Period of an object orbiting near the surface of the Earth

[sciencegem]

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!

 

[jedishrfu]

This may explain it better:

http://hyperphysics.phy-astr.gsu.edu/hbase/mechanics/earthole.html

 

[A.T.]

I'm trying to figure this out intuitively, any suggestions on the practical logic behind this interesting concept?

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.)