Question

Calculate the ratio of the angular momentum of the earth about its axis due its spinning motion to that about the sun due to its orbital motion. Radius of the earth=6400km and radius of the orbit of the earth about the sun is equal to $1.5\times {{10}^{8}}km$.

Answer 1

Angular momentum of a body about an axis of rotation is given as Substitute $\omega$ by using the formula $\omega =\dfrac{2\pi }{T}$. Then use the formula to find the angular momentums of earth in both the given cases.

Formula used:
$L=I\omega$
$\omega =\dfrac{2\pi }{T}$

Complete step-by-step answer:
Angular momentum of a body about an axis of rotation is given as $L=I\omega$ …..(i),
I is the moment of inertia and $\omega$ is the angular velocity of the body about the axis of rotation.
And the angular velocity of the body is related to its time period (T) of motion as $\omega =\dfrac{2\pi }{T}$.
Substitute the value of $\omega$ in equation (i).
$\Rightarrow L=I\left( \dfrac{2\pi }{T} \right)$.
Let us calculate the angular moment of earth about its axis of rotation.
For this, let us assume that earth is a uniform solid sphere. The moment of inertia of a solid sphere of mass M and radius R about an axis passing through its centre is given as $I=\dfrac{2}{5}M{{R}^{2}}$.
The time period of earth’s rotation about its axis is 24 hours. Therefore, T=24h.
Hence, the angular momentum of earth about its axis of rotation is ${{L}_{1}}=\left( \dfrac{2}{5}M{{R}^{2}} \right)\left( \dfrac{2\pi }{24h} \right)$
$\Rightarrow {{L}_{1}}=\left( \dfrac{4\pi }{5}M{{R}^{2}} \right)\left( \dfrac{1}{24h} \right)$ …. (ii)
Let us calculate the angular moment of earth about the sun due to its orbital motion.
For this, let us assume that earth rotates around the sun in a uniform circular motion. When a body of mass M rotates around an axis in a circular path of radius R’ with the axis passing through its centre and perpendicular to its plane, its moment of inertia is given as $I=MR{{'}^{2}}$.
The time period of earth’s rotation around the sun is 365 days. Therefore, T=365days=$365\times 24h$.
Hence, the angular momentum of earth about the sun is ${{L}_{2}}=\left( MR{{'}^{2}} \right)\left( \dfrac{2\pi }{365\times 24h} \right)$
${{L}_{2}}=\left( 2\pi MR{{'}^{2}} \right)\left( \dfrac{1}{365\times 24h} \right)$ …. (iii).
Divide (ii) by (iii)
$\Rightarrow \dfrac{{{L}_{1}}}{{{L}_{2}}}=\dfrac{\left( \dfrac{4\pi }{5}M{{R}^{2}} \right)\left( \dfrac{1}{24h} \right)}{\left( 2\pi MR{{'}^{2}} \right)\left( \dfrac{1}{365\times 24h} \right)}$
$\Rightarrow \dfrac{{{L}_{1}}}{{{L}_{2}}}=\dfrac{146{{R}^{2}}}{R{{'}^{2}}}$
Here, R=6400km and $R'=1.5\times {{10}^{8}}km$
$\Rightarrow \dfrac{{{L}_{1}}}{{{L}_{2}}}=\dfrac{146{{(6400)}^{2}}}{{{\left( 1.5\times {{10}^{8}} \right)}^{2}}}=2.65\times {{10}^{-7}}$.
Therefore, the ratio of the angular momentum of the earth about its axis due its spinning motion to that about the sun due to its orbital motion is $2.65\times {{10}^{-7}}$.

Note: Sometimes, you may find the formula for angular momentum as L = mvr. This formula is used for a point mass body when it is rotating about an axis perpendicular to its plane, in a circular motion with its centre on the axis. Here, it can be used for the angular momentum of earth about the sun.
L = mvr is the same as $L=m{{r}^{2}}\omega$ if we substitute $\omega =\dfrac{v}{r}$.