jsm6252 said:
not for class just general interest as i saw a documentery about the universe
Ah. Well, this isn't quite the right forum for this inquiry. Nevertheless, for the case that D H mentioned, where all the mass is concentrated at a point at the center of the Earth, consider a body in an orbit around that mass. If the body were in a circular orbit with a radius equal to that of the radius of the Earth (so it just skimmed the surface), then its period would be given by the formula
[tex]T = \frac{2 \pi}{\sqrt{\mu}}\right) r^{3/2}[/tex]
where [tex]\mu[/tex] is the gravitational parameter for the Earth, G*M.
Now, the same formula applies to an elliptical orbit, where r is replaced by the length of the semi-major axis of the ellipse.
This is where the clever bit comes in. It doesn't matter how broad or skinny the ellipse is, the period will remain the same if the length of the semimajor axis remains the same. So if we imagine that the orbit of the object is elliptical, and we let that ellipse become more and more eccentric (skinnier and skinnier), then eventually we will have an orbit that looks like a straight line, the object falling straight towards the center and bouncing back up again (the turn around the backside of the central mass point becoming arbitrarily close to the mass point). The semimajor axis of that "ellipse" is half the distance from the central point to the top of the "orbit", which is just half of the Earth's radius. This gives us for the trip to the center and back,
[tex]T = \frac{2 \pi}{\sqrt{G M}} \left( \frac{R_{earth}}{2}\right)^{3/2}[/tex]
Plug in the numbers for M, G, and R
earth and see what you get.
For another pleasant surprise, plug in the values to obtain the period of the Earth-surface-skimming satellite. That is, the period of time it wold take a satellite to travel to the far side of the Earth and back again, traveling around, say, the equator (ignoring terrain issues and air resistance).
