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183_notes:examples:earth_s_translational_angular_momentum [2014/11/20 00:05] – pwirving | 183_notes:examples:earth_s_translational_angular_momentum [2014/11/20 16:30] (current) – pwirving | ||
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Distance from the Sun: $1.5$ x $10^{11}$m | Distance from the Sun: $1.5$ x $10^{11}$m | ||
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Assume Earth moves in a perfect circular orbit | Assume Earth moves in a perfect circular orbit | ||
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+ | Assume main interaction is with the sun | ||
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{{183_projects: | {{183_projects: | ||
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+ | Circumference of a circle = $2\pi r$ | ||
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+ | $\vec{p} = m\vec{v}$ | ||
$v = s/t$ | $v = s/t$ | ||
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=== Solution === | === Solution === | ||
- | The Earth makes one complete orbit of the Sun in 1 year, so its average speed is: | + | 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: |
$v = \frac{2\pi(1.5 \times 10^{11}m)}{(365)(24)(60)(60)s} = 3.0 \times 10^4 m/s$ | $v = \frac{2\pi(1.5 \times 10^{11}m)}{(365)(24)(60)(60)s} = 3.0 \times 10^4 m/s$ | ||
- | At location A | + | 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 $ | $ \vec{p} = \langle 0, 6 \times 10^{24}kg \cdot 3.0 \times 10^{4} m/s, 0 \rangle $ | ||
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+ | Computing for momentum we get: | ||
$ \vec{p} = \langle 0, 1.8 \times 10^{29}, 0 \rangle | $ \vec{p} = \langle 0, 1.8 \times 10^{29}, 0 \rangle | ||
$\mid\vec{p}\mid = 1.8 \times 10^{29} kg \cdot m/s$ | $\mid\vec{p}\mid = 1.8 \times 10^{29} kg \cdot m/s$ | ||
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+ | We know that the magnitude of the Earth' | ||
$\mid\vec{L}_{trans, | $\mid\vec{L}_{trans, | ||
- | $\mid\vec{L}_{trans, | + | Compute for $\left|\vec{L}_{trans, |
- | At location $B, \mid\vec{r}\mid, | + | $\mid\vec{L}_{trans, |
+ | It turns out that at location $B, \mid\vec{r}\mid, |