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Basic relativity question (probably easy for everyone on here xD)

  1. Nov 26, 2011 #1
    1. The problem statement, all variables and given/known data

    a quasar is moving away from earth with a speed 0f 0.850c. It emits a proton that eventually reaches earth with a speed of 0.519c relative to the earth. How fast is the proton moving relative to the QUASAR.

    3. The attempt at a solution

    u = u' + v / 1 + u'(v) / c^2

    u = 0.578c relative to earth.

    How do I go about getting an answer relative to the object emitting the proton?

    thanks for help. appreciate it
     
  2. jcsd
  3. Nov 27, 2011 #2

    BruceW

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    Your equation is kindof hard to read, since it has several division signs. And are u,u' and v the absolute values of the speeds?

    Also, Is this homework? You should show your own working first, since we're only meant to help with homework.
     
  4. Dec 6, 2011 #3
    A quasar is moving AWAY from the earth with a speed of 0.850C. Itemits a proton that eventually reaches the earth, and is travelling at a speed of 0.519C relative to the EARTH. how fast is the proton moving RELATIVE to the QUASAR?

    my attempt: u = u' +v / (1 + u'v/c^2)
    u = 0.578c RELATIVE to EARTH.

    however, I am looking for the speed RELATIVE to the QUASAR. I do not know how to do that with that given formula!

    any ideaS? thanks
     
  5. Dec 6, 2011 #4

    BruceW

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    The equation is still hard to read, man. Does it mean:
    [tex]u= \frac{u'+v}{1+ \frac{u' v}{c^2}}[/tex]
    And you said that u=0.578c relative to earth, but that doesn't make sense, because you already know the speed of the proton and the quasar relative to earth.
    I think you should start by defining what u, u' and v mean in your equation.
     
  6. Dec 7, 2011 #5
    i don't know what it means? it's on the formula sheet there is an example using it. all i want to know is the speed of the proton relative to the quasar...
    QUASAR<-0.85c-<QUASAR "shoots" PROTON>0.519>PROTON <EARTH>
    the velocities are relative to earth.

    THERE! I DREW IT OUT :) Does that make sense?
     
  7. Dec 7, 2011 #6

    BruceW

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    Yes, I already understood that bit. But I'm trying to help you through the question. Since you have the equation, you just need to work out what the symbols mean. Try to work it out from your formula sheet.
     
  8. Dec 7, 2011 #7

    vela

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    Do you see the contradiction between the two things in red? That's why BruceW said it didn't make sense.

    In any case, how did you manage to come up with 0.578c? I don't get that when I plug in the numbers the way I assume you did.
     
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