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Homework Help: Pulsar Rotation Problem

  1. Aug 31, 2010 #1
    1. The problem statement, all variables and given/known data
    I have this problem for my homework:

    Time standards are now based on atomic clocks. A promising second standard is based on pulsars, which are rotating neutron stars (highly compact stars consisting only of neutrons). Some rotate at a rate that is highly stable, sending out a radio beacon that sweeps briefly across Earth once with each rotation, like a lighthouse beacon. Suppose a pulsar rotates once every 1.610 431 448 872 75 ± 4 ms, where the trailing ± 4 indicates the uncertainty in the last decimal place (it does not mean ± 4 ms).

    How many times does the pulsar rotate in 30.0 days?

    2. Relevant equations
    This should just be a basic conversion, but I'm doing something wrong here.


    3. The attempt at a solution

    I've attempted to convert the rotation period to seconds by dividing by 1000 first (1 ms = 1 x 10^-3 s, yes?). Then I calculate how many seconds are in 28 days:
    (30 x 24 (hours/day) x 60 (minutes/hour) x 60 (seconds/minute)) = 2592000 seconds
    I thought I could just take the number of rotations per second and multiply it by how many seconds are in 30 days.

    0.001610413... x 2592000 = 4174.19 rotations in 28 days.

    I submitted this answer to my online homework and it was marked wrong. It said the answer is 1.61 x 10^9.

    Can anyone please show me what I did wrong here?
     
  2. jcsd
  3. Aug 31, 2010 #2

    vela

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    Look at the units of your answer. You have 1.61x10-3 s/rotation times 2592000 s, so the units come out to be s2/rotation. How should you combine the numbers to get just "rotations" as the units?

    You should also check your answer to make sure it's reasonable. You found about 4000 rotations, and you were given that one rotation takes about 1 ms. That's equivalent to about 4 seconds, which is a far cry from a month. Something clearly went wrong.
     
  4. Aug 31, 2010 #3
    Thanks for replying, Vela. I didn't realize how off track I am. ack.
    So it could work to find the rotations/sec. Then that "sec" would cancel the one from the total seconds in 30 days, yes?
    Just to make sure I'm doing it right: (I realized I was multiplying wrong earlier)
    [and just to clarify: ms is millisecond, right? so .001 seconds?]
    1610.413 rotation/s x 259200 seconds (in 30 days) ... so that would boil it down to just rotations, but I'm still getting a different answer: 4.174 x 10^8.
    :\
    any ideas?
     
  5. Aug 31, 2010 #4

    vela

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    You're not flipping the number correctly. You have

    [tex]\frac{1.61\times10^{-3}~\textrm{s}}{1~\textrm{rotation}}[/tex]

    When you flip it over, you get


    [tex]\frac{1~\textrm{rotation}}{1.61\times10^{-3}~\textrm{s}} = 621~\frac{\textrm{rotations}}{\textrm{s}}[/tex]

    Make sense?
     
  6. Aug 31, 2010 #5
    AHA! Thank you so much! I had a feeling I was messing up something like that.
    Thanks again, Vela! :)
     
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