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Kinetic Energy of Satellite's Orbiting Earth

  1. Apr 4, 2007 #1
    One way to attack a satellite in Earth's orbit is to launch a swarm of pellets in the same orbit as the satellite but in the opposite direction. Suppose a satellite in a circular orbit 500 km above Earth's surface collides with a pellet having mass 4.0 g.

    a.) What is the kinetic energy of the pellet in the reference frame of the satellite

    b.) What is the ratio of this kinetic energy to the kinetic energy of a 4.0 g bullet from a modern army rifle with a muzzle velocity of 950 m/s?

    So for the kinetic energy, K = (1/2)m(v^2). I know that you need to find the velocity for that orbit's radius, therefore, I've done: Net Force = ma --> ((Gm1m2)/(r^2)) = m2((v^2)/r). With r = Radius of the Earth + 500 km --> ((6.37 x 10^6) + (5 x 10^6)). From there, I solved for the velocity and came up with (numerous times) = 7620 m/s. Then the kinetic energy, with 'm' being .004 kg, =116129 J. Which is the wrong answer.

    So...moving on to part B with the wrong answer in part A didn't make much sense. I just don't quite know what I've done wrong. I know what the right answers are ( a. (4.6 x 10^5) J, b. 260. ), I just can't seem to get to them.

    Any suggestions?
  2. jcsd
  3. Apr 4, 2007 #2


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    a) The pellet also has orbital velocity, which is equal in magnitude to the satellite's, but in the opposite direction. You need to account for this velocity too.

    b) It depends on what you're shooting at. If your target is still, then the velocity is 950 m/s. If your target is moving at you, this must be accounted for. I doubt they're talking about shooting the satellite. The rifle will never reach it.
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