Answer
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Hint: 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 $\omega =\dfrac{v}{r}$.
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 $\omega =\dfrac{v}{r}$.
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