# Alice in the tunnel.

## Recommended Posts

This one is a version of a problem due to Lewis Carrol, hence the thread title.

A and B are two points on the surface of the Earth which are connected by a straight tunnel.

Both A and B are on the same side of the centre of the Earth.

Alice rolls a ball into the tunnel at A.

How long will it take the ball to reach B, neglecting frictional resistances?

Take the radius of the Earth as 4000miles and g = 32 ft/sc2

Edited by studiot

##### Share on other sites

This one is a version of a problem due to Lewis Carrol, hence the thread title.

A and B are two points on the surface of the Earth which are connected by a straight tunnel.

Both A and B are on the same side of the centre of the Earth.

Alice rolls a ball into the tunnel at A.

How long will it take the ball to reach B, neglecting frictional resistances?

Take the radius of the Earth as 4000miles and g = 32 ft/sc2

Are you insinuating that the line connecting A and B is the diametre of the Earth? If so, Alice's buddy at the end of the tunnel will have better things to do than to wait for her ball to reach him.

If not, the same applies and the ball will most likely stop at a point which correlates with the perpendicular projection of the centre of the Earth on the line connecting A and B.

Basically, this is what I think ...

Knowing the honourable Lewis Carrol, this must be a trick question. It's a matter of perspective, though: the only thing moving the ball would be gravity and perhaps a femalish push (no sexism pun intended).

So imagine the ball rolling towards the Aussie guy. If we watch the world from his perspective, the ball can roll down again towards Alice.

Of course, this can't keep going on so that's why the ball must stop at the centre of A-B.

So this is the cycle of the ball:

If even those last 3 occur? I think image 4 (alice4) can still happen due to momentum? alice5 and alice6 would be a bit overkill, I think. If alice6 would be able to exist, then you should imagine being the Aussie. And suddenly, a ball would jump from a hole into the air. Which is very unlikely.

I think.

Edited by Function

##### Share on other sites

Sorry I didn't explain it very well but now you also make me feel ashamed as your picture is sooo much better than mine.

Interestingly the answer is the same as the key question in the Hitchikers Guide to the Galaxy.

##### Share on other sites

Interestingly the answer is the same as the key question in the Hitchikers Guide to the Galaxy.

42 what?

How would that be?

##### Share on other sites

1) 42 what?

2) How would that be?

1) 42.5 minutes actually, Douglas Adams got it slightly wrong.

2) That is the question of the puzzle.

##### Share on other sites

1) 42.5 minutes actually, Douglas Adams got it slightly wrong.

2) That is the question of the puzzle.

Well then, no time to lose! Let's build a computer that will provide us the correct question in some million years.

But now for the real question: did the honourable Lewis Carroll really use feet and miles?

On a second thought ... It cannot really be 42.5 minutes for all points A and B, can it? What if the distance between A and B is smaller? What if it becomes larger?

Edited by Function

##### Share on other sites

But now for the real question: did the honourable Lewis Carroll really use feet and miles?

Why not?

##### Share on other sites

Well then, no time to lose! Let's build a computer that will provide us the correct question in some million years.

But now for the real question: did the honourable Lewis Carroll really use feet and miles?

On a second thought ... It cannot really be 42.5 minutes for all points A and B, can it? What if the distance between A and B is smaller? What if it becomes larger?

A little mechanical analysis (you only need high school mechanics for this) will answer these side questions en route to the answer to the original question.

##### Share on other sites

A little mechanical analysis (you only need high school mechanics for this) will answer these side questions en route to the answer to the original question.

I have some fundamental questions regarding the gravitational acceleration ... It cannot be constant, can it? The deeper the ball rolls, the smaller the mass causing the ball to roll towards the centre of the earth and the larger the mass causing the ball to be attracted to the opposite direction. According to Newtonian gravity, that is.

So it'd be necessary to

• Determine, in function of the distance the ball has been rolling from A to B, the mass of the part of the earth "below" a line cutting the earth, perpendicular to the line connecting the ball and the centre of the earth
• Determine, in function of the distance the ball has been rolling from A to B, the mass of the part of the earth "above" a line cutting the earth, perpendicular to the line connecting the ball and the centre of the earth

Unless ... A projectile piercing through a body is still attracted to the centre of the whole body and not its individual parts ...

Which sould be the case, no? So my drawing is not correct.

Why is it incorrect? I think it's incorrect because, if you'd place an expandable, flexible, heat-resistant point at the centre of the earth, it will not expand (let's neglect heat causing expansion), correct? It will just float in place in the centre, right? On the other hand, one should ask himself the question that if indeed an expandable point would be the subject of an infinite amount of forces, defined by the gravitational forces of each time 2 half Earths and their respective centres of mass, defined by planes through the point, in all possible directions, it would either expand until it reaches the centres of gravity of all of these infinite amount of half Earths due to forces stretching it out in a sphere-like manner, or be at rest due to a net null force.

Does this even make sense to anyone here? Is it clear what I mean?

Basically the picture below is what I mean, but only in 2D, so in one plane. Imagine an infinite amount of these "half earths", defined by an infinite amound of 'parasaggital' planes through the centre and thus, an infinite amount of radial gravitational force vectors pulling the centre point to their respective centre of mass. So we have the earth, then we would have another sphere defined by the centres of mass of all half earths, and another sphere defined by the vectors defined by the gravitational force of each half earth on the point in the centre of the earth.

I simply guess this is not how it works?

Well ... This is too hard for me. Good luck, guys!

Edited by Function

##### Share on other sites

function

Well ... This is too hard for me. Good luck, guys!

Pity you are getting closer.

##### Share on other sites

Pity you are getting closer.

I'd have to know first if an object is indeed at all times attracted towards the same initial centre of mass?

##### Share on other sites

Hint

I posted this partly because we have had lots of arguments about clocks lately and this forms an unusual clock I wondered if our resident Dr Clock had heard of.

##### Share on other sites

Strictly, this only works if the tunnel is small.

In that case, yes, we can assume that the attraction is always to the centre.
You might also benefit from this.

https://en.wikipedia.org/wiki/Shell_theorem

##### Share on other sites

As always, Wiki is not quite up to high school maths

The value of gravity at distance a = (R - d) from the centre

is $g' = g\frac{a}{R}$

Where R is the radius of the Earth, a is the distance from the centre, g is the usual surface acceleration and d is the depth.

edited to improve formula.

Edited by studiot

##### Share on other sites

Sorry I didn't explain it very well but now you also make me feel ashamed as your picture is sooo much better than mine.

Interestingly the answer is the same as the key question in the Hitchikers Guide to the Galaxy.

Wait. By this diagram the tunnel is a random distance from the center of the earth.

How long is the tunnel? I mean, unless that picture is to scale.

##### Share on other sites

Wait. By this diagram the tunnel is a random distance from the center of the earth.

How long is the tunnel? I mean, unless that picture is to scale.

I feared interest in this delightful problem had died so thank you.

All the necessary information is in the OP and in post#14

I gave function the formula he requested, which is easily goolgable or derivable as John Cuthber said.

You will also need this.

Is the length of the tunnel or its depth relevant?

##### Share on other sites

I feared interest in this delightful problem had died so thank you.

All the necessary information is in the OP and in post#14

I gave function the formula he requested, which is easily goolgable or derivable as John Cuthber said.

You will also need this.

Is the length of the tunnel or its depth relevant?

Assuming that the tunnel is straight through the earth, and that both sides of the earth would be at the same level above sea level, and that it was all solid glass(picking a smooth near friction-less element.)

Rolling a ball into it would be equivalent as rolling it down a hole with a hill at the other end.

The effect of gravity acting on the ball would change, and would even itself out going both up and down.

So if I derive an equation from the conservation of energy(you said no friction, so I can do this without altering the problem) then this problem becomes elementary.

The equation I got was:

Edit: How the heck do you attach images???????

Edit 2: I'm now testing ways to insert equations. Do not quote this post.

$\large gh_1+(v_1^2)/2=gh_2+ (v_2^2)/2$

Edit 3:

$\large gh_1+(v_1^2)/2=gh_2+ (v_2^2)/2$


Edit 4:

$$gh_1+(v_1^2)/2=gh_2+ (v_2^2)/2$$


Edit 5:

$$gh_1+(v_1^2)/2=gh_2+ (v_2^2)/2$$

Edit 6:

$gh_1+(v_1^2)/2=gh_2+ (v_2^2)/2$
Edit 7:
$[gh_1+(v_1^2)/2=gh_2+ (v_2^2)/2\]$

Edit 8:
You are not allowed to use that image extension on this community.
Edit 9:
You are not allowed to use that image extension on this community.
What is the point of having the ability to imbed urls into images to put them in a post, if you're not allowed to do it .......
Edit 10:
CLOSE ENOUGH DAMN IT
Edit 11:
Edit 12:
You are not allowed to use that image extension on this community.
Edit 13:
Edit 14:
You are not allowed to use that image extension on this community.
I give up with this thread, on account of it being impossible to add images. The site doesn't trust me enough to add image extensions.
Edit 15:
$g' = g\frac{a}{R}$
You've got to be kidding me.............
Edit 16:
$$gh_1+(v_1^2)/2=gh_2+ (v_2^2)/2$$
Edit 17:
$gh_1+(v_1^2)/2=gh_2+ (v_2^2)/2\$
Almost..........
Edit 18:
$gh_1+(v_1^2)/2=gh_2+ (v_2^2)/2\$

YESSSSSS!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! YES YES YES!!!!!!!

Now, taking that equation, we can calculate the velocity when it reaches the center of the earth.

It's velocity would be 0 when it reaches the other side. Considering there's no friction.

I'm not doing this tonight.

Edited by Raider5678

##### Share on other sites

I would've gotten that far, but what about the time element.

studiot, I wasn't going to "abandon thread", but I'm having a 4-day internship so I'm short on time

Basically I got the same formula in a similar way by using Newton's F = G*m*m'/r² = m' * a so a = G*m/r²

##### Share on other sites

Let me try to clear up some questions.

The tunnel is straight.

It does not follow the curve of the Earth - it is not at constant depth.

It is not intended to pass through the centre of the Earth.

If you google the acceleration at depth d you will find the derivation of your choice of this formula

eg

or John Cuthber's correct but OTT link.

This is not about the shell theorem all you need is the formula.

Now I have amended the usually presented formula ( as in my link) to make things easier.

This is simply to replace the depth, d with the distance to the centre, a

Obviously a + d = R or a = (R - d)

This clearly shows the relationship between g at the surface with g' at any point p inside the Earth.

g' acts along the same line SPC as g

The shell theorem simply states at any depth d or distance from the centre a, we can remove the shaded annulus and consider gravity as though it were a new sphere of radius a.

Hope this helps

Edited by studiot

##### Share on other sites

So in order to find the correct amount of time, you'd have to integrate g' twice for dt?

##### Share on other sites

So in order to find the correct amount of time, you'd have to integrate g' twice for dt?

By figuring out the height, you can figure out the constant acceleration.

Except then the depth of the earth comes into play.........

##### Share on other sites

No integration is necessary.

Perhaps this definition will help as a hint

When particle moves in a straight line in such a way that its acceleration is always directed towards a fixed point of the line and is proportional to its distance from that point, the particle is said to move with simple harmonic motion

##### Share on other sites

Will it act like a pendulum?

##### Share on other sites

Will it act like a pendulum?

Almost let's see

Let P be any point on the straight tunnel AB and O be the right bisector of AB, through Earth centre C.

Then the geometrical relationships are as shown in the first sketch.

The acceleration due to gravity at P is g' and acts towards C as shown in the second sketch.

This therefore has a component acting along APOB, given by the triangle as also shown in the second sketch.

Substitution shows that this acceleration component is proportional to the distance of P along AB from O and by definition must be directed towards O.

This is the definition of simple harmonic motion.

So the ball executes SHM between A and B with a period

$2\pi \sqrt {\frac{a}{g}}$

Travel time from A to B is half this or

$\pi \sqrt {\frac{{4000*5280}}{{32}}} = 100\sqrt {66} {\rm{seconds}}$

Note that this is like the pendulum, independent of the Amplitude, (ie the distance along y) but depends upon the movement arm.

Perhaps you can also see why the use of a, the distance to the Earth's centre is better than the depth, d.

That is the High School derivation, if you really want to do a double integration then the equation of motion is

$\frac{{{d^2}y}}{{d{t^2}}} + \frac{g}{a}y = 0$

##### Share on other sites

I don't know about the answer but that tunnel will defenatelly solve the global CO2 problem xD since it's heavier then air it should theoretically collect in the tunnel. However due to this we might even experience a huge pressure drop xD Anyway Ill stop overthinking cuz itll lead to bad things