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Velocity of an asteroid from an infinite distance

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  1. May 8, 2015 #1
    1. The problem statement, all variables and given/known data

    An asteroid which strikes the earth starts from rest a very large distance r from the earth (say r = ∞). What will its speed be when it hits the earth? (Use M = 6.0×1024 kg, r= 6.4×106 m, G = 6.7×10-11 m3 kg-1 s-2)

    2. Relevant equations

    U=-GmM/r
    F=GmM/r^2

    3. The attempt at a solution

    At r=∞ the potential energy is 0.

    I replaced F with m*a to get this:

    m*a=GmM/r^2

    Dividing out an m and plugging in the numbers gets me 9.81 as the acceleration (which makes sense as that is the accepted value of acceleration for Earth). I don't know how I am supposed to find the velocity after this. The answer is 1.1*10^4 m/s. (I know the answer because it is a back-test, just need guidance on how it is done).

    Thanks!
     
  2. jcsd
  3. May 8, 2015 #2

    PeroK

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    Can you think of what remains constant during the asteroid's trajectory towards the Earth?
     
  4. May 8, 2015 #3
    Weight of both Earth/Asteroid, acceleration from Earth's gravity on asteroid. I don't know of anything else that may pertain to the problem though. Direction maybe?
     
  5. May 8, 2015 #4

    PeroK

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    I'm sorry to say, you're not correct on most of those. The Earth and the asteroid have constant mass - but the Earth has no weight as such. And, the acceleration of the asteroid is most certainly not constant.

    Regarding what else is relevant. Think about the physical quantities you know about.
     
  6. May 8, 2015 #5
    The physical quantities I know are what is given in the problem statement. Mass of earth, radius of earth, the gravitational constant, and the distance between Earth/asteroid (infinity). I know from the equation of potential energy that when it is at infinity the potential energy is 0. My natural thought on this is that if its infinite distance away there would be no attraction between the earth and asteroid meaning the asteroid wouldn't even go towards the earth. Given that there is in fact an answer means I am wrong. I don't think you can use kinematic equations on this either. I really just don't know where else to go.
     
  7. May 8, 2015 #6

    PeroK

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    Assume the asteroid starts to move towards the Earth. There's no such thing as infinity, so it's just a long way away. The clue is in your own post.
     
  8. May 8, 2015 #7
    Oh, a thing I just thought of is conservation of energy, but given that initial potential energy is 0 equating that to .5mv^2 would do me no good as v would be 0 to make that work.
     
  9. May 8, 2015 #8

    PeroK

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    You're on the right track now. Conservation of energy is one of the most important and useful laws of physics. Never forget it!

    Now, think carefully about what happens to the potential energy as the asteroid gets closer to Earth.
     
  10. May 8, 2015 #9
    Potential energy would decrease as it got closer, but if the starting potential energy is 0 does that make the potential energy become negative?
     
  11. May 8, 2015 #10

    PeroK

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    Absolutely correct! If you take the PE to be 0 at "infinity", then it's negative everywhere else.
     
  12. May 8, 2015 #11
    Alright, so eventually it would have to hit earth which is why it gives me the earth's radius as the stopping distance. So..

    .5mv^2=-GmM/r
    m's cancel
    .5v^2=-(6.7*10^-11)*6*10^24/(6.4*10^6)
    v^2=-125,625,000

    My only hiccup now is you can't take the square root of a negative number? Would it be negative of a negative or is the negative in the original equation already accounting for what we just arrived at as all potential energy being negative as the closer it gets to earth? Ignoring the negative gives me the correct answer so thank you very much on that part!
     
  13. May 8, 2015 #12

    PeroK

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    Now, you need to think more carefully about conservation of energy (KE + PE) in this case. What is total the (constant) total energy of the asteroid?
     
  14. May 8, 2015 #13
    Hmm, would it be 0? Since PE gets negative as KE becomes positive when we start moving towards Earth. Initially it wasn't moving and PE was 0 so 0+0=0
     
  15. May 8, 2015 #14

    PeroK

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    Yes. The total energy is 0.

    Note: That also shows that energy is arbitrary: the total energy can be anything you like. It's actually the changes in energy that are absolute.
     
  16. May 8, 2015 #15
    So KE+PE=0

    .5mv^2+(-GmM/r)=0
    .5mv^2=GmM/r

    Which would get rid of my negative allowing me to find the proper answer. Is that line of thought correct?
     
  17. May 8, 2015 #16

    PeroK

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    Yes.
     
  18. May 8, 2015 #17
    Awesome!! Thank you so much for your help. It was greatly appreciated.

    Have a great day!
     
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