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Homework Help: Centripetal force question(involves satellite going around earth.

  1. Apr 18, 2007 #1
    1. The problem statement, all variables and given/known data
    Determine how long it would take for the satellite to make 1 complete revolution around earth
    A 500 kg satellite is orbiting earth, radius of earth is 6.38 x 10^3 km, and acceleration of gravity as the orbital altitude of 160km is the same as earth basically 9.8.


    2. Relevant equations
    fc=mv^2/v
    fc= 4pi^2*m*r/ t^2


    3. The attempt at a solution
    I've tried using both above equations. usually i started with energy equations like eg= mgh and ek= 1/2mv^2 to find velocity. Then I would put it into the fc= mv^2/v and find force then i would put that info into fc= 4pi^2*m*r/ t^2 to find the time. But none of it came out with the right answer which is 5132.8 seconds, I have tried everything I could think of. If you could help me please do.:cry:
     
    Last edited: Apr 18, 2007
  2. jcsd
  3. Apr 18, 2007 #2
    BEWARE of your units. You need to use 6.38E6+160E3 for radius.


    Your first eqn is fine v^2/r=a
    Period=circumference/v
     
  4. Apr 18, 2007 #3
    I have used the right units, it's 6380 km + 160 km right?
     
  5. Apr 18, 2007 #4
    yes but should reduce to v=sqrt(ar) so T=2*pi*sqrt(r/a) where R is in meters
     
  6. Apr 18, 2007 #5
    can you plug the numbers in for me, im still confused as heck. Remember the answer has to be 5132.8 seconds, which i just cant find how to get to. With those I got to 114.74, actually i got that multiple times, but it's not the right answer.
     
    Last edited: Apr 18, 2007
  7. Apr 18, 2007 #6
    i did and verified correct answer, but again, don't use km!!! Your answer:

    6.28*sqrt(6540000/9.8)
     
  8. Apr 18, 2007 #7
    I don't think that this answers good enough though. It's about 2 off.
     
    Last edited: Apr 18, 2007
  9. Apr 18, 2007 #8
    well i can live with it, and its a lot better than 5000 off:uhh: seriously, think about sigfigs.
     
  10. Apr 18, 2007 #9
    Okay heres what i tried to do
    500kg*9.8^2/6540000 = .0073
    then i plugged it into
    t=sqrt of 4pi^2*500kg*6540000m/.0073

    This seems like it should be right but it just isn't! darnit. this is making me really angry I wonder if my teacher possibly posted the wrong answer?
     
    Last edited: Apr 18, 2007
  11. Apr 18, 2007 #10
    Really if you're woried about the result, give it a rest. i am sure if we didn't make the assumption that a=9.8 and instead did the more rigorous approach of using,

    G*M/r^2 ,

    with our revised r of 6540000, we'd nail it. It was your suggestion so I thought perhaps something in the problem suggested it was OK. and we need not not even know GM
    but simply mult 9.8*(6.38E6/6.54E6)^2 which of course could have been included in my expression 2*pi*sqrt(xxxxxx).

    But get the mass of 500kg out of there, cancels out everywhere!
     
    Last edited: Apr 18, 2007
  12. Apr 18, 2007 #11
    I still cannot get the answer of 5132.8...
     
  13. Apr 19, 2007 #12
    you're right there, it now becomes 5166 or so. This is the correct answer with all info provided. Is this online homework?
     
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