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--- 183_notes:examples:angular_momentum_of_halley_s_comet [2014/11/20 06:22] – pwirving
+++ 183_notes:examples:angular_momentum_of_halley_s_comet [2014/11/20 06:45] – pwirving
@@ Line 41: / Line 41: @@
 Direction: At both locations, the direction of the translational angular momentum of the comet is in the -z direction (into the computer); determined by using the right-hand rule.
-At location 1:
+Given this information we know at location 1 the translational angular momentum of the comet relative to the sun will be:
+$\left|\vec{L}_{trans,Sun}\right| = \left|\vec{r}_A\right|\left|\vec{p}\right|\sin \theta$
+We don't know the momentum but we do know the mass and velocity of the comet at $\vec{r}_{1}$ so our equation becomes:
+$\left|\vec{L}_{trans,Sun}\right| = \left|\vec{r}_A\right|\left|\vec{v}\right|\left|m\right|\sin \theta$
+Substituting in for the known variables we get:
 $\mid\vec{L}_{trans,Sun}\mid$ = $(8.77$ x $10^{10}m)(2.2$ x $10^{14}kg)(5.46$ x $10^4m/s)sin 90^{\circ}$
+Solving for $\mid\vec{L}_{trans,Sun}\mid$ we get:
 $= 1.1$ x $10^{30}$ $kg \cdot m^2/s$
+In vector form $\vec{L}_{trans,Sun}$ is:
 $\vec{L}_{trans,Sun}$ = $\langle{0, 0, -1.1 x 10^30}\rangle$ $kg \cdot m^2/s$
-At location 2:
+The same step by step process is used to solve for $\vec{L}_{trans,Sun}$ at location 2:
 $\mid\vec{L}_{trans,Sun}\mid$ = $(1.19$ x $10^{12}m)(2.2$ x $10^{14}kg)(1.32$ x $10^4m/s)sin 17.81^{\circ}$
@@ Line 57: / Line 69: @@
 $\vec{L}_{trans,Sun}$ = $\langle{0, 0, -1.1 x 10^30}\rangle$ $kg \cdot m^2/s$
+Even in the highly elliptical orbit, the comet's translational angular momentum is constant throughout the orbit, despite the fact that it's position, its momentum, and the angle between them change continuously implying that angular momentum is a conserved quantity.