Question

A planet of mass m is in an elliptical orbit about the sun $\left( {m < < {M_{sun}}} \right)$ ​ with an orbital period $T$. If $A$ be the area of orbit, then its angular momentum would be-
A.$2mAT$
B.$mAT$
C.$\;\dfrac{{mA}}{{2T}}$
D.$\;\dfrac{{2mA}}{T}$

Answer 1

Angular momentum can be defined as any rotating object’s property which is given by the moment of inertia times the angular velocity. We need to write this definition mathematically and we should keep on substituting the other formulas into this. So that we can get the resultant angular momentum.

Complete answer:
Mathematically angular moment of any rotating object is given as,
$L = I\omega$…….. (1)
Here $L$ is the angular momentum of the rotating object.
$I$ is known as the moment of inertia of the rotating body.
$\omega$ is known as the angular velocity of the rotating object.
Now we know that moment of inertia can be written as, $I = m{r^2}$
Therefore substituting this in equation (1) we get,
$L = m{r^2}\omega$…….. (2)
Here$m$is the mass of the object
$r$ is the radius of the object.
Given that$A$is the area of the orbit. Also, we know the area of the circle formula,
$A = \pi {r^2}$
Rearranging the above equation to get the radius on one side and the other variables on the other side we get,
${r^2} = \dfrac{A}{\pi }$ …….. (3)
Substituting equation (3) on (2)
$L = m\dfrac{A}{\pi }\omega$ …… (4)
The magnitude of the vector quantity of the angular velocity is the angular frequency. Therefore substituting the angular frequency formula in the above equation.
$\omega = \dfrac{{2\pi }}{T}$
Here $T$ is the orbital time period.
Substituting in the equation (4) we get,
$L = m \times \dfrac{A}{\pi } \times \dfrac{{2\pi }}{T}$
Simplifying,
$L = \;\dfrac{{2mA}}{T}$

Therefore the correct option is D.

Note:
Angular momentum is a vector quantity. The formula for the angular momentum varies according to two cases. The first case is the point object. A point object is defined as an object accelerating around a fixed point. For this case, the angular momentum formula is given as $\vec L = r \times \vec p$. The other case is the extended object which can be defined as the object revolving its axis. The formula for this case is given as, $\vec L = I \times \vec \omega$