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--- 183_notes:examples:earth_s_translational_angular_momentum [2014/11/16 20:27] – pwirving
+++ 183_notes:examples:earth_s_translational_angular_momentum [2014/11/20 00:35] – pwirving
@@ Line 6: / Line 6: @@
 === 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.
@@ Line 18: / Line 19: @@
 === Approximations & Assumptions ===
+Assume Earth moves in a perfect circular orbit
@@ Line 24: / Line 25: @@
-{{course_planning:projects:mi3e_11-006.jpg?400}}
+{{183_projects:mi3e_11-002.jpg?400}}
+Circumference of a circle = $2\pi r$
+$\vec{p} = m\vec{v}$
+$v = s/t$
+$\left|\vec{L}_{trans}\right| = \left|\vec{r}_A\right|\left|\vec{p}\right|\sin \theta$
@@ Line 31: / Line 39: @@
 === 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:
+$v = \frac{2\pi(1.5 \times 10^{11}m)}{(365)(24)(60)(60)s} = 3.0 \times 10^4 m/s$
+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.
-$v = \frac{2\pi(1.5 x 10^{11}m}{(365)(24)(60)(60)s} = 3.0 x 10^4 m/s$
+$\mid\vec{L}_{trans,Sun}\mid = 2.7 \times 10^{40} kg \cdot m^2/s$
+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.