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More Uniform Circular Motion problems

  1. Jul 7, 2007 #1
    An Earth satellite moves in a circular orbit 605 km above the Earth's surface. The period of the motion is 96.6 min.
    (a) What is the speed of the satellite?




    a=v^2/r and T=2*pi*r/v



    3. The attempt at a solution

    So far I've tried converting my km to m and min to sec. I then tried using the T equation as V=2*pi*r/T to get the speed, but that isn't correct. I suspect that the 605km isn't the right radius since it's from the earths surface, but I tried adding the radius of the earth to it and that wasn't correct either (not to mention that the radius of the earth isn't given, so that pretty much says it's not a factor). I don't really know where to go now.
     
  2. jcsd
  3. Jul 7, 2007 #2

    nrqed

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    You are doing it the right way. Provide th enumbers if you want me to double check. Btw, you must add the radius of the Earth to find th evalue of "r" to use in the equation. Even if they don't provide that value in the question, they assume that you can look it up. make sure you put everything in meters and in seconds as you said. Is the answer given in m/s?
     
  4. Jul 7, 2007 #3
    This is my calculation this time:

    [2*pi*(605km*1000m/1km)*(6356.75km*1000m/1km)]/(90.6min*60sec/1min)

    I got 4,445,196,121m/s

    edit: the answer is supposed to be in m/s
     
  5. Jul 7, 2007 #4

    nrqed

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    You must add the radius of the Earth to the 605 km, not multiply them together! (You could have noticed that this was incorect by writing out the units of your answer)
     
  6. Jul 7, 2007 #5
    - Hmm, that was a dumb mistake. So I tried that, but still got the wrong answer. This is what I did.

    [2*pi*((605*1000)+(6356.75*1000))/(96.6*60)

    That gave me 8046.72m/s, and the webassign application said it was incorrect. I'm down to my last submission before it's marked completely wrong, so I only have one more chance to get it right.

    Could it be that my earth radius is incorrect? I just did another google search and found another value of 6,378.1km. I used the smaller of the 2 that were given on wikipedia.
     
  7. Jul 7, 2007 #6

    nrqed

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    Mmm... The value I use is 6.37 x 10^6 m. yes, it sounds like the value you used is a bit off.

    Now, I have never used Webassign so I don't know how it works with sig figs. Maybe you entered your value with too many sig figs. since th etime was given with three sig figs, it seems that giving 6 sig figs is incorrect.

    So you should try 6.37 or 6.38 but I am not sure what the rules of webassign are for rsig figs in the answer
     
  8. Jul 7, 2007 #7
    Webassign doesn't seem to really care about how many sig figs you use, but I tried plugging your value for earths radius into the equation and got 8062.03m/s, but that was wrong as well. So now I'm stuck not knowing how to do this problem. I'll have to wait till after the assignment is ended and check the answer key.
     
  9. Jul 7, 2007 #8

    nrqed

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    I don't get the numerical value you gave.

    [itex] \frac{2 \pi (6370 + 605) \times 1000}{96.6 \times 60}= 7561 m/s [/itex]

    (which should be rounded off to 7.56 km/s).
     
  10. Jul 7, 2007 #9
    - I'm really not sure how I got the answer I got either. I tried again and got the same thing you did. Ah well, I'm more worried about learning how to do it than getting a single problem on the homework right. Thanks for the help.
     
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