Figuring out time to center of the Earth via a vacuum hole

ryank614
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Homework Statement



A hole is drilled from the North to South pole. THe air is evacuated. An object dropped from the surface would accelerate at a=-gs/R(e), where s is the distance of the object from the center of the Earth. Find the magnitude of velocity at the middle of the Earth and the time in seconds required for the object to fall to the middle.

R(e)=radius of the Earth given as 6370 km.
g=9.81m/s^2

2. The attempt at a solution

I have figured out the first part.

a(s)=dv/ds * ds/dt = v * dv/ds

Using this, since acceleration is given in terms of position,

a(s) ds = v dv

Integrating with the left side from s0=6370000 meters to s1=0 AND v0=0 and v1=v,
I found the magnitude of the velocity at the middle of the Earth to be 7905 m/s

Then comes the problem. I can't seem to figure out the time it takes. This is what I have so far.

From the above integral, I found velocity as a function of position.

Using v(s) = ds/dt, I put ds/v(s) = dt.

I tried integrating this which would be

ds/SquareRoot(g*s^2/R(e)) = dt.

I get t=12630 seconds. The answer is 1266 seconds. Where did I go wrong?
 
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Since a=-\frac{g}{R_e}s and a=\ddot{s} you get

\ddot{s}+\frac{g}{R_e}s=0

Do you recognize this equation?
 
no sorry I don't :(
 
It is the harmoning oscillator equation :smile:, i.e.

\ddot{x}+\omega^2\,x=0

with period

T=\frac{2\,\pi}{\omega}
 
Rainbow Child said:
Since a=-\frac{g}{R_e}s and a=\ddot{s} you get

\ddot{s}+\frac{g}{R_e}s=0

Do you recognize this equation?

Ok, so you are saying the double integral of s + gs/R(e) = 0. I can't completely see how that helps me the time.

Unless, are you saying that g/R(e) is equal to w?
 
Yes!

\omega^2=\frac{g}{R_e}

Now, can you figure out how much time needs the object to travel from the surface to the center?
 
Wow. Thank you very much. I just plugged in w and divided by 4 since it was asking for the time to the center.

Thanks!
 
Glad I helped! :smile:
 
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