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Asteroid Impact, Shorter Days

  1. Dec 2, 2014 #1
    1. The problem statement, all variables and given/known data

    Suppose that an asteroid traveling straight toward the centre 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, for the day to become 28.0%
    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


    3. The attempt at a solution

    I have completed this question, but I think my reasoning is rather questionable. It hasn't convinced me, which doesn't give me much hope...

    Moment on Inertia, [itex]I = \frac{2}{5}MR^{2}[/itex]
    Angular Momentum, [itex]L = I \omega[/itex]

    [itex]\omega = 7.27[/itex]x[itex]10^{-5}[/itex]rads[itex]^{-1}[/itex]

    [itex]M_{e} = 5.97[/itex]x[itex]10^{24}[/itex]kg

    [itex]R = 6371000m[/itex]

    [itex]L = 7.05[/itex]x[itex]10^{33}[/itex]

    So that is the angular momentum of the earth before the impact, and I thought perhaps I can keep this constant, and using an angular velocity 28% slower, calculate the extra mass that must be added. The difference between this mass and the initial mass being the mass of the asteroid.

    Through some algebra I arrive at an asteroid that has a mass [itex]M = 0.39M_{E}[/itex] or [itex]2.33[/itex]x[itex]10^{24}[/itex]kg.

    This sounds very big, but it's hard to judge if that's a sensible number.

    Anyway, my assumption of constant angular momentum is where I think a problem lies,

    what do you think?

    Thanks!
     
  2. jcsd
  3. Dec 2, 2014 #2

    SteamKing

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    Consider that the mass of the moon is only about 1.2% of the earth's mass, and I would say that your calculation of the mass of the 'asteroid' lies somewhere between the masses of Mars and Venus.

    I think that if angular momentum is used, assuming I = (2/5) MR2, then both M and R would need some adjustment after the collision, given the quantity of material added to the earth.

    Perhaps you should look at this problem from an energy standpoint. As the earth turns on its axis each day, a certain amount of rotational energy is generated. If the earth slows down its rotation, for whatever reason, the amount of energy generated is reduced. This approach may produce an asteroid of more reasonable size.
     
  4. Dec 2, 2014 #3
    I actually tried to solve this question using rotational kinetic energy before trying angular momentum, and I still get an absurd answer. I will carefully check all of my algebra, but both methods of working rely on this idea that the angular momentum and the rotational kinetic energy remain the same after the collision.

    Is this not problematic?
     
  5. Dec 2, 2014 #4
    I checked my algebra for the kinetic energy solution, and got the same result that I first got; [itex]0.93M_{E}[/itex] :/
     
  6. Dec 2, 2014 #5

    SteamKing

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    I think you mean 0.39 Me.

    Still, I think that with so much material being added to the earth, the mass moment of inertia of the planet undergoes a significant change in its value, and since this quantity is used to calculate both angular momentum and rotational kinetic energy, it must be adjusted somehow to account for the mass added by the asteroid.
     
  7. Dec 2, 2014 #6
    Actually, I don't - I found the answer to be 0.93ME, but 0.39ME in my angular momentum calculation.

    The only relevant shape that I can think of for this modified moment of inertia is treating the earth as a hollow sphere, since so much mass will be added 'just below the surface', but that's simply not true.
     
  8. Dec 2, 2014 #7

    SteamKing

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    I don't think a hollow sphere is the correct shape to use for the MOI calculation after impact. After all, you still have the original mass of the earth spinning away, albeit slower, and its distribution has not changed. You do, however, have a lot of new mass which is either lumped onto the original surface of the earth, so that the new planet has a shape resembling a dumbbell, or the new mass has been spread evenly over the entire surface, increasing R, which in turn increases the MOI.
     
  9. Dec 2, 2014 #8

    NTW

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    By other way, that is, assuming a linear inelastic collision of a body of the Earth's mass at a velocity of 293,7 m/s [at that velocity, its linear KE is the same as the Earth's rotational KE] with a immobile body of a certain mass, and calculating that mass for a reduction of velocity of 28%, I get practically the same results as you: 2,312 * 1024 kg or 38,72% of the Earth's mass... The difference may be in rounding here and there...
     
    Last edited: Dec 2, 2014
  10. Dec 2, 2014 #9

    haruspex

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    That will get you nowhere. There is a huge and unknown loss of mechanical energy as heat. The only way to find it is by first applying conservation of angular momentum.
    Please post that algebra.
    I get less than half your number.
     
  11. Dec 3, 2014 #10
    Ok,

    [itex]I = \frac{2}{5}M_{E}R_{E}^{2}[/itex]

    [itex]L = I \omega[/itex]

    [itex]\omega = \frac{d \theta}{dt} = \frac{2 \pi}{86400}[/itex]

    The angular momentum of the earth prior to collision;

    [itex]L = \frac{2}{5}M_{E}R_{E}^{2} \omega = 7.05[/itex]x[itex]10^{33} Kgm^{2}s^{-1}[/itex]

    A 28% slower day corresponds to [itex]0.72 \omega[/itex]

    <Insert questionable logic of assuming angular momentum remains unchanged after the impact>

    [itex]M_{E + A} =[/itex] mass of the earth + mass of asteroid

    [itex]L = \frac{2}{5}M_{E+A}R_{E}^{2} (0.72 \omega)[/itex]

    [itex]M_{E+A} = \frac{5L}{2 R_{E}^{2} (0.72 \omega)} = 8.29[/itex]x[itex]10^{24}Kg[/itex]

    [itex]M_{A} = M_{E+A} - M_{E} = 2.33[/itex]x[itex]10^{24}[/itex]kg or [itex]0.39 M_{E}[/itex]
     
  12. Dec 3, 2014 #11

    gneill

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    Can you explain why everyone is assuming that the asteroid's mass is going to be spread throughout the volume of the Earth? After all, it seems that you're applying the same moment of inertia formula before and after collision, just assuming a change in mass.

    The problem states that the asteroid buries itself just below the surface. If you assume that it remains in one concentrated location at one Earth radius from the center of rotation, what's the mass moment of inertia associated with it?
     
  13. Dec 3, 2014 #12
    Could I use the moment of inertia for a dumbbell and use the parallel axis theorem to make it rotate about one of the spheres?

    The only other shapes we've really looked at are disks, cylinders, and rods.
     
  14. Dec 3, 2014 #13

    gneill

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    You've got a sphere (the Earth) and what is essentially a point mass (the asteroid) located at the equator. What's the moment of inertia of a point mass rotating at a distance R from the center of rotation?
     
  15. Dec 3, 2014 #14
    [itex]I = MR^{2}[/itex]

    But I also need to take into consideration the moment of inertia of the earth, and I don't know how I combine these quantities.
     
  16. Dec 3, 2014 #15

    gneill

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    Moment of inertia is the rotational analog of mass. They are combined in the same way: just sum them.
     
  17. Dec 3, 2014 #16
    Great, thanks!

    I'm still unsure about my method of keeping angular momentum constant. I can't think of a good reason why this would be the case.
     
  18. Dec 3, 2014 #17

    gneill

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    Think about what comprises the system under consideration. What's the total angular momentum prior to impact? (The nature of the trajectory of the asteroid is important!). Are there any external forces or torques involved or is the system isolated?
     
  19. Dec 3, 2014 #18
    Ah ha!

    The question states that the asteroids trajectory is at the equator and straight towards the center - Therefore the torque is zero and the angular momentum does remain constant.

    Thanks a lot!
     
  20. Dec 3, 2014 #19

    SteamKing

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    I don't think it's that simple.

    For a sizable mass located a distance away from the axis of rotation, the moment of inertia of the earth combined with the asteroid should be calculated by applying the parallel axis theorem.

    For

    Me - mass of the earth
    Ma - mass of the asteroid

    Je - moment of inertia of the earth about its axis of rotation
    Ja - moment of inertia of the asteroid about its centroid
    Je+a - resultant moment of inertia of the earth and asteroid

    Da - distance of the center of the asteroid from the rotation axis of the earth

    Je+a = Je + Ja + Ma * Da2

    The complete analysis would entail calculating the distance of the centroid of the combined earth-asteroid body from the original axis of rotation, and then using that centroid to calculate the moment of inertia.

    Of course, if Ma is small compared to Me, then you can neglect the additional moment due to transfer. However, it appears that Ma is a sizable fraction of Me, so the additional term should be evaluated.
     
  21. Dec 3, 2014 #20

    gneill

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    Work it out the simple way first and see if it's a significant mass. The level of the problem and understanding demonstrated by the OP suggests that a simple approach may be warranted.
     
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