Question

The angular momentum of the earth revolving around the sun, is proportional to ${r^n}$, where $r$ is the distance between the centers of earth and the sun. The value of $n$ is :
(A) 1
(B) -2
(C) -1
(D) $\dfrac{1}{2}$

Answer 1

Hint: From the formula angular momentum of the Earth in its orbit around the sun we can find out the condition for the revolution of the Earth in its orbit. From there we have to find the equation in terms of the angular momentum and find how it varies with $r$. From the variation of angular momentum with the radius, we can find the answer.

Formula used: In the solution, we will be using the following formula,
$\Rightarrow L = mvr$
where $L$ is the angular momentum of the Earth
$m$ is the mass of the Earth and $r$ is the radius of the Earth in its orbit around the Sun.
$\Rightarrow F = G\dfrac{{Mm}}{{{r^2}}}$
Where $F$ is the force due to gravity between the Earth and the Sun
$G$ is the universal gravitational constant.
and $M$ is the mass of the sun.

Complete step by step answer:
The Earth revolves in an orbit around the sun. So the angular momentum of the earth in this orbit is given by,
$\Rightarrow L = mvr$
Now the earth keeps revolving around the Sun because the centripetal acceleration of the earth is provided by the force of gravitation between the Earth and the Sun.
Therefore, we can write
$\Rightarrow \dfrac{{m{v^2}}}{r} = {F_g}$
where $\dfrac{{m{v^2}}}{r}$ is the centripetal force and ${F_g}$ is the force due to gravitation and is given by ${F_g} = G\dfrac{{Mm}}{{{r^2}}}$
So by substituting the value we get,
$\Rightarrow \dfrac{{m{v^2}}}{r} = G\dfrac{{Mm}}{{{r^2}}}$
Now on the L.H.S of this equation, to bring the numerator in the terms of the angular momentum $L = mvr$, we multiply $m{r^2}$ on both the numerator and denominator.
Hence, we get
$\Rightarrow \dfrac{{m{v^2}}}{r} \times \dfrac{{m{r^2}}}{{m{r^2}}} = G\dfrac{{Mm}}{{{r^2}}}$
So the numerator becomes, ${m^2}{v^2}{r^2} = {L^2}$
Substituting we get,
$\Rightarrow \dfrac{{{L^2}}}{{m{r^3}}} = G\dfrac{{Mm}}{{{r^2}}}$
Now by taking the denominator to the L.H.S to the R.H.S we get,
$\Rightarrow {L^2} = G\dfrac{{Mm}}{{{r^2}}} \times m{r^3}$
by cancelling ${r^2}$ from the numerator and the denominator we get,
$\Rightarrow {L^2} = GM{m^2}r$
Taking square root on both the sides,
$\Rightarrow L = \sqrt {GM{m^2}r} $
Therefore the angular momentum is directly proportional to the square root of $r$.
Hence, $L \propto \sqrt r $
We can write this as,
$\Rightarrow L \propto {r^{\dfrac{1}{2}}}$
Therefore, from the question, the value of $n$ is $\dfrac{1}{2}$
So the correct option is (D); $\dfrac{1}{2}$.

Note:
The angular momentum is considered the rotational analog to linear momentum. The Earth has a very large angular momentum because of its huge inertia. We can also solve this problem by taking,
$\Rightarrow L = mvr = m \times r\omega \times r = m{r^2}\omega $
The value of $\omega $ is given by,
$\Rightarrow \omega = \sqrt {\dfrac{{GM}}{{{r^3}}}} $
By substituting we get,
$\Rightarrow L = m{r^2}\sqrt {\dfrac{{GM}}{{{r^3}}}} $
$\Rightarrow L = \sqrt {\dfrac{{GM}}{{{r^3}}}{m^2}{r^4}} $
By cancelling the $r$ we get,
$\Rightarrow L = \sqrt {GM{m^2}r} $
Hence, $L \propto \sqrt r $ and $n$ is $\dfrac{1}{2}$.