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Speed of Impact of meteoroid

  1. Apr 15, 2014 #1
    1. The problem statement, all variables and given/known data

    Q: "Two meteoroids are heading for earth. Their speeds as they cross the moons orbit are 1km/s.
    The first meteoroid is heading straight for earth, what is it's speed of impact?"

    Velocity = 1km/s
    Rearth = 6.37E6 m
    Dmoon = 3.84E8 m
    G = 6.673E-11 N*m^2/kg^2


    2. Relevant equations
    E = 1/2mv[itex]^{2}[/itex] - √G*Mearth*m/r


    3. The attempt at a solution
    PE = (G * Mearth * m)/Dmoon
    KE = -PE
    1/2mv[itex]^{2}[/itex] = -PE

    From there can cancel the masses (lower-case m) of the meteoroid and solve for v. I don't feel like that is the right track. Shouldn't it be as the PE goes to 0? Any help would be appreciated. Thanks!
     
  2. jcsd
  3. Apr 15, 2014 #2

    gneill

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    Staff: Mentor

    Hi MEwill, Welcome to Physics Forums.

    I'm not sure where you got the square root of G in the energy formula in your Relevant Equations, you might want to check that.

    The standard zero reference for gravitational potential energy is taken to be at infinite distance, so PE will not be going to zero as the meteoroid approaches the planet's surface.

    Instead, determine the total energy E at the given location (just crossing the Moon's orbit) and assume that that energy is conserved, so it's the same when it reaches the planet's surface.
     
  4. Apr 15, 2014 #3
    The square root was a typo! I wasn't using it in my calculations. Figured it out I'm pretty sure. I wasn't adding the PE's. It should be the sum of the potential energy when crossing the moons orbit and the potential energy just before is reaches the earth's surface (using radius of the earth) then set that = to the KE and solve for v? Just to make sure my logic is on the right track.
     
  5. Apr 15, 2014 #4

    gneill

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    Staff: Mentor

    Not sure what you mean by "adding the PE's". You want to employ a conservation law, which in this case is conservation of mechanical energy (KE + PE = constant).

    When the body in orbit is of insignificant mass with respect to the primary (here the Earth is the primary massive body) then you can work with what's called "specific" quantities. That's energy quantities given in "per unit mass" amounts. Essentially that means you just disregard the mass of the projectile when writing your conservation of energy equation. This is fine, since its mass always cancels out in the process of solving for the velocity anyways :smile:

    In this problem you find what's known as the "specific mechanical energy" of the body and use the fact that this quantity is conserved:
    $$\xi = \frac{v^2}{2} - \frac{\mu}{r}$$
    where ##\mu = G M##, and ##M## being the mass of the Earth.

    ##\xi## is conserved along the trajectory of the projectile (meteoroid). So write the expression for ##\xi## for the two locations of interest and set them equal...
     
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