183_notes:examples:angular_momentum_of_halley_s_comet

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183_notes:examples:angular_momentum_of_halley_s_comet [2014/11/20 06:37] pwirving183_notes:examples:angular_momentum_of_halley_s_comet [2014/11/20 16:30] (current) pwirving
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 === Approximations & Assumptions === === Approximations & Assumptions ===
  
 +No other interactions the rest of the solar system.
  
 +Assume main interaction is with the sun.
  
  
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 $\left|\vec{L}_{trans,Sun}\right| = \left|\vec{r}_A\right|\left|\vec{v}\right|\left|m\right|\sin \theta$ $\left|\vec{L}_{trans,Sun}\right| = \left|\vec{r}_A\right|\left|\vec{v}\right|\left|m\right|\sin \theta$
 +
 +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 $10^4m/s)sin 90^{\circ}$ $\mid\vec{L}_{trans,Sun}\mid$ = $(8.77$ 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 $10^{30}$ $kg \cdot m^2/s$ $= 1.1$ x $10^{30}$ $kg \cdot m^2/s$
 +
 +In vector form $\vec{L}_{trans,Sun}$ is:
  
 $\vec{L}_{trans,Sun}$ = $\langle{0, 0, -1.1 x 10^30}\rangle$ $kg \cdot m^2/s$ $\vec{L}_{trans,Sun}$ = $\langle{0, 0, -1.1 x 10^30}\rangle$ $kg \cdot m^2/s$
  
  
-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 $10^4m/s)sin 17.81^{\circ}$ $\mid\vec{L}_{trans,Sun}\mid$ = $(1.19$ x $10^{12}m)(2.2$ x $10^{14}kg)(1.32$ x $10^4m/s)sin 17.81^{\circ}$
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 $\vec{L}_{trans,Sun}$ = $\langle{0, 0, -1.1 x 10^30}\rangle$ $kg \cdot m^2/s$ $\vec{L}_{trans,Sun}$ = $\langle{0, 0, -1.1 x 10^30}\rangle$ $kg \cdot m^2/s$
 +
 +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.
  
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  • by pwirving