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Centripetal Acceleration

  1. Jun 4, 2007 #1
    1. The problem statement, all variables and given/known data
    Calculate the centripetal acceleration, in units of m/s2, of the Earth in its orbit around the Sun. Assume that the Earth's orbit is a circle of radius 148,022 thousand km.

    2. Relevant equations
    v= 2 pi r/ T
    Ac = V^2/ r

    3. The attempt at a solution
    I solve for V and get 29.48 m/s . For T, I use 3.15 x 10^7 seconds, and for r I use 1.48022 x 10^9

    When I solve for Ac, I get 5.87 X 10-6 m/s2
    I square 29.48 m/s and divide by 1.48022 x 10^9

    The answer is 5.85 x 10-3 m/s2. I have no idea where I'm going wrong! I'm more concerned about why my answer is 10-6 when it should be 10-3 . Any ideas?
  2. jcsd
  3. Jun 4, 2007 #2

    Chi Meson

    User Avatar
    Science Advisor
    Homework Helper

    Do you really think the earth is travelling 30 m/s? 66 mph? :biggrin: That's not even speeding on a US highway. Might want to check your v again.
    Last edited: Jun 4, 2007
  4. Jun 4, 2007 #3
    That's true, it doesn't sound right does it?

    However, if the answer is supposed to be m/s2 in the end, wouldn't my calculation for V still be correct? Taking 148022 KM and converting to meters becomes 148022000 M. When multiplied by 6.28, I then get 9.3 X10^8. The number of seconds in a year is 3.156 x 10^7.

    So: 9.3 X10^8 / 3.156 x 10^7 = 29.5 m/s2

  5. Jun 4, 2007 #4

    D H

    Staff: Mentor

  6. Jun 4, 2007 #5
    I just copied the question from the book. Didn't see anything about the moon. Do I need to consider that for this question?
  7. Jun 5, 2007 #6

    D H

    Staff: Mentor

    No. I highlighted 148022 km because that number is the source your error. It is wrong. I simply used the Earth-moon distance to illustrate that you have the wrong value.
  8. Jun 5, 2007 #7
    Thank you so much!!!

    I see my mistake.
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