# A trampoline-esque mix of holiness and darkness......maybe

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Hey kids,

Not sure if I have raised this 'query' before or if anyone else has on this forum (astucally i think i have but can't be bothered to look for it)........anyways here it is:

Say that all of a sudden a hole appeared in the Earth, a hole big enough to stretch from the good old U.K. right through to the Pacific, right through the core of our lovely planet.....

Now, (i know that all you people with adenoids will be going "ooh, thats scientifically incorrect, you can't do that...."etc etc)...relax, its a 'what if' scenario, now say one were to jump into the hole......the big question and crux of this post is:

would you get stuck in the middle, in say a kind of floating manner?

My own opinion is that after the initial fall (lets say from the northern hemisphere down to the southern) that you would make some indentation into the southern hemisphere and then be repelled back in the other direction and so forth......now due to loss of energy (bit like those swinging balls) you would eventually end up soemwhere around the middle just rotating at some unknown speed in a non-gravity manner...bit like an iron filing in a magnetic field.

Anyways, enjoy.

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bit like those swinging balls

Should say like a Newton's cradle ....... just to stop the "ooh-err Matron brigade"

but that sums it up for me

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ha ha, i prefer swinging balls!

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assume the earth to be a perfect sphere.

I assume you are meaning that the fall is radial...ie on a vector passing from you through the earths centre.

the spherical symmetry implies that the mass at a lower radius than you can be considered a point source.

you would feel a gravitational acceleration given by g= GM®/r^2

note that M® is the mass contained between r=0 and you....this changes as you fall to ward the centre.

The acceleration produces a velocity v= S adt and noting that dt= dt/dr * dr, and dt/dr =1/v we can substitute, I think and find

v^2= S a*dr where S is taken to be an integral sign

ie v^2= S (GM®/r^2)*dr

once a relation for M® is found the integral can be done.

but basically you accelerate reaching higher and higher speeds until the centre is reached. At this point there is no more mass interior to you and the acceleration stops.

Youre velocity remains constant so you pass straight through the core and start to climb back toward the surface. You are now doing work against gravity, so you are constant being decelerated, until, if the earth is a perfect sphere, and your journey perfectly symmetric, you would reach the surface again with a perfect zero velocity.

I suppose you would then start to be sucked back, repeating this motion for eternity in a perfect frictionless journey. If friction is present you would end up in the place of lowest potential...the centre, like you say.

With no forces here, the motion with which you enter the centre might be maintained.

Thats my point of view.

Perhaps to mathematical?

ps you can solve the integral via..knowing the density profile of the earth....I dont know that.

but stars behave somewhat as....p®=p0*(1-r^2/R^2) where R is the earths radius, and p0 the core density.

then dm= 4pi*r^2*p® dr

thats only for the insane, or physicists amongst us! ed, ally, me?

this you could use to work out the speed.....which isnt particularly interesting.

Paul

hope we all followed that?!

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i am sad. I have done the integrals to see how fast would would be moving as you passed through the centre.

first thing knowing mass of the earth, assuming this profile, gives us the density of material at the core.

M= S 4pi*r^2*p0*(1-r^2/R^2) dr......this is evaluated over the entire radius of earth. this yields the mass of the eart (since the whole sphere was covered by the integral. But we know the mass from measurement. It is 6x10^24 kg

then after integrating p0= 15M/( 8pi*R^3)....equal to 13660 kg/m3. Ie 1 cubic metre weighs 13.66 tonnes...thats the core density

we need to know how the mass varis with distance. this has been found already via the density relation I arbitrarily invoked.

m® = 4pi*p0*(r^3/3 - r^5/5R^2)

and plugging in to the equation for v^2 in the previous post, we find, after integrating once more.

v^2 = 8pi/15 * G* p0*R^2

which implies that, at the centre of the earth, you would be moving at 7.907km/s...about 17,800 mph...a little less than the escape velocity of earth, 11km/s

once you have picked your density distribution or profile...it could be constant, or could vary as 1/r, or 1/r^2 (like mine) or strange like sinusoidal, the integrals are actually starightforward....seriously. A bit of calculus and these can be done, and the 1/r^2 one mimicks stars reasonably well. meaning the core density and pressure can be found, and temp...and luminosity profile. Good astrophysics!

sorry Ed, hope this overly mathematical appoach doesnt detract from your post....

Cheers

Paul

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er...yeh, i agree with that lot Paul.

(just left u a msg on faceybuk)

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just to throw a spanner in the works, wouldn't the "hole" be filled with atmosphere? and if so wouldn't it's density increase pretty substantially with an extra 4000 miles aprox of preasure on it? i'd have thought that a human's terminal velocity in those densities would be far lower than the normal sky diving speed, after all when whatever his name was jumped from that balloon at the edge of space he was going over 600 mph in the thin air before slowing as he got lower. you'd still end up at the center but it might take longer to get there.

(sorry for being vague on names but i'm knackered)

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dont agree with me.... challenge me. its how we learn. After all we are scientists. We love challenging others theories, and defending our own to the death!

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absolutely, but cant deal with air friction. But youre right.

my answer holds in a vacuum?!!!

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This happened in Sci Fi - the object ended up in the centre after bouncing a lot. Given the 'ringworld feedback effect' one assumes the maths was correct.

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As stated by Paul, you go right through the Earth to the other side, accelerating towards the centre then decelerating as you rise back to the surface on the other side. Then you fall back down and the process continues. Furthermore if you work out the period of oscillation it comes out the same as the time for Earth orbit. It is of course not very realistic.

A similar idea was taken up by Lewis Carroll in Sylvie and Bruno, as a way of running trains without fuel. Imagine a perfectly horizontal tunnel (ie one perpendicular to Earth's radius). The midpoint of the tunnel is closer to the Earth's centre than the ends, so objects will roll towards this midpoint, then continue on towards the end and roll back again.

The period of oscillation (as a function of tunnel length and depth) is left as an exercise for the reader...

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acey, i didnt know that the period of your oscillation would be equal to the period of earths orbit.

the second bit sounds trickier, but i may be bored later!

paul

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now say one were to jump into the hole......

Wouldn't you hit the side of the hole due to the Coriolis effect?

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now say one were to jump into the hole......

Wouldn't you hit the side of the hole due to the Coriolis effect?

It's hypothetical!

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Better make the hole go straight down Earth's axis of rotation. Otherwise yes, I guess you'd hit the side!

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• 3 months later...

Haha, interesting and crazy question. I like it! If there is a hole from one side of the Earth surface through the core to another surface of the Earth, maybe we can assume that the gravitational force in the hole same as the between of 2 similar planets which have half of the mass of Earth with same density, and others feature of the Earth. The distance between this 2 planets might be dependent on how big the "imagined hole" is.

Kenny

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Without sounding stupid could this be why things don't escape black holes, they just get stuck in limbo in the middle????

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Also which way would water go, clockwise or anti clockwise as it drained through the hole???

Sorry I know,,,,,another silly question....

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Great question. We did this problem in high school while studying Simple Harmonic Motion.

Again interesting quetions raised by

Yeah, you would hit the side of the hole.

2.And tinvek about terminal velocity if the hole was filled with atmosphere:

Challenge to this view - Air pressure will not increase as you would expect. Gravity reduces as you go deeper. So air pressure may actually decrease as you approach the centre and will fall to zero at the centre. So you may not have a terminal velocity, but rather non-uniform acceleration.

Challenge my view please. Its what we do best :-)

Thanks,

Vincent.

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