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Centripetal acceleration

  1. Dec 4, 2005 #1
    :confused: Given: r(Earth)=6.37*10^6 m
    m(earth)=5.98*10^24 kg
    r(moon)=1.74*10^6m
    g=9.8 m/s^2
    G=6.67259*10^-11 Nm^2/kg^2

    Calculate the speed of a 97.4kg person at the equator.
    do u use a=v^2/r v=sqr root (a*r) ?
     
  2. jcsd
  3. Dec 4, 2005 #2
    Hello????!!!!!!

    Can Anybody Help Me?!
     
  4. Dec 4, 2005 #3
    [tex] \Sigma F = ma_c [/tex]

    ... you know that gravity is the force acting towards the center, you know the radius, so you can find the v.
     
  5. Dec 4, 2005 #4
    .....

    ok, taking what you said i got that Gravity=m*v^2/r

    v=(gr/m)^1/2 correct?
     
  6. Dec 4, 2005 #5
    which radius do u use? the earth's or the moon's?
     
  7. Dec 4, 2005 #6
    where is the person?
     
  8. Dec 4, 2005 #7
    equartor

    he is at the equator
     
  9. Dec 4, 2005 #8
    of which? the earth or the moon? which radius does it make sense to use?
     
  10. Dec 4, 2005 #9
    at the earth's equator
     
  11. Dec 4, 2005 #10
    yeah, so use the earth's equator..
     
  12. Dec 4, 2005 #11
    do u use 9.8m/s^2 or 6.67*10^-11
     
  13. Dec 4, 2005 #12
    those are two different things... you can use either, if you know what you're doing.
     
    Last edited: Dec 4, 2005
  14. Dec 4, 2005 #13
    [tex] F_g - N = m \times \frac {v^2}{r} [/tex]
    [tex] \frac{GMm}{r^2} - mg = \frac {mv^2}{r} [/tex]
    [tex] \frac{GM}{r^2} - g = \frac{v^2}{r} [/tex]

    now solve for v.




    do you understand what i did?
     
    Last edited: Dec 4, 2005
  15. Dec 4, 2005 #14
    v=(GM/r^2-g)^1/2


    =(((6.67*10^-11*5.98*10^24)/(6.37*10^6)^2)-9.8)^1/2 correct?
     
  16. Dec 4, 2005 #15
    there should be an "r"

    v=((GM/r^2 - g)*r)^1/2


    more importantly, do you understand how i got that?
     
    Last edited: Dec 4, 2005
  17. Dec 4, 2005 #16
    yes thank you. i knew it had something to do with F=GMm/R^2 i just didnt know what to do. thank for all your help and your patience.
     
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