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Finding Acceleration for Object on Equator

  1. Nov 21, 2004 #1
    An object lying on Earth's equator is accelerated in the following three directions.
    (a) toward the center of Earth because Earth rotates
    (b) toward the Sun because Earth revolves around the Sun in an almost circular orbit
    (c) toward the center of our galaxy because the Sun moves about the galactic center
    For the latter, the period is 2.5 * 10^8 y and the radius is 2.2 * 10^20 m. Calculate these three accelerations as multiples of g = 9.8 m/s2.

    Ok so this shouldn't be hard, I'm probably making it harder than it really is, I'm having trouble working with gravity and circular motion for some odd reason. So any help?
  2. jcsd
  3. Nov 21, 2004 #2
    Find the centripetal acceleration and then divide by g.
  4. Nov 21, 2004 #3
    well I found part C but I can't figure out teh answers to part a and b. :blushing: This should be easy!
  5. Nov 21, 2004 #4
    A and B work exactly the same way as C. Just use the period eqn (T=2(pi)r/v) to find velocity and then plug into the centripetal accel. eqn and divide by g. For A, you'll use the earth's radius for r and 24 hours for the period, but convert it to seconds. For B, you'll use the distance from the earth to the sun for r and a period of 365 days, once again converted to seconds.
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