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Armageddon homework

  1. Dec 18, 2009 #1
    Armageddon movie question

    Hello, my teacher gives me a homework/test with the relevant parameters of the movie armageddon. i need some help with the relevant equations and presumptions of that case.

    1. The problem statement, all variables and given/known data

    The problem said that a asteroid with 1,45e^6 meters , travels to the earth to 10 km/s. in this situation, the NASA send a ship to put a nuclear bomb in the asteroid and that this explodes in 2 pieces and turning aside itself next to the Earth. the explotion must be place after that the asteroid goes to the called "zero barrier". four hours after the impact with the earth.

    Determines a expresion for L( distance from the center of the earth to the "zero barrier", in terms of the relevant variables.

    -It does not make consideration with respect to the power of a bomb and its influence in sepracion of two parts of asteiorde
    -despises the attraction of the sun and the moon on the asteroid
    - is relevant the mass of the asteroid? why?
    -density of the rock = 3000kg/m^3.

    2. Relevant equations
    i supossed that the relevant equations are:

    gravitacion its not include>

    3. The attempt at a solution
    I suppose that the momentum its conservated and other suposition is that the asteroid goes with velocity to the earth after the explotion.

    if someone helps me would be fine :)

    Attached Files:

    Last edited: Dec 19, 2009
  2. jcsd
  3. Dec 18, 2009 #2
    With variable aceleracion. because the force of gravity depends of r. the momentum its not conservative. what can i do when the asteroid its divide in two pieces? about the momentum.
  4. Dec 19, 2009 #3
    This is an interesting problem. The solutions to it depend on level of math you are expected to know.

    The very simplest solution would be to ignore the effects of gravity altogether and figure out the initial velocity in the direction perpendicular to d so as to to just graze the earth.

    The second simplest would be maybe have sufficient velocity at a distance R from the earths center so that both pieces would just avoid being captured into orbit, in other words: 0.5 m(V^2)/R = -G Mm/R^2 Note m is 1/2 the mass of the original asteroid. Velocity here would just be the resultant of combining the initial 10m/s and the perpendicular velocity caused by splitting the asteroid.

    A more accurate solution would be to assume the asteroid parts are on a curved course, but this would assume knowledge of much more than you have stated, and maybe not what is expected. The picture below depicts the first two approaches.


    I would try the second approach and see what you get.
    Last edited: Dec 19, 2009
  5. Dec 19, 2009 #4
    thanks for respond.
    can you be more clear in this part: Velocity here would just be the resultant of combining the initial 10m/s and the perpendicular velocity caused by splitting the asteroid.
    because my english its not very good, sorry me.

    and other note:for effects in this case. the asteroid don't have rotation, and mi teacher says me now, that the problem isn't include, terms of Gravitation.

    So the first aproach i will try.
    Last edited: Dec 19, 2009
  6. Dec 19, 2009 #5
    Ok, then you know the horizontal velocity is 10m/s and say other velocity = ?

    Your drawing shows a right triangle where distance to asteroid is d * cos (a) where a is small angle in triangle. We also know tan (a) = ?/10 m/s.

    Can you finish?
  7. Dec 19, 2009 #6


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    josftx: it's 10 km/s and not 10 m/s, right?
  8. Dec 19, 2009 #7
    Yes, right, 10km/s..

    thanks denverdoc
  9. Dec 19, 2009 #8
    D is the distance from the center of the earth to a distance to the asteroid. right?, but de distante to the barrer zero to the asteroid is dsin(a) , not cos(a). or for the distante d you want mean "L" ?.

    The horizontal velocity of the asteroid is 10km/s, but when it explodes, part in two pieces. and i formule this presumpsionts for the velocities:
    Momentum when the asteroid travel:
    [tex]\vec{p}= -P\hat{i}+P\hat{j}[/tex]
    when the asteroid its divide en two half with the same weight is:
    \vec{p_{A}}= -p_{A}\hat{i}+p_{A}\hat{j}[/tex]
    [tex]\vec{p_{B}}= -p_{B}\hat{i}-p_{B}\hat{j}[/tex]

    in fact:
    [tex]\vec{p}= -m_{asteroid}v_{x}\hat{i}+m_{asteroid}v_{y}\hat{j}[/tex]

    and when its separeted:
    \vec{p_{A}}= -\frac{m_{asteroid}}{2}v_{ax}\hat{i}-\frac{m_{asteroid}}{2}v_{ay}\hat{j}[/tex]
    \vec{p_{B}}= -\frac{m_{asteroid}}{2}v_{bx}\hat{i}+\frac{m_{asteroid}}{2}v_{by}\hat{j}[/tex]

    for conservation of momentum in explosion:
    -p\hat{i}= -p_{A}\hat{i} -p_{B}\hat{i}
    p\hat{j}= -p_{A}\hat{j} p_{B}\hat{j}

    2v_{x}= v_{ax}+ v_{bx}
    and for y-axis, component:
    v_{ay}= v_{by}
    Last edited: Dec 19, 2009
  10. Dec 19, 2009 #9
    that all looks good, but really doesn't tell us much about the velocity, Vy needed to avoid contact with earth.

    I also am confused by your wording and I know you are trying--are we to assume that time to impact is 4 hrs? And yes, I meant L, not d.
  11. Dec 19, 2009 #10
    the "barrer zero", is located 4 hours, before the impact. sorry for the confusion.

    that i do, its only that i suppose. and i don't know if i'm going in the right way.
    Last edited: Dec 19, 2009
  12. Dec 19, 2009 #11


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    4 hours to impact? I don't understand. If it's four hours to impact, isn't L simply (10 km/s)(4 hours), minus the radius of the Earth?
  13. Dec 19, 2009 #12
    I don't know either--the question is vague. But we know that impact occurs at the surface of the earth, so d must be:

    4 hr* 3600 sec/hr *10,000 m/s + R which is the radius of the earth.

    We also need to deflect the asteroid by a distance not quite R in the 4 hours . That distance is L sin (theta), and again tan (theta) would be Vy/Vx = Vy/10 Km/s.

    Unless gravity is called in or we are given more information the problem, thats about all I can see to do-- you should solve for Vy. Beyond that, ??
  14. Dec 19, 2009 #13

    yes, is a vague problem from a vague professor. but that is irelevant right now :cry:.

    denverdoc, why you put the Vx like the velocity initial of the asteroid?, because the velocity when the asteroid is broken, is diferent. and with my suppositions y find a expresion for the velocity of Va(one part of the asteroid).
  15. Dec 19, 2009 #14
    You are right. I am assuming something very unlikely--that a nuclear bomb can be made like a shaped charge so that only the vertical velocity is changed. But maybe it is a long and thin asteroid and can be "cut" into two. In one of your posts you suggest that the momentum is conserved. This is true. The parts may split along an axis at that is not perpendicular to L. In one case, the horizontal velocity will be increased, while the other half will be slowed. I haven't seen the movie, but just now read a bit about the best driller on Earth and his crew are the team recruited, so if a nuclear explosion can be shaped, I'm assuming these guys can do it like your diagram shows.
  16. Dec 19, 2009 #15
    Observations in the systems, and suppositions:

    [tex]V_{ay} > V_{ax}[/tex]
    [tex]\frac{V_{ay}}{V_{ax}} > 1[/tex]
    [tex]arctan(\frac{V_{ay}}{V_{ax}}) > arctan(1)[/tex]
    [tex]arctan(\frac{V_{ay}}{V_{ax}}) > \frac{\pi}{4}[/tex]

    [tex] 2V_{x} = V_{a}+V_{b} [/tex] consecuense of conservation of momentum.

    [tex] Lcos\theta = V_{a}t [/tex] where t is the time to impact, because if the asteroid isn't desviate, cos(theta) = 0. L is the surface of the earth.
    [tex] tan\theta = d/V_{a}t[/tex] d= distance near of the earth, when the asteroid have a desviation.

    are good that suppositions or are redudant?
  17. Dec 19, 2009 #16
    The distance to the center of earth is 4*3600*10,000 m/s -4000m. The height about 4000m.

    The angle is so small that all kinds of approximations are true which simplify the solution. In radians, sin (theta) = theta = tan (theta)
  18. Dec 20, 2009 #17
    the distance to the center of earth , from where?
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