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Advanced Engineering Problem: Asteroid on a collision course

  1. Oct 9, 2015 #1
    1. The problem statement, all variables and given/known data

    A start to solving this question would be much appreciated.

    Question: An asteroid from the Kuiper belt (approximately 30 AU from the Sun) is knocked out of its circular orbit and begins to fall straight inwards towards the Sun. We want to know how long it will take to get here and how fast it will be moving when it does.

    Using the conservation of energy we have:

    E =1/2(mv^2) - (GMm/r)

    where E is the total energy of the system, m is the mass of the asteroid v is its velocity, G is Newton's constant, M is the mass of the Sun, and r is the distance from the asteroid to the Sun. As the asteroid begins its journey very far away (compared to where it ends up) we may approximate its initial potential energy by zero. Also, assuming (rather unrealistically) that the asteroid starts falling from rest, we may take its initial kinetic energy to be zero as well, so that E = 0

    a) Write down the differential equation for r(t) giving the position of the asteroid as a function of time. Pay close attention to signs.

    b) Solve the differential equation for t(r), the time as a function of position.

    c) Plug in the numbers to find out how long it will take the asteroid to reach Earth orbit. Take G = 6.67 x 10^-11 (Nm^2) / (kg^2) and M = 2 x 10^30 kg.

    d) How fast will it be moving when it crosses Earth's orbit (in km/s)?

    e) Assuming the asteroid has a mass m = 9 x 10^20 kg (approximately equal to that of Ceres), compute the kinetic energy released if it were to collide with the Earth. Remark: For comparison the energy released by the asteroid that wiped out the dinosaurs was estimated to be around 4 x 10^23 J.

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    3. The attempt at a solution


    My attempt to question #1:

    E = 0

    r = (v^2) / 2GM (conservation of energy equation)

    r = r1 ((v+v1) / 2) t (constant acceleration)

    r = ((v^2) / GM(v+v1) (1./t))

    dr/dt = ((v^2) / GM(v+v1) (1/t^2))

    Am I on the right track?



    My attempt to question #2:

    dr/dt = ((v^2) / GM(v+v1) (1/t^2))

    by separable variables

    (t^-2)dt = (GM (v2 + v) / v^2)dr

    by integration I get

    * = (GM (v2 + v) / v^2) r

    t = cube root ((v^2/(GM(v2+v)r) +c)

    Am I still on the right track?




    My attempt to question #3:

    t = 0

    I know I'm lost now.
     
  2. jcsd
  3. Oct 9, 2015 #2
    [itex] v = \dot{r} [/itex]. The energy conservation equation itself is a differential equation for [itex] r [/itex].
     
  4. Oct 9, 2015 #3
    Am I on the right track now?

    t - t0 = ∫ (1 / sqrt((2/m)(E-v(x))) dx
     
  5. Oct 9, 2015 #4
    From
    [tex] E =\dfrac{1}{2} m v^2 - G\dfrac{Mm}{r} = 0, [/tex]
    the velocity is given by
    [tex] v = \sqrt{\dfrac{2GM}{r}} = \dfrac{dr}{dt}. [/tex]

    Now, just integrate it with suitable initial conditions, e.g. when [itex] t=0 [/itex], the radius [itex] r = R (50 \text{AU}) [/itex] .
     
  6. Oct 9, 2015 #5

    SteamKing

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    where does R = 50 AU come from? The OP mentions 30 AU.
     
  7. Oct 9, 2015 #6
    You are correct. It's terrible mistake. :wideeyed:
     
  8. Oct 9, 2015 #7
    I'm getting t = 211 years until the asteroid strikes the Earth. With initial conditions, R = 31 AU; t0 = 0

    Is this correct?
     
  9. Oct 9, 2015 #8

    SteamKing

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    Why did you take R = 31 AU? The OP said the Kuiper Belt is approximately 30 AU from the sun.
     
  10. Oct 9, 2015 #9
    I added an additional AU because the Earth is an AU from the Sun.
     
  11. Oct 9, 2015 #10

    SteamKing

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    The OP specifically stated the Kuiper Belt was approximately 30 AU from the sun, not from the earth.
     
  12. Oct 9, 2015 #11
    Thank you, SteamKing. I guess I was interested in the impact of Earth. I'll recalculate.
     
  13. Oct 9, 2015 #12
    If I've calculated correctly my t = 2430s or 40.5 min; and the velocity of the asteroid would be 1512 mph.
     
  14. Oct 9, 2015 #13

    SteamKing

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    Is this the time it takes the asteroid to fall all the way from the Kuiper Belt to the earth? What happened to 211 years?
     
  15. Oct 9, 2015 #14
    Yes. The time from the Kuiper Belt to Earth's orbit. I'm still asking myself the same question about 211 years. Twice I've calculated r = 30AU and I determined t = 2430 s. I'll crunch the numbers again and see if I can derive a t of approx. 211 years.
     
  16. Oct 9, 2015 #15
    I have to get ready for class. I'll recalculate several times for t. My first answer for t is now 65.9 years.
     
  17. Oct 9, 2015 #16

    SteamKing

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    2430 sec. is only about 40 minutes. If an object took that little time to travel 30 AU, it would be zipping by earth at considerably more than 1500 mph.

    Remember, it takes light about 8 minutes to travel 1 AU from the sun to the earth.
     
  18. Oct 9, 2015 #17
    Thank you, Daeho Ro and Steamking. My final answer for t is 66.1 years, v is 2.2 km/s, K = 2.18x10^21 J.

    Mark
     
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