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Asteroid colliding with a planet

  1. Mar 14, 2015 #1
    1. The problem statement, all variables and given/known data
    Suppose that an asteroid traveling straight toward the center of the earth were to collide with our planet at the equator and bury itself just below the surface.

    What would have to be the mass of this asteroid, in terms of the earth's mass M , for the day to become 26.0% longer than it presently is as a result of the collision? Assume that the asteroid is very small compared to the earth and that the earth is uniform throughout.

    2. Relevant equations
    Conservation of angular momentum: Linitial = Lfinal

    3. The attempt at a solution
    ωfinal = (0.74)ωinitial (This makes the final speed 26% slower than the original)


    Linitial = Lfinal
    Iinitialωinitial = Ifinalωfinal
    Iinitialωinitial = Ifinal(0.74)ωinitialinitial cancel out)
    (2/5)MR2 = ((2/5)MR2 + mR2)(0.74) (R2's cancel out)
    (2/5)M = ((2/5)M + m) (0.74) (Distribute the 0.74)
    (2/5)M = ((1.48/5)M + 0.74m)
    0.74 m = (2/5)M - (1.48/5)M
    0.74m = M((2/5) - (1.48/5))
    0.74m = 0.104M
    m = 0.141M

    This answer is m = 0.104M which is on the right side of the second to last line (0.74m = 0.104M), but I still have the 0.74 on the left side. Where is my mistake?
     
    Last edited: Mar 14, 2015
  2. jcsd
  3. Mar 14, 2015 #2

    gneill

    User Avatar

    Staff: Mentor

    The rotation period is given by ##\frac{2 \pi}{ω}## for whatever the current value of ω is. For the new period to be 26% larger, then
    $$\frac{2 \pi}{ω_f} = 1.26 \frac{2 \pi}{ω_o} $$
     
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