- #1

gary350

- 274

- 66

Diameter if Earth is 7926 miles.

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

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- Thread starter gary350
- Start date

In summary, assuming no wind resistance and a uniform Earth, an elevator that falls freely down a shaft after passing the center of the Earth will reach the exact same height on the other side of the Earth from which it was dropped. This takes approximately 38 minutes, with a slight loss due to gravitational radiation. However, since the Earth's density increases with depth, the actual time may be slightly less than 38 minutes. A simplified version of this calculation can be found on the hyperphysics website.

- #1

gary350

- 274

- 66

Diameter if Earth is 7926 miles.

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

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- #2

danR

- 352

- 4

gary350 said:

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.

- #3

cjl

Science Advisor

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- 612

- #4

Subductionzon

- 172

- 2

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.

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- #5

Shoku Z

- 18

- 0

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

- #6

Superstring

- 120

- 0

[tex]g = -k^2r[/tex]

where [itex]k^2 = \frac{4}{3} \pi G \rho[/itex],

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

[tex]\ddot{r}+k^2r=0[/tex]

The solution to which is:

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

[tex]T=\frac{2\pi}{k}=\sqrt{\frac{3\pi}{G\rho}}[/tex]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.

- #7

RK1992

- 89

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Subductionzon said:

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")

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The formula for calculating elevator free fall after passing the Earth's core is **h = (R - d) * 2**, where h is the height, R is the radius of the Earth, and d is the distance from the surface to the core.

The mass of the Earth has no effect on the elevator's free fall. The gravitational force between two objects only depends on their masses and the distance between them. Therefore, the mass of the Earth does not affect the acceleration of the elevator.

The acceleration of the elevator during free fall after passing the Earth's core is **9.8 m/s ^{2}**. This is equivalent to the acceleration due to gravity on the surface of the Earth.

The accuracy of the calculated elevator free fall can be affected by factors such as air resistance, the shape and size of the elevator, and the presence of any external forces acting on the elevator.

No, it is not possible for the elevator to reach the other side of the Earth after passing through the core. This is due to the effects of air resistance, which would slow down the elevator's speed and eventually bring it to a stop before reaching the other side.

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