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Another gravity problem Please help

  1. Apr 18, 2008 #1
    1. The problem statement, all variables and given/known data
    While visiting Planet Physics, you toss a rock straight up into the air at 11m/s and catch it 2.5 s later. While you visit the surface, your cruise ship orbits at an altitude equal to the radius of the planet every 230 min. What are the mass & radius of Planet Physics?


    2. Relevant equations



    3. The attempt at a solution

    0m=0m + (11m/s)(2.5s) + .5a(2.5s)^2
    ac=-8.8

    230 min = 13800s angular velocity = 2pi rad/ 13800 = 4.55E-4 rad/s

    ac= angular velocity^2 * r
    8.8= (4.55E-4)* r
    r=4.25 * 10 ^7 m This is not the correct answer. Could someone please explain what I am doing wrong?
     
  2. jcsd
  3. Apr 18, 2008 #2

    Hootenanny

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    Note that your altitude is equal to the radius of the planet, which means that your cruise ship is 2r from the centre of the planet.
     
  4. Apr 18, 2008 #3
    ohhhh.... of course. Thanks
     
  5. Apr 18, 2008 #4

    Hootenanny

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    A pleasure :smile:
     
  6. Apr 18, 2008 #5
    wait a second, If I divide my answer by 2, I still don't get the right answer, which is 1.33E6 m
     
  7. Apr 18, 2008 #6

    Hootenanny

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    The acceleration that you calculated earlier is the acceleration on the surface, i.e. at r, whereas your cruise ship is orbiting at 2r.
     
  8. Apr 18, 2008 #7
    Would the gravity be one fourth of what it is at the surface? Nope... that doesn't work either.
     
  9. Apr 18, 2008 #8
    I really am not sure what to do
     
  10. Apr 18, 2008 #9

    Kurdt

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    You're equating two things that are incompatible. The acceleration at the height of the ship is not the same as at the surface of the planet. If you consider setting up two equations in terms of the mass of the planet and the radius, then you can eliminate a variable.
     
  11. Apr 18, 2008 #10
    Ok, here's what I tried:

    since at the surface of the planet
    g= GM/R^2
    R= sqrt[ GM/g]

    I calculated g at the surface to be 8.8 m/s^s

    so I substituted that into the equation:
    R= sqrt[GM/8.8]

    Then I move on to the orbit which is 2R from the center

    So the g of the ship would be

    gship= GM/(2*sqrt[GM/8.8])^2

    I get gship= 2.2 m/s^2

    But using the equation a= angular velocity^2 * r
    I again get a wrong value for r.
     
  12. Apr 18, 2008 #11

    Kurdt

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    This part is good.

    The ship is still at [itex]2r[/itex].
     
  13. Apr 18, 2008 #12
    That's why I multiplied my expression for R by 2
     
  14. Apr 18, 2008 #13

    Kurdt

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    It was that bit to which I was referring in my previous post. Sorry for not making that clear.
     
  15. Apr 18, 2008 #14
    2.2 m/s^2 = (4.55E-4)^2*r
    r= 1.06E7
    this r is the distance between the ship and the center of the planet

    So then I divide by 2 to get the radius of the planet: 5.31E6 m

    But that's not the correct answer
     
  16. Apr 18, 2008 #15

    Kurdt

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    Is this entered online? I get that as the answer for the radius.
     
  17. Apr 18, 2008 #16
    no, it's the answer in the back of my textbook
     
  18. Apr 18, 2008 #17

    Kurdt

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    Are you sure the question is as it is stated the book? Failing that I'll get someone else to check that I haven't missed something by mistake. What is the answer in the book out of curiosity?
     
  19. Apr 18, 2008 #18
    no, It's from the back of my textbook
     
  20. Apr 19, 2008 #19

    Shooting Star

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    I get the radius as ~ 5.31E6 m.

    EDIT: Sorry, bcjochim07. I didn't see that you've already given the answer in post #5.
     
    Last edited: Apr 19, 2008
  21. Apr 19, 2008 #20

    alphysicist

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    Hi Shooting Star,

    In post #5 he said the book's answer was 1.33e6 m. I got the same answer that you and he did.
     
    Last edited: Apr 19, 2008
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