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Kinetic energy of a meteor

  1. Apr 14, 2013 #1
    How can I calculate kinetic energy of a meteor? On what variables it depends?

    Thanks and sorry for my bad English.
     
  2. jcsd
  3. Apr 14, 2013 #2

    mfb

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    You can use the usual formula for kinetic energy: ##E=\frac{1}{2}mv^2## with the mass m and the velocity v (relative to earth). Typical velocities are 10 to 30 km/s.
     
  4. Apr 14, 2013 #3
    I thought meteors were faster. :smile:
     
  5. Apr 14, 2013 #4

    mfb

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    Well, ~11km/s is the minimal velocity, and corresponds to the escape velocity of earth. ~30km/s is the orbital velocity of earth, but most objects nearby have similar orbits, so their relative velocity is lower. Objects from the outer solar system can have higher relative velocities, up to ~75km/s should be possible (but really rare). To get even more speed, you need objects not bound to the solar system.
     
  6. Apr 14, 2013 #5
    30 kms/second is pretty damn fast. That would get them from space to ground in about 3 seconds, if they came in vertically and were not slowed by air resistance.
     
  7. Apr 14, 2013 #6
    as mbf said 'relative velocity' . comets can have either a prograde or retrograde orbit. so the approach speed of a retrograde is going to be pretty high....
    also taking in to account keplers second(?) law ....the comet will be most likely at its perihelion when approaching the Sun and so will have an increased speed.
     
    Last edited: Apr 14, 2013
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