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Calculate the speed of the rock as it enters Earth's atmosphere

  1. Nov 10, 2012 #1
    1. The problem statement, all variables and given/known data
    In Robert Heinlein's The Moon is a Harsh Mistress, the colonial inhabitants of the moon threaten to launch rocks down onto Earth if they are not given independence (or at least representation). Assuming that a gun could launch a rock of mass m at twice the lunar escape speed, calculate the speed of the rock as it enters Earth's atmosphere


    2. Relevant equations
    KE = 0.5mv^2
    PE = -G M1 M2 / R

    3. The attempt at a solution
    I found the lunar escape speed to be 2374 meters per second. Beyond this, I can't get any farther. I've tried using the lunar escape speed and initial velocity—4749 m/s—to use conservation of gravitational potential energy, but I'm not getting the correct answer, which is 11.9 km/s. Any help would be appreciated!
     
  2. jcsd
  3. Nov 10, 2012 #2
    Hello,
    what is conserved? Is the gravitational potential energy conserved?
     
  4. Nov 10, 2012 #3
    By the way did you take the Earth's gravitational potential into account?
     
  5. Nov 10, 2012 #4
    Yes. What I tried to do was say that energy is conserved (it's asking for an approximate solution) and find that, if the exit velocity is twice the escape speed, then the final velocity should be the escape speed. Using that, and the distance from the moon to the earth, I tried to use the formula G m M /r. However,I'm not getting the correct answer.
    I initially thought my distances were off, but I've tried changing them, and still not working.
     
  6. Nov 10, 2012 #5
    I don't see what you mean by final velocity. If it's the velocity at infinite distance, the velocity is squared in the kinetic energy. If you double the escape velocity you would get sqrt(3) this velocity.
    How do you use it?
     
  7. Nov 10, 2012 #6
    Perhaps a better question: what is the total energy at the launch?
     
  8. Nov 11, 2012 #7
    At the launch, it would be the kinetic energy 0.5 (Escape velocity*2)^2 * mass of the satellite, plus the gravitational potential energy due to the earth, minus the magnitude of the gravitational energy due to the moon.
     
  9. Nov 13, 2012 #8
    Ah ok, if you pick up a stone you do work and the stone gains energy. If the stone is farther from a massive
    body, its gravitational energy is higher. Minus the magnitude of the gravitational energy = gravitational energy.
    If r increases (-GmM/r) increases too(its absolute value deacreses but it's a negative number, as r increases
    it goes to 0=max gravitational energy)(The energy can be negative). Then you add up all the energies.
     
  10. Nov 13, 2012 #9
    It's "plus the magnitude of the gravitational energy due to the moon" not minus, but both the energy due to the earth and the energy due to the moon are negative... hope this helped.
     
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