA = \text{\tt Area of Triangle(Figure)}}}$

$\large{\tt \leadsto dA = \dfrac{1}{2} \times Base \times Height}$

Substituting,

$\large{\tt \leadsto dA = \dfrac{1}{2} \times (SP) \times (P' M)}$

Here P' M = v sin θ × dt (Applying Trigonometric ratios)

Substituting the values,

• SP = r
• P'M = (v sin θ). dt

$\large{\tt \leadsto dA = \dfrac{1}{2} \times r \times (v sin \theta).dt }$

$\large{\tt \leadsto dA = \dfrac{1}{2} . rv sin \theta \times dt}$

$\large{\tt \leadsto \dfrac{dA}{dt} = \dfrac{1}{2}. rv sin \theta}$

Multiplying with Mass on Both Numerator & Denominator.

$\large{\tt \leadsto \dfrac{dA}{dt} = \dfrac{1}{2} . \dfrac{rv sin \theta \times m}{m}}$

⇒ P = mv

Substituting,

$\large{\tt \leadsto \dfrac{dA}{dt} = \dfrac{r \times P sin \theta}{2m}}$

From Cross Product of Vectors we Know,

$\large{\boxed{\vec{r} \times \vec{P} = r \: P \: sin \theta}}$

Substituting,

$\large{ \leadsto \dfrac{dA}{dt} = \dfrac{\vec{r} \times \vec{P}}{2m}}$

But r × P = Angular momentum = J

Substituting,

$\large{ \tt \leadsto \dfrac{dA}{dt} = \dfrac{J}{2m}}$

$\huge{\boxed{\boxed{\tt \dfrac{dA}{dt} = \dfrac{J}{2m}}}}$

So, the Areal velocity per second will be J/2m (Option- 2)

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