183_notes:examples:earth_s_translational_angular_momentum [Projects & Practices in Physics]

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===== Example: Earth's Translational Angular Momentum =====

Calculate the magnitude of the Earth's translational (orbital) angular momentum relative to the Sun when the Earth is at location A and when the Earth is at location B as displayed in the representation of the situation in the representations. The mass of the Earth is $6$ x $10^{24}kg$ and its distance from the Sun is $1.5$ x $10^{11}m$.


=== Facts ===

Mass of the Earth: $6$ X $10^{24}$kg

Distance from the Sun: $1.5$ x $10^{11}$m


=== Lacking ===

The magnitude of the Earth's translational (orbital) angular momentum relative to the Sun when the Earth is at location A on the representation and when it is at location B on the representation.



=== Approximations & Assumptions ===

Assume Earth moves in a perfect circular orbit

Assume main interaction is with the sun


=== Representations ===


{{183_projects:mi3e_11-002.jpg?400}}

Circumference of a circle = $2\pi r$

$\vec{p} = m\vec{v}$

$v = s/t$

$\left|\vec{L}_{trans}\right| = \left|\vec{r}_A\right|\left|\vec{p}\right|\sin \theta$



=== Solution ===

The Earth makes one complete orbit of the Sun in 1 year, so you need to break down 1 year into seconds and know that the distance the Earth travels in that time is $2\pi r$ in order to find its average speed is:

$v = \frac{2\pi(1.5 \times 10^{11}m)}{(365)(24)(60)(60)s} = 3.0 \times 10^4 m/s$

With this average velocity we can find the momentum of Earth at location A as we know the mass of the Earth and now know the velocity of the Earth.

$ \vec{p} = \langle 0, 6 \times 10^{24}kg \cdot 3.0 \times 10^{4} m/s, 0 \rangle $

Computing for momentum we get:

$ \vec{p} = \langle 0, 1.8 \times 10^{29}, 0 \rangle  kg \cdot m/s$

$\mid\vec{p}\mid = 1.8 \times 10^{29} kg \cdot m/s$

We know that the magnitude of the Earth's translational angular momentum relative to the sun is given by  $\left|\vec{L}_{trans,Sun}\right| = \left|\vec{r}_A\right|\left|\vec{p}\right|\sin \theta$

$\mid\vec{L}_{trans,Sun}\mid = (1.5 \times 10^{11} m)(1.8 \times 10^{29} kg \cdot m/s) sin 90^{\circ}$

Compute for $\left|\vec{L}_{trans,Sun}\right|$ by inputting the known values for the variables.

$\mid\vec{L}_{trans,Sun}\mid = 2.7 \times 10^{40} kg \cdot m^2/s$

It turns out that at location $B, \mid\vec{r}\mid, \mid\vec{p}\mid$, and $\theta$ are the same as they were at location A, so $ \mid\vec{L}_{trans,Sun}\mid$ also has the same value it had at location A.