Calculating Elevator Free Fall After Passing the Earth's Core

[gary350]

Lets assume there is an elevator shaft all the way through the center of earth. If the elevator free falls down the shaft after it passes the center of the Earth how much height will the elevator gain before coming to a complete stop assuming no wind resistance in the elevator shaft?

Diameter if Earth is 7926 miles.

I reality how much height will the elevator really gain before it comes to a stop?

 

[danR]

Lets assume there is an elevator shaft all the way through the center of earth. If the elevator free falls down the shaft after it passes the center of the Earth how much height will the elevator gain before coming to a complete stop assuming no wind resistance in the elevator shaft?

Diameter if Earth is 7926 miles.

I reality how much height will the elevator really gain before it comes to a stop?

A bit less than 3463 miles. The system will lose a tad to gravitational radiation. But an expert will give you an exact answer shortly.

 

[cjl]

Ignoring friction and air resistance, it will go up to exactly the same height as it started, on the other side of the earth. This will take around 40 minutes.

 

[Subductionzon]

It would take about 38 minutes to go through the Earth. If you follow http://hyperphysics.phy-astr.gsu.edu/hbase/mechanics/earthole.html" you will find a simplified version of it. It assumes a uniform Earth. But since the Earth's density increases with depth it would take about 4 minutes less than the 42 minutes that they calculated.

 

[Shoku Z]

Ha,OP google gravity train.Interesting concept.And the result shocking!

 

[Superstring]

The acceleration due to gravity at any radial displacement "r" from the center of the Earth (while inside of it) along the diameter is given by:

$$g = -k^2r$$

where $k^2 = \frac{4}{3} \pi G \rho$,

$G$ is the gravitational constant, and $\rho$ is the average density.From this, you get the simple harmonic motion differential equation:

$$\ddot{r}+k^2r=0$$

The solution to which is:

$$r=r_0~cos(kt+\phi )$$So the elevator would oscillate in simple harmonic motion with a period of:

$$T=\frac{2\pi}{k}=\sqrt{\frac{3\pi}{G\rho}}$$In short (to answer your question), it would reach the exact same height on the other side of the planet from which it was dropped.

 

[RK1992]

It would take about 38 minutes to go through the Earth. If you follow http://hyperphysics.phy-astr.gsu.edu/hbase/mechanics/earthole.html" you will find a simplified version of it. It assumes a uniform Earth. But since the Earth's density increases with depth it would take about 4 minutes less than the 42 minutes that they calculated.

interesting - a very similar question came up in a 1st year cambridge physics paper (http://www-teach.phy.cam.ac.uk/dms/dms_getFile.php?node=5735")