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--- 184_notes:examples:week3_spaceship_asteroid [2018/02/03 20:13] – [Solution] tallpaul
+++ 184_notes:examples:week3_spaceship_asteroid [2021/05/19 15:08] (current) – schram45
@@ Line 1: / Line 1: @@
-=====Preventing an Asteroid Collision=====
+[[184_notes:pc_energy|Return to Electric Potential Energy]]
+=====Example: Preventing an Asteroid Collision=====
 Suppose your friend is vacationing in Italy, and she has lent you her spaceship for the weekend. You have gathered together a group of friends and you are currently cruising through the heavens together and having a great time. You are surrounded by nothingness in all directions. Suddenly, the radar starts beeping ferociously. The ship is on a collision course with an asteroid. You are not too worried about survival -- the ship is practically indestructible. However, you know your friend would be devastated if you returned her spaceship with a scratch or dent from the asteroid. You need to prevent the collision.
@@ Line 17: / Line 19: @@
   * The central component can be charged using charge from the asteroid.
   * The electric potential energy of a point charge in the electric field of another point charge is  $U_r=\frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r}$  This was derived in the notes [[184_notes:pc_energy#General_Relationship_-_Energy_and_Force|here]].
 ===Goal===
  * Prevent the asteroid collision using the long-distance wiring setup.
-===f===
-  * The current distance between the ship and the asteroid.
-  * The distribution of charge on the asteroid.
-  * The distribution of charge on the central component and on the ship itself.
 ===Representations===
@@ Line 62: / Line 59: @@
 </WRAP>
-The change in electric potential energy will depend on how close the asteroid gets to the ship, and how we choose to charge the central component. Currently, its distance is $4000 \text{ m/s}\cdot 60 \text{ minutes/second}\cdot 10 \text{ minutes}=2.4\cdot 10^6 \text{ m} $. We also know$ q_{comp}+q_{ast}=50 \text{ C} $, and we don't want the asteroid to approach the central component at a distance any closer than$ 30 \text{ m} $(this is half the width of the ship, and the distance from the central component to the wall). For simplicity of calculation, we'll orient our coordinates so that the central component is at the origin, and the asteroid lies on the$ x$-axis. We'll keep the initial and final positions of the asteroid as variables:
+The change in electric potential energy will depend on how close the asteroid gets to the ship, and how we choose to charge the central component. Currently, its distance is $4000 \text{ m/s}\cdot 60 \text{ seconds/minute}\cdot 10 \text{ minutes}=2.4\cdot 10^6 \text{ m} $. We also know$ q_{comp}+q_{ast}=50 \text{ C} $, and we don't want the asteroid to approach the central component at a distance any closer than$ 30 \text{ m} $(this is half the width of the ship, and the distance from the central component to the wall). For simplicity of calculation, we'll orient our coordinates so that the central component is at the origin, and the asteroid lies on the$ x$-axis. We'll keep the initial and final positions of the asteroid as variables:
 \begin{align*}
 \Delta U &= \frac{1}{4\pi\epsilon_0}\frac{q_{ast}q_{comp}}{x_f} - \frac{1}{4\pi\epsilon_0}\frac{q_{ast}q_{comp}}{x_i} \\
@@ Line 81: / Line 78: @@
  $q_{comp}(50 \text{ C}-q_{comp}) = 50 \text{ C} \cdot q_{comp} - q_{comp}^2 > 530 \text{ C}^2$
-A simple guess of  $q_{comp}=q_{ast}=25\text{ C}$  yields  $q_{comp}q_{ast} = 625 \text{ C}^2 > 530 \text{ C}^2$ , which is enough to save the ship from cosmetic damage. To still save the ship while charging the central component minimally, one simply needs to solve the quadratic equation based on the inequality above:  $50 \text{ C} \cdot q_{comp} - q_{comp}^2 = 530 \text{ C}^2$ . An application of the quadratic equation or a quick query to Wolfram-Alpha gives a minimum charge of  $q_{comp}\approx 15 \text{ C}$ , which of course means  $35\text{ C}$  remains on the asteroid. Notice that if we transfer all the charge from the asteroid to the central component,  $q_{comp}q_{ast}=0$ , since  $q_{ast}=0$ . If we do this, the asteroid will collide with the ship! It's worth convincing yourself that this result makes sense. \\
+To still save the ship while charging the central component minimally, one simply needs to solve the quadratic equation based on the inequality above:  $50 \text{ C} \cdot q_{comp} - q_{comp}^2 = 530 \text{ C}^2$ . An application of the quadratic formula or a quick query to Wolfram-Alpha gives a minimum charge of  $q_{comp}\approx 15 \text{ C}$ , which means  $35\text{ C}$  remains on the asteroid. Notice that if we transfer all the charge from the asteroid to the central component,  $q_{comp}q_{ast}=0$ , since  $q_{ast}=0$ . If we do this, the asteroid will collide with the ship! It's worth convincing yourself that this result makes sense. \\
 <WRAP TIP>