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Final velocity of an asteroid

  1. Feb 15, 2009 #1
    1. The problem statement, all variables and given/known data

    A NASA satellite has just observed an asteroid that is on a collision course with the Earth. The asteroid has an estimated mass, based on its size, of 5 × 10^9 kg. It is approaching the Earth on a head-on course with a velocity of 611 m/s relative to the Earth and is now 5.0 × 10^6 km away. With what speed will it hit the Earth's surface, neglecting friction with the atmosphere?


    2. Relevant equations

    .5(m*v1^2) + mgh = .5(m*v2^2) + mgh

    Fg =G(m*me/re^2) where me = mass of earth and re = radius of earth or in this case distance of the object. I believe the standard excepted value for mass of the earth is 5.89*10^24 kg. *edit* and the excepted value for G = 6.67*10^-11 *edit*

    3. The attempt at a solution

    First off I've converted the distance into meters.
    The approach I've been trying is to find the kinetic energy at a speed of 611 m/s (incredibly easy) and find the potential energy at a distance of 5*10^9 m away. Then add them together and solve for v2 in the first equation. Unfortunately the gravitational pull at this distance is not 9.8 m/s. Because of this I've been trying to use the second equation to determine the gravitational pull at that distance and then substitute it into the first equation as g. But try as I might I can not get the right freaking answer. Any help would be greatly appreciated.
     
  2. jcsd
  3. Feb 15, 2009 #2

    LowlyPion

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  4. Feb 15, 2009 #3
    Unfortunately that gives me a final velocity that's smaller than the initial, which is definitely not correct. Then again I can't rule out the possibility of me doing something incredibly stupid XD.
     
  5. Feb 15, 2009 #4

    LowlyPion

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    Check your signs. That (-) is there for a reason.
     
  6. Feb 15, 2009 #5
    If i'm not mistaken the (-) sign is only relevant if I have the potential energy on the opposite side from the initial velocity. in my case I do it as follows.

    (.5*(5*10^9)*611^2) + ((6.67*10^-11))*(5*10^24)*(5*10^9))/(5*10^9) = (.5*(5*10^9)v2^2) + 0

    (9.333025*10^14) + (3.335*10^14) = (2.5*10^9)v2^2

    v2 = 711.8433817 m/s

    And estimating off an answer to the same problem with a slightly different initial velocity the answer should be somewhere in the 900 to 1000 range
     
    Last edited: Feb 15, 2009
  7. Feb 15, 2009 #6

    LowlyPion

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    Your answer looks ok.

    Discarding m from all the terms, and using the Standard gravitational parameter μ as 4*105 (in km)

    Vi2 + 2*μ *(1/6.73 - 1/5*106) = Vf2

    6112 + 126,000 = Vf2 = 500,000

    Vf = 707
     
  8. Feb 15, 2009 #7
    Well unfortunately the assignment is past due already, but on the plus side I know the correct answer. apparently its 11200 m/s. not sure how but that's what it says in the answer key.
     
  9. Feb 15, 2009 #8

    D H

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    No, it doesn't.

    The Earth's standard gravitational parameter is 398,600 km3/sec2.

    The Earth's radius is 6738 kilometers, not 6.73 kilometers.

    The initial velocity is 611 m/sec, not 611 km/sec.

    Which is completely wrong.


    That minus sign is very important.

    This is wrong for several reasons. First off, there is the sign error. Secondly, the asteroid will hit the Earth (6378 kilometers). A lesser error: The Earth's mass is 5.97×1024 kg.

    t is a good idea to always carry the units with the numbers. This is part of what got Lowly Pion in trouble.

    The total energy of the asteroid is

    [tex]E_{tot} = KE + PE = \frac 1 2 m_a v^2 - \frac {GM_e m_a}{r}[/tex]

    This is a conserved quantity, so this expression describes the asteroids total energy initially and at the moment it hits the Earth.
     
  10. Feb 15, 2009 #9

    D H

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    That is the correct answer.
     
  11. Feb 15, 2009 #10

    LowlyPion

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    Oops. Right you are D H. Thanks.

    I scratched it out too quickly and dropped the units - twice.

    Sorry for any confusion. My haste made waste.
     
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