Suppose we had a straight tunnel, through Earth's center, to a point on the opposite side of the planet, and used it to deliver mail to the other side. With what speed would our packages pass through Earth's center.

Question

So:
ag = (GM)/R2
Mins = (4/3)πR3*ρ

So, I'm assuming (but not completely sure) that the question is asking for the man's speed at the center (but again, I could be interpreting the question completely wrong. Either way, I'm still stuck on how to solve it.)

I figured that I could take the integral of ag with respect to R (since the distance from the center of the earth is continuously changing as you pass through the center.)

So: ag = ∫[G*ρ*(4/3)π *R3]/R2
(from 0 to R)

My problem here is that ρ is just M/V which is 3M/(4πR3), which basically cancels out a whole bunch of R's and leaves you with:

ag = GM∫[1/R2
(from 0 to R...I'll worry about signs later)

AND:
ag = GM/R, which doesn't actually work out since you end up having m2/s2, which is not acceleration.

I know I probably messed up the integral, but I'm not sure how else you would solve this problem.

Answer 1

I think you might get a clearer picture of the situation by reading the discussion here:

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