# Gravity train

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671264.  Tue Feb 16, 2010 5:22 am

soup wrote:
 Sophie.A wrote: which increases as L increases, tending to infinity towards the Poles.

Eh? I may be being thick (probably am) but as L increases would the value for

not decrease?

Yes, simplify it to the square root of 1/x

If x = 9 then the square root of 1/9 is the answer ie 1/3

If x = 16 then the square root of 1/16 is the answer ie 1/4

1/4 is less than 1/3 so as L increases the value decreases, except that between <E> and <E> degrees cos L decreases in value so under those circumstances as L increases the value of cos L decreases (and actually becomes -ve between <E> and <E> degrees).

<E> soup is right but when I Googled for 'cos graph' for some reason I got 4 sin graphs and didn't bother checking closely.

Last edited by Celebaelin on Tue Feb 16, 2010 6:11 am; edited 1 time in total

 671284.  Tue Feb 16, 2010 5:57 am Thinking it through :-Ah, cos 0° = 1 whilst cos 90° = 0 so as L increases from 0° to 90° cos L decreases from 1 to 0 so yes Sophie was right all along (you can use that in your sig if you want :-) )the equation does increase as L increases (Cos L decreases as L increases). You will need to forgive me it is a long time since I did any maths.

671537.  Tue Feb 16, 2010 1:29 pm

 Sophie.A wrote: Note also that the tunnel does not have to be straight – it can be in two straight halves, each connected to the centre of the Earth. Thus, in order to get from one place on earth to anywhere else in exactly 42 minutes 10 seconds, you only have to travel in a straight tunnel to the centre of the Earth, and then in another straight tunnel from the centre of the Earth to your destination point.

I don't understand this bit. Don't you arrive with a large directional momentum at the centre, and this momentum can only be used to continue in the same direction - ie a straight tunnel?

671547.  Tue Feb 16, 2010 2:11 pm

 gruff5 wrote: I don't understand this bit. Don't you arrive with a large directional momentum at the centre, and this momentum can only be used to continue in the same direction - ie a straight tunnel?

Not the only bit that I don'tunderstand. I thought Steven said that you didn't need to go via the centre, but also said that such a journey wasn't possible because of the molten core. ???

 671565.  Tue Feb 16, 2010 2:42 pm I suppose that a flat straight bridge of sufficient length would work as well as a tunnel.

671585.  Tue Feb 16, 2010 3:19 pm

 BondiTram wrote: Not the only bit that I don'tunderstand. I thought Steven said that you didn't need to go via the centre,...

That's true in the same sense that you would accelerate down a 'vertical mineshaft' type arrangemnet then start slowing once you reached the middle and just make it to the other side the same thing would happen with a tunnel cut along a chord, you'd need a low friction track system though. You'd accelerate from A to D then deccelerate from D to B.

As regards the molten 'core' that starts a lot closer to the crust than you'd think at first. You couldn't tunnel through the mantle for example.

671847.  Wed Feb 17, 2010 7:20 am

Since Sophie's first formula only depends on the density of the sphere (assumed constant) and the gravitational constant (which is, as the name implies, constant), the only explanation for the discrepancy of two seconds in the time is that people are assuming different values for the density.
Sophie's second formula leads to an infinite time for the gravity train to get from one side of the north pole to the other side of the north pole. I would respectfully suggest that there is something wrong with the formula. It would conclude that to go an infinitely short distance would take an infinite time.
Almost the whole point of the mathematics is that it takes the same time to get from one point on the Earth's surface to any other point on the Earth's surface via a straight tunnel. The "reason" for this is that, although the length of a chord is shorter than the pole-to-pole route, only a component of the gravitational force is acting in the direction the tunnel. So the force in the direction of the tunnel is less, but so is the distance, so the time stays the same. I hope this simplified explanation may help people to "get their head round this". Obviously, you would need to show mathematically that the decrease in force and the decrease in distance varied so as to leave the time unchanged.
Sophie also wrote:
.
 Quote: Note also that the tunnel does not have to be straight – it can be in two straight halves, each connected to the centre of the Earth. Thus, in order to get from one place on earth to anywhere else in exactly 42 minutes 10 seconds, you only have to travel in a straight tunnel to the centre of the Earth, and then in another straight tunnel from the centre of the Earth to your destination point.

If you took the gravity train to the centre of the Earth, and somehow managed to leap off while it was travelling at its maximum velocity, you would then need to leap onto another travelling in a different direction while it too was travelling at its maximum velocity. If there were a train waiting on another platform for you to get into, pointing in another direction, there would be no force to make it leave the platform.
Finally, what is the sideways component of gravity, AFB? Are you thinking that, because the Earth is spinning, there is a tendency for everything to be dragged "sideways", in the opposite direction to the Earth's spin? That's not the case. Gravity at the equator is less than at the poles, partly because the distance from the Earth's centre is greater, and partly because some of the gravitational force is needed to keep anything at the equator rotating with the Earth.

 671873.  Wed Feb 17, 2010 7:57 am One assumes some sort of switching system, 'points' as it were.

671940.  Wed Feb 17, 2010 11:22 am

 Quote: Note also that the tunnel does not have to be straight – it can be in two straight halves, each connected to the centre of the Earth.

Hmm, thought we were talking about a "gravity train" that was un-powered and could just be dropped down a single "bent" tunnel" to re-emerge, rather than a passenger who would be splattered by the huge G-forces involved in leaping among different speeding trains, each travelling along their own straight tunnels....

It is neat the way Sophie's formula just depends on density, though. I thought at first she'd made a mistake - "but it must take longer to drop through a bigger planet!".

 731162.  Mon Aug 02, 2010 11:15 pm Sorry - we've overlooked the ficticious forces since this "tunnel" would need to be considered in a rotational frame of reference (with the exception of the single case going through the poles). Can someone factor this in? I assert that the time taken would vary by more than two seconds, but wouldn't be enough to cover National Rail's sorry ass...

 817380.  Wed May 18, 2011 7:55 pm I must admit to not understanding at all how you could travel through a tunnel from one side of the Earth to the other in 42 or 43 minutes. (Naturally ignoring the heat from the molten core.) At the surface of the Earth our planet's mass keeps us all in our place with an acceleration towards the centre of the Earth of 9.8 m/s/s. As we pass along a tunnel deeper into the Earth we will be closer to the centre, which would increase the pull of gravity. But we will also find that some of our planet is now above us, and exerting a gravitational pull away from the centre o he Earth. (The amount of the planet to the right and left exactly balancing out and having no effect.) As we pass through the tunnel the gravitational pull towards the centre will weakened, and the gravitational pull upwards towards the beginning of our tunnel will increase. At the centre of the Earth all the planet will be around us, pulling us equally in all directions. We will be weightless. And because the decrease in the downwards pull has been so gradual we will arrive with zero velocity, and zero momentum to carry us through this weightless zone. (Recent modelling with our grandson's plasticine has confirmed this expectation.) We are now trapped, forever. How smooth are the sides of the tunnel? Can we gain enough purchase to climb out? Perhaps a friend could drop a ladder down to us, though I am not sure where to stand the base of the ladder. A rope would probably be best. A long one. 42 minutes? 43 minutes? Trapped in a weightless void at the centre of our tunnel. I just hope you remembered to bring some sandwiches. Anthony Wheeler

 817431.  Thu May 19, 2011 8:10 am Anthony, i think the scenario you are describing there might apply in a tunnel that was open to the atmosphere and therefore filled with air, so that your fall speed is reduced to terminal velocity (about 140 mph near sea level for a human body). But I think this 42-minute journey only applies for an imagined tunnel that has a vacuum inside. Without doing the maths, the continual and unimpeded acceleration would lead to a very high velocity at the centre of the Earth (I would guess it would be many thousands of miles per hour). This velocity & momentum would sufficient to carry him thru the airless void back to the other side, where he would arrive with all that high velocity spent. HTH gruff5

818282.  Sun May 22, 2011 6:29 am

 Wheelera wrote: As we pass along a tunnel deeper into the Earth we will be closer to the centre, which would increase the pull of gravity.

This is not quite correct. You forget that you are already inside the Earth when you approach the centre; when you are inside the Earth and getting closer to the centre, gravity is due to an increasingly smaller mass of the Earth "below" you (the shell of the Earth "above" you does not contribute any net gravitational force on you); therefore the pull of gravity actually decreases.

Now, when you are outside the Earth, gravity does increase in strength the closer you are the centre. But this only applies when you are wholly outside the Earth; once you get inside the Earth itself, things are different.

Let's do a little calculation. Suppose you are halfway to the centre of the Earth from the surface. Thus if the radius of the Earth is R, suppose you are R/2 away from the centre. If the density of the Earth is ρ, the mass of the Earth that is contributing the gravitational force on you is volume × density = (4/3) × π × (R/2)³ × ρ = πR³ρ/6. If your mass is m, then the gravitational force on you (by Newton's law of gravitation) is GπR³ρ/6 × m / (R/2)² = 2GπRρm/3. (Remember the mass of the Earth forming a shell "above" you does not exert any net gravitational force on you.) On the other hand, the gravitational force on you on the surface of the Earth is 4GπRρm/3 (as you can verify by a similar calculation). This shows that gravity actually decreases in strength when you are halfway to the centre of the Earth inside the Earth itself.

 818285.  Sun May 22, 2011 6:44 am This whole discussion recalls to me The Algebraist, by Iain Banks. --tom

 818328.  Sun May 22, 2011 9:50 am It recalls to me a migraine that I had a few years ago. ; )

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