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Circular Motion and Centripetal Acceleration

  1. Oct 1, 2004 #1
    I’m having trouble with ch3 #73 from Tipler/Mosca Physics for scientists and engineers 5th edition. I’ve tried a few things and I keep getting stuck. Am I taking the right steps?

    An object resting on the equator has an acceleration toward the center of the earth due to the earth’s rotational motion about its axis, and an acceleration toward the sun due to the earth’s orbital motion. Calculate the magnitudes of both of these accelerations and express them a fraction of the magnitude of free-fall acceleration g. Use values from the physical-data table in the textbook.

    So we have centripetal acceleration = v^2/r (Ac = v^2/r)
    g = 9.81 m/s^2
    earth’s radius @ equator = 6370 km


    Ac = g^2/6370 = (9.81 m/s^2)^2/6370 km = .015 m^2/Kms^4? (what units/reductions do I need to make? Is this correct?

    Period formula
    v = 2PIr/T
    I have 2 PI and r, but what do I use for T?

    Thanks for your help!
     
  2. jcsd
  3. Oct 1, 2004 #2

    Tide

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    No, you're not taking the right steps. You need to understand what the factors are in your formula for centripetal acceleration. v is the speed about the axis of rotation and r is the distance from the axis.

    The speed is total distance traveled divided by time. Standing at the equator of the rotating Earth you travel a distance equal to the circumference of the planet in a day.
     
  4. Oct 1, 2004 #3
    Ok, so V = (2PI(6370))/86400(secs in a year) = .46 m/s

    Ac = (.46)^2/6370 = 3.32 x 10 to the -5 m/s^2? The back of the book has 3.44 x 10 to the -3g.

    Help..
     
  5. Oct 1, 2004 #4

    Gokul43201

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    Should you not convert km into m ? And you are asked to find out what fraction of the value of g, this number is.
     
  6. Oct 1, 2004 #5

    Doc Al

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    The Earth's radius is given in kilometers, not meters. Convert.
     
  7. Oct 1, 2004 #6
    OOOPS, my bad…Thanks so much!!!

    6370 km = 6370000 m
    so V = (2PI(6370000m))/86400s = 463.24 m/s

    (463.24)^2/6370000 = .0337

    .0337/9.81 = .003434 which I can label as 3.43 x (10 to the -3)g which is REAL close to the 3.44 x (10 to the -3)g in the back of the text.

    What formula do I use to get the acceleration toward the sun due to the earth’s orbital motion?

    I’m assuming it’s At = dv/dt but what is my change in velocity and/or time?

    Thanks for your time!
     
  8. Oct 1, 2004 #7

    Tide

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    You should have learned something from doing the first part of the problem. It's exactly the same problem with different input. You know how far the Earth travels around the Sun in a year so you can calculate the speed and use the very same centripetal acceleration formula.
     
  9. Oct 4, 2004 #8
    I didn't understand how to figure out the orbital speed & distance, but I found them in the back of the book so that made it easy to plug it into the Ac formula. Thanks for your time!
     
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