# Gravity Train

1. Sep 12, 2011

### -Castiel-

A few months ago I was surfing idly and found this little thing called a Gravity Train. For those who do not know what it is, see http://en.wikipedia.org/wiki/Gravity_railroad" [Broken].

After reading about it I thought maybe I should try and find out how much time it would take (42.2 minutes) but anything I tried did not work. Finally, when I decided to give up and sneak a peek I saw that they calculated it considering it as an oscillation (which in a way it is), but what I have been trying to do is calculate that time using kinematics.

I know the relation of 'a' with displacement (x), a = g((R-x)/2R) where g = 9.8, r = radius of earth, x = the depth to which the body is inside the earth.

Whatever approach I take from here leads me to a dead end. Any ideas how I should proceed?

Last edited by a moderator: May 5, 2017
2. Sep 12, 2011

### Ken G

Start with what is wrong with your expression for a. If I put in x=0, I get a = g/2, which doesn't make a lot of sense.

3. Sep 12, 2011

### A.T.

.
I think you mean http://en.wikipedia.org/wiki/Gravity_train" [Broken].

Last edited by a moderator: May 5, 2017
4. Sep 13, 2011

### -Castiel-

Sorry about that, it should be

a = g((R-2x)/R) where g = 9.8, r = radius of earth, x = the depth to which the body is inside the earth.

I am not used to writing formulae like this so...>_<

Yes, I do.

On another note: No edit button?

Last edited by a moderator: May 5, 2017
5. Sep 13, 2011

### nasu

What do you mean by calculating time "using kinematics"?
First you need to find the (kinematic) equations of the specific motion. You cannot use the ones for uniform accelerated motion. The acceleration is not constant.
You can start with Newton second law which will provide a differential equation. You need to solve it to get the equation of the motion (position as a function of time).
This above mentioned equation happens to be the equation of a simple harmonic oscillator.
So the kinematic equations are these of the harmonic oscillator.