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An asteroid flying past the moon

  1. Mar 27, 2013 #1
    I am not that great in physics, but I put a lot of effort in trying to learn it. I would really appreciate any help on this problem, since I am not even sure where to start.

    An asteroid is about to fly past the moon (about 384,000km away) and is on a collision course with the earth. The mass of the asteroid is about 1.45e20 kg. It approaches earth on a direct impact course with a velocity of about 2.0m/s. You are part of an elite team of experts assigned to find a way to deflect the path of the asteroid. Note: Mass of the earth is about 6e24 kg, and the radius is 6400km.
    1) Ignoring the effects of air-resistance, if the total kinetic energy of the asteroid was released in an explosion when it impacts the surface of the earth, what is the TNT equivalent of that energy? 1 ton TNT = 4.184e9 J. 1 megaton TNT = 1Mt = 1e6 t of TNT = 4.184e15 J.
    2) If all of this energy can be used to power 7W LED light bulbs for a year, how many such bulbs can you power in total?
    3) Your team devised a plan to insert and detonate a nuclear device in the center of the asteroid. Suppose this was done in a timely fashion, so that the asteroid has just past the moon. Your team had calculated that about 60% of the energy will go into splitting the asteroid into two halves of the same mass and that the two halves will begin to move apart along a direction perpendicular to the initial velocity vector with the hope that the two parts will miss the earth. What is the total energy needed in the detonation? How many TNT equivalents do we need? If a typical nuclear bomb yields 50Mt of energy, how many will you need? (For this part, ignore the effects of gravity.)

    I guess I need to use kinematics and gravitation equations.

    Any help will be appreciated.
    Last edited: Mar 27, 2013
  2. jcsd
  3. Mar 27, 2013 #2
    Hello *Jessica*

    You could start with calculating the mechanical energy possessed by the asteroid. What would be the total energy it possesses given its position relative to earth and its speed?

    By the way I need to clarify this; is the speed 2.0 m/s when it just passes the moon or when it is just about to strike the Earth?
  4. Mar 27, 2013 #3
    I find energy using this equation, right?

    E=1/2 mv^2 + (-(Gm_1m_2)/r^2)

    I am not sure about the velocity. This is the exact wording.
  5. Mar 27, 2013 #4
    Check your potential energy expression.

    I guess you'll need to assume that speed is near the moon and see whether we get the prescribed answer. If you don't then try the other assumption.
  6. Mar 27, 2013 #5
    Oh yes! The radius is not squared in the potential energy equation. So I get E= -9.067e27 J. How do I find now the kinetic energy it released?
  7. Mar 27, 2013 #6
    Yes that is one correction. And I hope that you have taken the difference in the potential energy (near the moon and near the ground) instead of only the potential energy at the position.

    Well, once you get the total energy, use law of conservation of it. What would be the total kinetic energy near the ground?
  8. Mar 27, 2013 #7
    Uhm no, I didn't take the difference. So my potential energy near the Earth will have the radius of the Earth in the denominator and the potential energy near the moon will have 384e6 m (or this number plus the radius of the Earth)?
    The conservation of energy equation is K1 + U1=K2 + U2. Why did I have to take the difference in the potential energy though?
  9. Mar 27, 2013 #8
    Well, if you want to find K2, you need to add K1 and U1-U2. Hence you would require the difference in the potential energy.
  10. Mar 27, 2013 #9
    That means that the mechanical energy equals to the kinetic energy, since the equations are the same. Are the equations for potential energy correct?
  11. Mar 27, 2013 #10
    Or do these potential energies cancel out each other and all we are left with is 1/2(mv^2)?
  12. Mar 27, 2013 #11
    Find U1-U2 and see for yourself.

    So then,what would be the final kinetic energy?
  13. Mar 28, 2013 #12
    1/2(mv^2). Thank you!
    Can you help with part 3, please?
  14. Mar 28, 2013 #13
    Hey could you show your working to obtain the final KE to just be sure?

    As for the third part, Since they have told us to ignore gravitational effects of the earth, Just imagine what the trajectories of the two pieces look like. Have you got the picture?
  15. Mar 28, 2013 #14
    Since we need to find the kinetic energy when the asteroid hits the Earth, the potential energy is 0, leaving us with only the equation for kinetic energy.

    The two pieces will just move in the opposite directions. Do we use the conservation of momentum here?
  16. Mar 28, 2013 #15
    No, the potential energy at the surface is not zero. Just calculate U1-U2. It is not zero. (think about it, according to you argument, an asteroid falling to Earth won't speed up)

    Draw the trajectory of the pieces to get an idea.
  17. Mar 28, 2013 #16
    K2=0.5(1.45e20)(2.0)^2 + (-(G(1.45e20)(6e24))/6.4e6)-(G(1.45e20)(6e24))/3.84e8)
  18. Mar 28, 2013 #17
    Shouldn't the change in potential energy be Umoon-Uground?
  19. Mar 28, 2013 #18
    Yes, you are right. I was just super tired, now I see it.
    And for part 2 I simply multiply 7W by the number of seconds in a year and then divide the number of joules by that number.
  20. Mar 28, 2013 #19
    The answer to part 2 seems correct
    Hope you understood what to do with part 3
  21. Mar 28, 2013 #20
    Not really.. I don't understand the 60% part. It's 60% of what energy? The kinetic one I calculate in part 1?
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