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Homework Help: Finding speed from acceleration through a potential difference

  1. Oct 2, 2006 #1
    I need help finding the speed of a proton in terms of c (speed of light) when all I have is an acceleration through a potential difference of 77 MV.

    I started by saying this was 77,000,000 Volts, and then using PE = qV to solve for energy. I used E = mc^2 to find mass and used this and rest mass of a proton to find gamma by using m = mo(gamma). I found gamma to be less than one. How is this possible? What did I do wrong? Any help is greatly appreciated.
     
  2. jcsd
  3. Oct 2, 2006 #2

    Hootenanny

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    One should use the expression for relativistic kinetic energy;

    [tex]E_{k}=m_{0}c^2(\gamma - 1)[/tex]
     
  4. Oct 2, 2006 #3
    but I need to find V, velocity.
     
  5. Oct 2, 2006 #4
    Oh wait, I see what you're saying, I will try this and post again if I have not done it correctly again. Thank you
     
  6. Oct 2, 2006 #5
    thanks very much, I got the question right. I have one more question if you are still around.
     
  7. Oct 2, 2006 #6

    Hootenanny

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    No problem, I'm still here.
     
  8. Oct 2, 2006 #7
    Great, I was going to post a new topic but I didn't want to flood the board too much.

    What is the Schwarzschild radius for a typical galaxy (with mass on the order of 10^41 kg)?

    I do not understand the "with mass on the order of 10^41 kg"

    I tried plugging (1e41) with the Schwarzchild radius formula but I got the wrong answer when I entered it into the test. I have one more response left so I need to get it right, thanks so much for the help so far, I'd be grateful if you could work with me on this one.
     
  9. Oct 2, 2006 #8

    Hootenanny

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    On the order of means exactly what you did. Off the order usually means powers of ten. If this is a webassign homework you probably know already, but you need to be very careful with the way you enter your answers (significant figures, units metres of kilometres etc). I've just done a quick calc myself and arrived at a radius of about 1.48x1014m.
     
  10. Oct 2, 2006 #9
    Ah! that must be why! I got the same exact answer, but they want it in km. I did not realize the Schwarzchild radius formula is worked to give answers in meters. So, since my webassign quiz wants it in km, it should be, 1.48x10^11 km, correct?
     
  11. Oct 2, 2006 #10

    Hootenanny

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    If your unsure check your units, assuming you entered everything into the equation in SI units;

    [tex]R = \frac{2GM}{c^{2}} =\frac{(m^{3}kg^{-1}s^{-2})(kg)}{(ms^{-1})^2} = m[/tex]

    Assuming webassign wants it to 3sf then yes, your answer should be correct.
     
  12. Oct 2, 2006 #11
    got it right! 100% score wahoo

    thanks a lot Hootenanny, thanks verryyy much for your time.
     
  13. Oct 2, 2006 #12

    Hootenanny

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    Well done mate. :approve:
    My pleasure
     
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