1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook