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Period of an object orbiting near the surface of the Earth

  1. May 5, 2015 #1
    During a physics lecture, the professor demonstrated how to find the period of an object that was dropped through a hole drilled straight from one end of our planet to the other. He finished by saying "an object orbiting the earth near the surface will have a period of the same length as that of the transport tunnel". He backed this up with a=omega^2 * R => 9.81=omega^2 * 6.38(10)^6 => omega=0.00124 s^-1 => T=2pi/omega=84 mins (the period through the transport tunnel). I'm trying to figure this out intuitively, any suggestions on the practical logic behind this interesting concept?

    Thanks for reading this!
     
  2. jcsd
  3. May 5, 2015 #2

    jedishrfu

    Staff: Mentor

  4. May 5, 2015 #3

    A.T.

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    Science Advisor
    Gold Member

    Decompose the orbit in x,y components. What is the gravity component along x as function of position along x? Compare to gravity inside a uniform density sphere as function of radial position. (Note that the Earth is not really of uniform density.)
     
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