183_notes:examples:earth_s_translational_angular_momentum

Differences

This shows you the differences between two versions of the page.

Link to this comparison view

Both sides previous revision Previous revision
Next revision		 Previous revision
183_notes:examples:earth_s_translational_angular_momentum [2014/11/16 20:44] – pwirving183_notes:examples:earth_s_translational_angular_momentum [2014/11/20 16:30] (current) – pwirving
Line 6: Line 6:
 === Facts === === Facts ===
  
 +Mass of the Earth: $6$ X $10^{24}$kg
  
- +Distance from the Sun: $1.5$ x $10^{11}$m
  
  
 === Lacking === === 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 === === Approximations & Assumptions ===
  
 +Assume Earth moves in a perfect circular orbit
  
 +Assume main interaction is with the sun
  
  
Line 24: Line 27:
  
  
-{{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 41:
 === 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 $
 +
 +Computing for momentum we get:
  
 $ \vec{p} = \langle 0, 1.8 \times 10^{29}, 0 \rangle  kg \cdot m/s$ $ \vec{p} = \langle 0, 1.8 \times 10^{29}, 0 \rangle  kg \cdot m/s$
  
 $\mid\vec{p}\mid = 1.8 \times 10^{29} kg \cdot m/s$ $\mid\vec{p}\mid = 1.8 \times 10^{29} kg \cdot m/s$
 +
 +We know that the magnitude of the Earth's translational angular momentum relative to the sun is given by  $\left|\vec{L}_{trans,Sun}\right| = \left|\vec{r}_A\right|\left|\vec{p}\right|\sin \theta$
  
 $\mid\vec{L}_{trans,Sun}\mid = (1.5 \times 10^{11} m)(1.8 \times 10^{29} kg \cdot m/s) sin 90^{\circ}$ $\mid\vec{L}_{trans,Sun}\mid = (1.5 \times 10^{11} m)(1.8 \times 10^{29} kg \cdot m/s) sin 90^{\circ}$
 +
 +Compute for $\left|\vec{L}_{trans,Sun}\right|$ by inputting the known values for the variables.
  
 $\mid\vec{L}_{trans,Sun}\mid = 2.7 \times 10^{40} kg \cdot m^2/s$ $\mid\vec{L}_{trans,Sun}\mid = 2.7 \times 10^{40} kg \cdot m^2/s$
  
 +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.
  • 183_notes/examples/earth_s_translational_angular_momentum.1416170695.txt.gz
  • Last modified: 2014/11/16 20:44
  • by pwirving