Question

Calculate the angular momentum of the Earth: (a) Calculate the magnitude of the translational angular momentum of the Earth relative to the center of the sun. See the data on inside back cover. (b) calculate the magnitude of the rotational angular momentum of the Earth. How does this compare to your result in part (a)?

(The angular momentum of the Earth relative to the center of the sun is the sum of the translational and rotational angular momenta. The rotational axis of the Earth is tipped $23.5°$away from a perpendicular to the plane of its orbit.

Short answer

The magnitude of the translational angular momentum of the Earth relative to the center of the Sun-$2.686\times {10}^{40}kg·m^2/s$.

The magnitude of the rotational angular momentum of the Earth is-$7.146\times {10}^{33}kg·m^2/s$

Step-by-step solution

Definition of Angular Momentum.

The rotating analogue of linear momentum is angular momentum (also known as moment of momentum or rotational momentum). Because it is a conserved quantity—the total angular momentum of a closed system remains constant—it is a significant quantity in physics.

The given Data is-

Mass of Earth,$M_{Earth}=6\times {10}^{24}kg$

Radius of Earth,$R_{Earth}=6.4\times {10}^{6}m$

Distance from Earth to Sun,$d=1.5\times {10}^{11}m$

Find the magnitude of the translational angular momentum of the Earth relative to the center of the Sun.

(a) Period of rotation of Earth about the Sun, $T=1$year

$=1year\left(\frac{365day}{1year}\right)\left(\frac{24hr}{1day}\right)\left(\frac{3600s}{1hr}\right)$

$=3156000s$

The angular speed of the Earth rotation about the Sun is calculated as

$\omega =\frac{2\pi rad}{T}$

$$\begin{gathered} =\frac{2\pi rad}{31536000s} \\ =1.99\times {10}^{-7}rad/s \end{gathered}$$

We can treat the Earth as a solid sphere, so the moment of inertia of the Earth about its axis of rotation is

$$\begin{gathered} I_{rot}=\frac{2}{5}{M}_{Earth}{R}_{Earth}^2 \\ =\frac{2}{5}\left(6\times {10}^{24}kg\right){\left(6.4\times {10}^{6}kg\right)}^2 \\ =9.83\times {10}^{37}kg·m^2 \end{gathered}$$

The moment of inertia of the Earth relative the center of the Sun is

$$\begin{gathered} I=M_{Earth}{d}^2 \\ =\left(6\times {10}^{24}kg\right){\left(1.5\times {10}^{11}kg\right)}^2 \\ =1.35\times {10}^{47}kg·m^2 \end{gathered}$$

The rotational angular momentum of an object rotating with angular speed $\left(\omega \right)$in terms of its moment of inertia is

$L_{rot}=I\omega ......\left(1\right)$

The magnitude of the Earth's translational angular momentum in relation to the Sun's center is provided by the sum of the translational and rotational angular momenta.

$$\begin{gathered} L_{rot}=I_{rot}\omega +I\omega \\ =\left(I_{rot}+I\right)\omega \\ =\left(9.83\times {10}^{37}kg·m^2+1.35\times {10}^{47}kg.m^2\right)\left(1.99\times {10}^{-7}rad/s\right) \end{gathered}$$

$=2.686\times {10}^{40}kg·m^2/s$

Find the magnitude of the rotational angular momentum of the Earth.

Period of rotation of Earth,

$$\begin{gathered} =24h\left(\frac{3600s}{1h}\right) \\ =86400s \end{gathered}$$

The angular speed of the Earth rotation calculated as

$$\begin{gathered} \omega =\frac{2\pi rad}{T} \\ =\frac{2\pi rad}{86400s} \\ =7.27\times {10}^{-5}rad/s \end{gathered}$$

The magnitude of the Earth's rotational angular momentum is computed as follows:

$$\begin{gathered} L_{rot}=I_{rot}\omega \\ =\left(9.83\times {10}^{37}kg·m^2\right)\left(7.27\times {10}^{-5}rad/s\right) \\ =7.146\times {10}^{33}kg·m^2/s \end{gathered}$$

On comparing the result in part (b) with the result in part (a), we get

$$\begin{gathered} \frac{L_{trans}}{L_{rot}}=\frac{2.686\times {10}^{40}kg·m^2/s}{7.146\times {10}^{33}kg·m^2/s} \\ =3.758\times {10}^6 \\ {L}_{rot}=\left(3.758\times {10}^6\right)L_{rot} \end{gathered}$$