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Homework Help: Centripetal acceleration on Earth

  1. May 24, 2010 #1
    1. The problem statement, all variables and given/known data
    An object orbits the earth at a constant speed in a circle of radius 6.38 x 106 m, very close to but not touching the earth's surface. What is its centripetal acceleration?

    2. Relevant equations
    a = v2/r = 4[tex]\pi[/tex]2v/T2
    v = 2[tex]\pi[/tex]r/T

    3. The attempt at a solution
    I plugged in r = 6.38 x 106 m and T = 24.0 h = (24.0 h x 3600 s / 1 h) = 86,400 s into the equation above and found a = 3.37 x 10-2 m/s2. However, I looked up the centripetal acceleration on the earth's surface and found out it is 0.006 m/s2. I can't understand why my answer is wrong. Can anyone point out the error in my logic?

  2. jcsd
  3. May 24, 2010 #2
    Try using this formula- 2[tex]\pi[/tex]/[tex]\omega[/tex]

    [tex]\omega[/tex]= 360/24*3600

    Let me know if you got the answer.
  4. May 24, 2010 #3
    Thanks for the suggestion RoughRoad but I don't really understand how I am supposed to use this equation.

    I found that my calculation for the angular velocity (2piR/T) is approximately equal to the wikipedia value for angular velocity, so I really am confused now because I can't see what I'm doing wrong in just squaring that value and dividing by 6.38 x 106.
  5. May 24, 2010 #4
    Is the mass of the satellite given?
  6. May 24, 2010 #5
    No, it isn't.
  7. May 24, 2010 #6
    try using the equation GMm/r^2=mv^2/r
    m-mass of satellite
    M-mass of earth
    r-radius of orbit
    v^2/r is the centripetal acceleration
  8. May 24, 2010 #7
    gravitational frce of earth is utilised for centripetal force
  9. May 24, 2010 #8
    You are right. But what about the velocity?
  10. May 24, 2010 #9
    you are asked to find the centripetal acceleration you just have to find the value of v^2/r which gives the centripetal acceleration
  11. May 24, 2010 #10

    Oh yeah! How can I be so foolish. Thanks for helping!
  12. May 24, 2010 #11
    hope you got the answer
  13. May 24, 2010 #12
    u r welcome
  14. May 24, 2010 #13
    And reply to my visitor msg pls
  15. May 24, 2010 #14
    I didn't, though, and that's the source of my confusion. I know I have the right value for v but when I plug it in to v^2/r, I get 3.37 x 10^-2 m/s^2. This answer is not equal to the answer I found for the actual acceleration, which is 0.006 m/s^2. My question is why is my answer different from the real value?
  16. May 24, 2010 #15
    what is the value of v?
  17. May 24, 2010 #16
    465.1 m/s
  18. May 24, 2010 #17
    there is some error in the velocity value
  19. May 24, 2010 #18
    we get acceleration value almost equal to g
  20. May 24, 2010 #19
    velocity in dis orbit must b approximately equal to 7.9*10^3m/s
  21. May 24, 2010 #20
    I don't understand why that is, especially since Wikipedia says it is 451 m/s
  22. May 24, 2010 #21
    it wud b easy if u jus show me dat particular part from d article
  23. May 24, 2010 #22
    It's on the sidebar, under "Physical Characteristics".
  24. May 24, 2010 #23
    it is equatorial rotation velocity of earth while that mentioned here is the velocity of a satellite of earth in an orbit very close to it both r different
  25. May 24, 2010 #24
    The centripetal acceleration must balance out the acceleration due to gravity (it is essentially at ground level, so 9.8 m/s^2).

    *Force of gravity = mg
    *Centripetal acceleration = a (not v^2/r, because you are solving for this total value, not any sub-values within a)

    The centripetal acceleration when traveling in a constant circular orbit at ground level on earth should be equal to gravity, or 9.8. You don't necessarily need velocity, radius, mass, time, etc.
  26. May 25, 2010 #25
    Try this-

    Centripetal acceleration= [tex]\omega[/tex]^2*r= (2[tex]\pi[/tex]/T)*r

    Let me know if you got the answer.
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