Voyage to The Bottom of the Earth [and back]

[bonanova]

Suppose you dug a frictionless tunnel straight through the Earth.

Then suppose you jumped into the tunnel.

Ignoring the inconvenient fact that this might turn you into a large cinder,

and ignoring wind drag [you evacuated the tunnel and encapsulated yourself]

when would you next see the light of day?

Would it matter whether the tunnel went through the Earth's center?

To keep it simple, assume the Earth is spherical and homogeneous.

Except of course for the hole you dug through it.

Assume if you like that gravity's acceleration at the surface is 9.8 m/s²,

and Earth's circumference is 40 million meters.

[EventHorizon]

One useful result is that inside a hollow shell, you feel no gravity. Here's a description of a proof of it (taken from

http://www.merlyn.de...vity1.htm#FiSSh):

Consider an arbitrary point X inside a uniformly-charged spherical surface, and consider both an arbitrary small element of "charged" surface area A and that second small element of area B which is marked out by straight lines from the edge of the first area through the point to meet the sphere again on the opposite side of the point.

The solid angles subtended by A and B at the point X are equal, and the directions XA and XB are opposite. Simple geometry makes it clear that the areas A and B (which have the same "tilt") and hence the charges are proportional to the squares of their distances from the point X. Their fields therefore cancel at that point; so there will be no net field from the whole set of such areas which covers the whole sphere, for any point X.

The argument is correct; I think I have expressed it correctly. Ramsey (s.3.2) says that Newton used it, in Principia Proposition LXX; indeed, it is in Hawking's cited book, p.880.

This means that you can completely ignore the parts of the earth further than you from the center of the earth, so effectively you feel the gravity of standing on a smaller earth.

The effective mass of this smaller earth would be d*(4/3)pi*r₂^3, where d is the density of the earth (assumed in the OP as homogeneous) and r₂ is your distance from the center of the earth. Notice that this effective mass is equal to the mass of the earth multiplied by r₂^3/r^3.

Plugging the new mass and radius into the equation to calculate gravitation gives

f₂ = Gm_E(r₂^3/r^3)/r₂^2 = (r₂/r)*Gm_E/r^2 = (r₂/r)*9.8m/s^2

So now we've got the gravity we can plug into some calculus.

Here's the setup for a tunnel through the center of the earth.

d²x/dt²= -9.8*x/r

dx/dt (0) = 0

x(0) = r

where r = (40*10^6)/2pi.

(time to pull out my old diff eq textbook... since I remember essentially nothing from that class.)

x''+(9.8/r)x=0

roots of the characteristic equation are +/- sqrt(9.8/r) * i.

x = c1* cos(sqrt(9.8/r)*t) + c2* sin(sqrt(9.8/r)*t)

The equation goes through half an oscillation (the answer we're looking for) in pi/(sqrt(9.8/r))

=pi*sqrt®/sqrt(9.8).

=pi*sqrt(40*10^6/(2pi)))/sqrt(9.8)

=2532.0769 seconds.

=42.2 minutes

I think I'll wait a bit to look at a tunnel not going through the center of the earth.... because I'm not sure I have much confidence in what I just did.

But if it is right, falling through a vacuum to the center of the earth would take 20 minutes. That's kinda interesting.

Does that sound plausible to anyone?

[EventHorizon]

One useful result is that inside a hollow shell, you feel no gravity. Here's a description of a proof of it (taken from

http://www.merlyn.de...vity1.htm#FiSSh):

This means that you can completely ignore the parts of the earth further than you from the center of the earth, so effectively you feel the gravity of standing on a smaller earth.

The effective mass of this smaller earth would be d*(4/3)pi*r₂^3, where d is the density of the earth (assumed in the OP as homogeneous) and r₂ is your distance from the center of the earth. Notice that this effective mass is equal to the mass of the earth multiplied by r₂^3/r^3.

Plugging the new mass and radius into the equation to calculate gravitation gives

f₂ = Gm_E(r₂^3/r^3)/r₂^2 = (r₂/r)*Gm_E/r^2 = (r₂/r)*9.8m/s^2

So now we've got the gravity we can plug into some calculus.

Here's the setup for a tunnel through the center of the earth.

d²x/dt²= -9.8*x/r

dx/dt (0) = 0

x(0) = r

where r = (40*10^6)/2pi.

(time to pull out my old diff eq textbook... since I remember essentially nothing from that class.)

x''+(9.8/r)x=0

roots of the characteristic equation are +/- sqrt(9.8/r) * i.

x = c1* cos(sqrt(9.8/r)*t) + c2* sin(sqrt(9.8/r)*t)

The equation goes through half an oscillation (the answer we're looking for) in pi/(sqrt(9.8/r))

=pi*sqrt®/sqrt(9.8).

=pi*sqrt(40*10^6/(2pi)))/sqrt(9.8)

=2532.0769 seconds.

=42.2 minutes

I think I'll wait a bit to look at a tunnel not going through the center of the earth.... because I'm not sure I have much confidence in what I just did.

But if it is right, falling through a vacuum to the center of the earth would take 20 minutes. That's kinda interesting.

Does that sound plausible to anyone?

Lets see what constant gravity would be required to get to the center of the earth in the 21.1 minutes I figured it would take to get there.

40*10⁶/2pi=.5g(2532/2)²

g=7.944 m/s²

You'd be moving slow through areas of higher gravity so you'd be accelerated quicker and be effected less by the lower gravity near the center of the earth since you'll be going quickly at that point. I'd think the equivalent constant gravity would be somewhat significantly more than half of surface gravity. ~80% doesn't seem to me like it would be too far off.

[psykomakia]

wouldn't gravity then keep you at the center of the earth? It's not like you'd fall all the way through the earth to the other side.

And since the question is

when would you next see the light of day?

The answer is: you wouldn't. Until you climbed back out.

Which is why it doesn't matter if

the tunnel went through the Earth's center

[fabpig]

Many many years ago, a friend asked me the very same question. When I 'passed', he gave an answer pretty-well the same as EventHorizon's.

His explanation was along the lines:

Travelling through the earth governed by gravity is much the same as orbitting the earth: the same principles apply. As Yuri Gagarin's first orbit was only about 200km (ave) above the surface, it should take about the same time to go through the earth as he took to do a half orbit.

I've checked the times out, and they do seem to tally.

Comments anybody?

[TheChad08]

wouldn't gravity then keep you at the center of the earth? It's not like you'd fall all the way through the earth to the other side.

And since the question is

The answer is: you wouldn't. Until you climbed back out.

Which is why it doesn't matter if

Here's the thing, no friction, no resistance, no etc. So assuming just gravity, the force on pulling you for the first half is the same force acting on you in the second half. With no resistance you can go from surface to surface, then you'll get pulled down again. This would be perpetual. In order to determine the time, simply figure out 9.81m/s^2 to the center of the earth, then double that.

[EventHorizon]

Many many years ago, a friend asked me the very same question. When I 'passed', he gave an answer pretty-well the same as EventHorizon's.

His explanation was along the lines:

Travelling through the earth governed by gravity is much the same as orbitting the earth: the same principles apply. As Yuri Gagarin's first orbit was only about 200km (ave) above the surface, it should take about the same time to go through the earth as he took to do a half orbit.

I've checked the times out, and they do seem to tally.

Comments anybody?

It takes the space shuttle 90 minutes to orbit the earth. Half of this orbit is 45 minutes, which would be the time to get to the other side of the earth. Removing the slight atmosphere encountered in LEO may make up for the 3 minute disparity (Edit: not to mention the other assumptions and imprecise numbers used).

As an aside, I read that because the density of the earth increases as you go down, you'd actually experience more gravity for a little while before it decreases, but we assumed homogeneity.

"when would you next see the light of day?"

It just came to me that if it was day on one side of the earth, it would be night on the other, so you'd need to make the return trip too. Unless it was 12 noon and the sun is directly over the tunnel... then it would be shining all the way through . It would be interesting to see a tunnel like that at night with the sun shining from beneath you.

Or it could just be the 42.2 minutes if the tunnel is perpendicular to the direction to the sun.

Or you could just say you always see the light of day, but in the form of the "light at the end of the tunnel."

So now we just need to determine if it matters if the frictionless tunnel goes through the center of the earth or not.

[EventHorizon]

Let x be the distance to the center of the tunnel.

Let z = the angle between the direction to the center of the earth and the center of the tunnel.

This means cos z = x / r₂, where r₂ is the distance to the center of the earth.

x / cos z = r₂, by just moving things around.

The gravity at r₂ is r₂*9.8/r in m/s², where r is the radius of the earth.

So when you are x away from the center of the tunnel, the gravity downwards (towards the center of the earth) is (9.8/r)* x / cos z.

To get the gravity in the direction of the tunnel, multiply by cos z, resulting in (9.8/r)*x.

So, same characteristic equation, same roots, and the resulting function's period is the same as before.

So it doesn't matter that the tunnel doesn't go through the center of the earth. It'll still take the same time.

What this means is that in a quazi-futuristic world where there are (essentially frictionless and perfectly straight) tunnels through the earth everywhere, it'll take about 42.2 minutes to get from anywhere to anywhere else via a tunnel (assuming the presence of the many tunnels doesn't effect gravity). So even if you want to move 3 feet to the left... if just by gravity, it'll take 42.2 minutes.

Also, if you sit on the edge of a completely flat and frictionless ice rink (that is exactly perpendicular to the direction to the center of the earth at the center), you will move from one side to the other side of the rink in 42.2 minutes by just sitting there. Or if you get the right initial momentum... you could orbit the center of the rink in the same time (Edit:42.2 minutes to get to the other side, 84.4 for one whole orbit) (as is obvious from the math so far, the gravity in the x direction simply depends on the distance away from 'center' of the x axis, which also applies to a frictionless orbit of the whole earth).

Kinda cool.

[bonanova]

fabpig, you're right on.

The same equations govern low/earth orbits

Just as EventHorizon showed the tunnel doesn't need to be an Earth diameter.

A surprising, non-intuitive result.

1. It is related to the find the number thread whose answer is 420.

Or as Y-San put it, ten times the answer to life, etc.

Which of course came from the hitch hiker's guide to the galaxy.

So 42 is also related to hitch hiking through the Earth, so to speak

2. It certainly did not originate with me.

It was first proposed about 400 years ago to Newton, no less!

Forgotten and rediscovered a while back by a mathematician.

Read more by Googling "Gravity Train."

And now that this wonderful number is out of the bag, so to speak,

Anyone want to take a guess what all the rows and major diagonals

of a 3 x 3 x 3 magic cube [using the numbers 1-27] sum to?

[Yoruichi-san]

fabpig, you're right on.

The same equations govern low/earth orbits

Just as EventHorizon showed the tunnel doesn't need to be an Earth diameter.

A surprising, non-intuitive result.

1. It is related to the find the number thread whose answer is 420.

Or as Y-San put it, ten times the answer to life, etc.

Which of course came from the hitch hiker's guide to the galaxy.

So 42 is also related to hitch hiking through the Earth, so to speak

2. It certainly did not originate with me.

It was first proposed about 400 years ago to Newton, no less!

Forgotten and rediscovered a while back by a mathematician.

Read more by Googling "Gravity Train."

And now that this wonderful number is out of the bag, so to speak,

Anyone want to take a guess what all the rows and major diagonals

of a 3 x 3 x 3 magic cube [using the numbers 1-27] sum to?

I like this problem...I recall running into it in a physics class . But I think the answer is intuitive...if you think of gravity as, like, invisible rubber bands pulling on the object, from all particles or chunks of the earth, then when you're on the diameter, there is equal pull from both sides, which balance each other out, but if you're closer to one edge, there is more pull from one side, and more pull -> more force -> more acceleration.

As for the magic cube...no idea .

Edited July 1, 2012 by Yoruichi-san

[fabpig]

In theory, then, an object placed at one end of the ice rink (assuming zero friction) will oscillate every ~84 minutes. Now that IS kinda cool, and I think somebody should set up the experiment.

[bonanova]

In theory, then, an object placed at one end of the ice rink (assuming zero friction) will oscillate every ~84 minutes. Now that IS kinda cool, and I think somebody should set up the experiment.

Just a couple things to arrange, in principle.

It would have to be very level, no wind currents, loud talking, foot stomping, and ice made from an inviscid liquid. That should do it. Oh, to eliminate that troublesome coriolis effect, stop the Earth from rotating for about an hour and a half. All most as much fun to think of what things would disturb the experiment. I'm sure there are others.

Nice idea.

[fabpig]

Great. How does Thursday suit you?

[ujjagrawal]

As you fall through a hole through Earth, your acceleration will progressively decrease as you approached the geometric center (your will be momentarily weightless there) with speed close to 7900 m/s. And the gravity force will cause you to oscillate back and forth through the center of Earth like a mass bobbing up and down on a spring. The period for this oscillation are

T = 2 pi sqrt(m/k) , where K = spring constant = mg/R_EARTH (F_Gravity = - mg x/R_EARTH = -Kx , x = displacement from center)

T = 2 pi sqrt (R_EARTH/g)

Substituting, R = 40 x 10⁶ /2 pi = 6.37 x 10⁶ m and g = 9.8 m/s²

T = 5063 sec = 84.4 mins

Thus, you will pop up on the opposite side of the Earth after little more than 42 mins, but unless you grab something to hold on .

Edited July 2, 2012 by ujjagrawal

[bonanova]

Also, if you sit on the edge of a completely flat and frictionless ice rink (that is exactly perpendicular to the direction to the center of the earth at the center), you will move from one side to the other side of the rink in 42.2 minutes by just sitting there. Or if you get the right initial momentum... you could orbit the center of the rink in the same time (Edit:42.2 minutes to get to the other side, 84.4 for one whole orbit) (as is obvious from the math so far, the gravity in the x direction simply depends on the distance away from 'center' of the x axis, which also applies to a frictionless orbit of the whole earth).

Kinda cool.

What a great Freshman physics problem that would make!

Trying to think of a simple pendulum that would emulate that same motion.

By analogy and reasoning, and not calculation, what might its length be?

=============

Also:

Suppose a New Yorker wanted to get to Los Angeles, say it's 3000 miles distant, by gravity ASAP.

We know she can get there in 42+ minutes.

Is that the optimal time?