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Homework Help: Nudging an Asteroid. Conservation of Momentum

  1. Feb 24, 2009 #1
    1. The problem statement, all variables and given/known data


    2. Relevant equations

    1.) m1*v1 + m2*v2 = m1'*v1' + m2'*v2'

    Where the ' denotes the velocity/mass after the event/whatever.

    2.) 0.5m1(v1)^2 + 0.5m2(v2)^2 = 0.5m1'(v1')^2 + 0.5m1'(v1')^2

    3. The attempt at a solution

    I know its a conservation of momentum style of problem, but my professor does an awful job of teaching IMO.

    Give me a nudge in the right direction, and I'll see what I can do from there.
    Last edited: Feb 24, 2009
  2. jcsd
  3. Feb 24, 2009 #2


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    I would think about it in terms of energy.

    Apophis has kinetic energy. You would attach the rocket gently and fire the rocket to change its kinetic energy and hence its velocity.

    Looks like they are asking how much the ΔKE of Apophis can be changed.
  4. Feb 24, 2009 #3
    Ok, so 1 will denote the asteroid and 2 the rocket.
    m1= 2*10^10kg
    v2= 2,500m/s

    These would be the initial values, correct?

    For the post whatever values, I'm not sure what to use. Should I assume that the rocket burns through all its fuel and has no mass?

    I have to variables if I do it this way though. The asteroids initial velocity, and its velocity after the rocket.

    edit: Browsing through my notes I see that I have v'=(v+[tex]\Delta[/tex]v)) and m'=(m+[tex]\Delta[/tex]m)

    I'm guessing I should use this. But how would I find the derivative? should I sub this various stuff in and solve for [tex]\Delta[/tex]v and m and take the derivative?
    Last edited: Feb 24, 2009
  5. Feb 24, 2009 #4


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    You're right. I see no initial velocity, so you have to work a little harder.

    What Δ(mv) can you achieve in the asteroid with the rocket burn?

    Won't that be the mass and velocity of the solid fuel?

    That will translate I think into a Δv of the asteroid.

    Now then consider how long until Δv * t > 10,000 km
  6. Feb 24, 2009 #5
    I'm confused about what you're saying will translate into the change in mass and the change in velocity of the asteroid.

    Burning the rocket fuel is how you change it. What equation do you use to go from the mass and velocity of the rocket fuel, to delta V of the asteroid?
  7. Feb 24, 2009 #6


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    20*103 << 2*1010
  8. Feb 24, 2009 #7
    Those are the two different mass values.

    What do you mean by "< <"
  9. Feb 24, 2009 #8


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    << means much less than

    < means less than

    What I meant was that you can ignore the burning time because you can consider that the burning of the fuel would all go to slowing the mass of the asteroid, even though some of that mass during the burn will also be the fuel on board.

    Hence consider that all the burn was a change in momentum of the asteroid alone.
  10. Feb 24, 2009 #9
    So... multiplying the burn speed by the mass of Apophis results in 5*107

    What good is this number?

    edit: Do I set the initial side equal to this value to find the initial velocity of the asteroid?
  11. Feb 24, 2009 #10


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    What good is the number?

    Plenty good.

    In terms of the mass of Apophis what is the Δv?

    Using that Δv how long will it take that Δv to make the position at earth greater than 10,000 km.
  12. Feb 24, 2009 #11
    Wait, so is 50million the answer to question A?

    I feel like I'm so close to understanding, but I have no idea how to get the change in velocity in terms of the mass.
  13. Feb 24, 2009 #12
    In my notes for this class, the only related thing I have is a rocket burning fuel.

    m(dV/dt)= -vex(dm/dt)

    where vex is the exhaust velocity. The problem i have with this though, is that the mass does change (rocket burning fuel), but you said that was negligible.

    Then after that step, after deriving some stuff.

    (v-v0) = vex(ln(m0/m)

    but... I don't know how to apply any of this.
  14. Feb 24, 2009 #13


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    If you have a certain P = mv for the a-roid and you now have a -Δ(mv) that means then that your Δv is the Δ(mv)/m

    Take your number 5*107 / 2*1010 = 2.5*10-3 m/s

    How many seconds then will it take to change the Δx when it reaches the vicinity of earth?
  15. Feb 24, 2009 #14
    So a.) is 5*10^7, b.) is 2.5*10^-3

    and c.) is the amount of time to get 10,000 km

    well if the velocity is 2.5*10^-3

    10,000 km = 10,000,000 m

    10 million divided by the velocity is the amount of time to move that far.

    So 10million/2.5*10^-3= 4*10^9 seconds.

    Am I correct with which answers go where? And I have to give a good explanation on how I did the work, so my grade isn't solely based on what my answers are.

    Thank you for your help, again. I'm definitely consider changing my major from engineering. This stuff is ridiculous.
  16. Feb 24, 2009 #15
    I have the same problem. I also got the time to be 4 x 10^9 seconds. This comes out to be 126.84 years. Could this be correct? That would mean our time is already up if the asteroid actually did head to us. It just feels like somewhere something is done wrong.
  17. Feb 24, 2009 #16
    Or it just means we don't have the technology yet to alter the course of something so massive.
  18. Feb 24, 2009 #17
    I found an error. The mass of the asteroid times rocket velocity is 5*10^13, NOT 5*10^7

    5*10^7 is the mass of the rocket times the rocket velocity.

    And why would I divide 5*10^13 by 2*10^10, when 2*10^10 is what I multiplied the velocity by in the first place?

    Or is the rocket mass times velocity what I actually need?
  19. Feb 24, 2009 #18


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    Wait a minute. The rocket fuel - 20K kg of it is what is accelerated to 2500 m/s. The asteroid is a dumb piece of cold iron getting pushed against.

    It's the 20K kg * 2500 m/s that constitutes the change.
  20. Feb 24, 2009 #19
    Ok cool. I said something different earlier, but you went with the correct numbers.
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