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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
@@ Line 9: / Line 9: @@
 Distance from the Sun:  $1.5$  x  $10^{11}$ m
@@ Line 26: / Line 20: @@
 Assume Earth moves in a perfect circular orbit
+Assume main interaction is with the sun
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 {{183_projects:mi3e_11-002.jpg?400}}
+Circumference of a circle =  $2\pi r$
+ $\vec{p} = m\vec{v}$
  $v = s/t$
-$\left|\vec{L}_{trans}\right| = \left|\vec{r}_A \times \vec{p}\right| = \left|\vec{r}_A\right|\left|\vec{p}\right|\sin \theta$
+ $\left|\vec{L}_{trans}\right| = \left|\vec{r}_A\right|\left|\vec{p}\right|\sin \theta$
@@ Line 41: / Line 41: @@
 === 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:
  $v = \frac{2\pi(1.5 \times 10^{11}m)}{(365)(24)(60)(60)s} = 3.0 \times 10^4 m/s$
-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.
  $\vec{p} = \langle 0, 6 \times 10^{24}kg \cdot 3.0 \times 10^{4} m/s, 0 \rangle$
+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   $\left|\vec{L}_{trans,Sun}\right| = \left|\vec{r}_A\right|\left|\vec{p}\right|\sin \theta$
  $\mid\vec{L}_{trans,Sun}\mid = (1.5 \times 10^{11} m)(1.8 \times 10^{29} kg \cdot m/s) sin 90^{\circ}$
-$\mid\vec{L}_{trans,Sun}\mid = 2.7 \times 10^{40} kg \cdot m^2/s$
+Compute for $\left|\vec{L}_{trans,Sun}\right|$ by inputting the known values for the variables.
-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.
+$\mid\vec{L}_{trans,Sun}\mid = 2.7 \times 10^{40} kg \cdot m^2/s$
+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.