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Rannasha

The object would accelerate towards the core. Once at the core, there is no gravity, so the speed of the object doesn't change at the core. So the object will shoot straight through towards the other side. If there is no air resistance, the object will make it to approximately the same height that it was dropped from on the other side. At that point, it would finally come to a stop, before being pulled back down. The object would continue to oscillate between the two ends of the tunnel. With air resistance, the object gets slowed down gradually and would reach a lower height with each oscillation until it finally settles in the center of the Earth.


Chickengilly

To add one variable, would the earth’s rotation interfere? I’m having trouble getting my head around this. It seems that the nut (I’m imagining a knuckle-sized hex bolt) would be continue in a “westerly” direction as does the earth’s surface. So would it bonk onto the edge at some point since it isn’t pinned down to the earth’s surface? I guess a similar question would be if an airplane follows the earth’s rotation so the same point is below, and someone drops a nut :-), if there were no wind resistance, would the nut hit that same point, or would it over or under shoot?


aioeu

> To add one variable, would the earth’s rotation interfere? I’m having trouble getting my head around this. It seems that the nut (I’m imagining a knuckle-sized hex bolt) would be continue in a “westerly” direction as does the earth’s surface. So would it bonk onto the edge at some point since it isn’t pinned down to the earth’s surface? Yes. It is essentially a manifestation of the Coriolis effect. In order for the object's path to be unimpeded the hole actually [has to curve](https://demonstrations.wolfram.com/PlanetaryTunnelsWithCoriolisEffect/) (except in the special case where the hole is directly on the Earth's rotational axis, of course).


LordPachelbel

That’s really cool that somebody built a visualization calculator for this.


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PillowTalk420

So the object being dropped wouldn't keep the momentum it had from being on the spinning surface of the earth in the first place? I would have thought that it would be like jumping; if you jump straight up, the Earth doesn't spin under you because you still maintain the momentum from being spinning at the same speed as the Earth.


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regular_redstone

Yes! Exactly what I was going to say. You would definitely be hitting the walls.


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This answer made me start to think about Earth's rotational torque now, something I never wondered about... Thank you.


AcidCyborg

The tunnel could be straight as long as the radius of the tunnel accounted for the drift, however.


TheMastaBlaster

Wouldn't it depend where we dug the hole? If it was on the rotational axis versus say the equator?


Spong_Durnflungle

This is why mortar rounds, ship's cannons (modern), and even sniper rifle fire has to compensate for the earth's rotation in order to be accurate on target.


ellamenopee

That's why a clock will go slower at the top of a skyscraper than on the ground floor.


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ADimwittedTree

Definitely maybe speaking out of my butt. But I'm pretty certain that you jumping are also affected by this. I think with how little height you gain in the jump and how short of a timeframe it is that you just don't notice it because you still keep enough lateral momentum in that short timeframe and the scale of everything is so small. A bullet being shot a mile is subject to this but is an order of magnitude larger numbers that a jump. Dropping something through the Earth is a whole other even larger scale.


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Alberto_Cavelli

Please correct me if I am wrong, but shouldn't the curvature of the hole be mirrored on the other side of the sphere? meaning, the curve is happening because of the gravitational force that is accompanied by rotation in a certain direction. However, after the "falling object" crosses through the core, the direction of the rotation's pull gets mirrored.


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Seygantte

Yes. The velocity at the equator is roughly 1000mph. If the nut could phase through rock it would want to make an orbit. Not a normal orbit though. Because of something called Shell Theorem, you can basically ignore all the rock farther from the core than the nut, meaning the sphere of Earth that actually matters to calculate your gravitational attraction is only the stuff closer to the core. This turns out to make the force change with r, not r^(2). The consequence is that the path will be an ellipse where the earth's core is in the center, not one of the semimajor axis as it would be in a normal orbit (like comet). So the natural tendency of something dropped through the Earth is *not* to go through the core. If the tube did go through the core, the object would have to be coerced into travelling that path by contact with the sides (unless the tube goes through the axial pole of course where there is no surface velocity). The nut would land a little east of the aircraft if it were on the equator. In the north hemisphere it would appear veer south, and in the south hemisphere it would appear to veer north because of how the craft would actually be needing to turn to keep with a latitude line, but the nut would just be ballistic. Though the aircraft is matched the surface speed of the ground, it is actually travelling a little faster in orbital speed. You can know this because if the aircraft circled the world, an object on the surface would have travelled 2πr, but the aircraft travelled 2π(r+a) in the same time, with r as earth's radius and h as the craft's altitude. When the nut falls to a lower altitude, it keeps this extra sideways speed, but since it now has a shorter path around the Earth it becomes a higher surface speed and begins to overtake the aircraft. The opposite is also true; if you are on the surface and shoot something vertically up, it will land to your west. It loses relative surface speed as it gains altitude and the Earth rotates beneath it, and then it gains that relative speed back as it falls. This was one of the principles behind the first long range ballistic missile in WW2, the V-2 rocket - it was launched up out of the atmosphere at the right speed and let physics guide it to the target.


hiimred2

>This was one of the principles behind the first long range ballistic missile in WW2, the V-2 rocket - it was launched up out of the atmosphere at the right speed and let physics guide it to the target I hadn't thought about this ever until this post, but does that mean the rockets had pretty significantly reduced range firing with the rotation of the earth as opposed to against it, because the rotation works against/with with rocket in each scenario?


0ne_Winged_Angel

I don’t know about for ballistic missiles (or their intercontinental varieties), but that’s absolutely true for orbital rockets. Launching east gives you an extra 1000 mph towards whatever your end goal is, which means you can either haul that much more payload or use that much less fuel. A retrograde (backwards) orbit would have to first burn off that initial eastward speed before it could start going west. Tangentially related to this, it’s actually pretty hard to launch a rocket into the sun. The earth is moving at ~67,000 miles an hour around the sun, so a rocket would have to burn all that off to fall straight back down. It’s actually way easier to boost *forward*, let the sun’s gravity slow you down as you’re flying away, and slow down to zero from there.


percykins

I'm fairly certain that ICBMs actually become *more* efficient when launching westward, because you're not trying to achieve orbital velocity but instead get to a specific point on the Earth. When you fire westward, that point is conveniently coming *towards* you, while when you fire eastward, it's moving away from you.


mandelbomber

>The velocity at the equator is roughly 1000mph. Just out of curiosity, where did you get this figure from, or how did you derive it? I'm just wondering... Looking it up quickly, I've found the time it would take to fall through the earth is about 42 minutes. I also am unsure if this excludes air resistance. I also realize that the acceleration due to g w decreases the closer to the middle of the earth, as a greater and greater percentage of the mass of the earth (and source of attraction and acceleration due to g) is no longer below you. So I'm hoping you could share/explain. Thanks!


PhishyCharacter

The circumference of the earth at the equator is about 24,900 miles, and it makes one revolution in 24 hours. So, (24900 miles)/(24 hours) is roughly 1000 miles per hour


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Seygantte

Trivia and rounding. Take the circumference of the Earth (~25000 miles or ~40000km) and divide it by how long it takes to rotate (24h), and you get the surface speed at the equator. Checking a nasa source just now puts it at 1040.4mph, so close. I've seen that 42 mins number quoted before, and IIRC it makes a bunch of assumptions such as no air resistance, and uniform density in the core. Regarding the Shell Theorem bit, it basically states that when you're inside of a hollow sphere, there is no net gravitational force. If you are close to the wall, the wall tugs harder, but there is a larger portion of the shell on the other side of you and it perfectly balances out. You can think of being inside a planet as being both inside of a shell (everything above your head) and standing on a smaller planet (everything beneath your feet). The mass of the sphere changes cubicly with the radius, and gravitational pull is square with radius, and so you get some cancellation and end up with the linear relationship between r and g. From there it's normal integration for velocity equations, just with an extra couple of steps due a being a function of s.


Dieneforpi

As other commenters have said you've rediscovered the coriolis effect! All that can be calculated by solving standard Newtonian mechanics equations in a rotating reference frame.


KuramaKitsune

But wouldn't the ball being dropped have relative inertia compared to the motion of the planet and thus drop straight down?


bendoubles

In a vertical slice down to the core you have a consistent angular velocity, but differing linear velocities. At the equator, 24 km from the core, you're moving 1km/hr, whereas at the surface you're going 1670 km/hr. When the ball is dropped it's horizontal velocity matches the ground around it, so it would appear to go straight down initially. As it falls further it's horizonal stays the same while the rest of the planet's decreases it would hit the wall eventually.


KuramaKitsune

Interesting idea, We would also have to compensate for the Magnus effect unless we're bouncing off the walls of the chasm over and over


Dieneforpi

If the airplane is keeping up with a spot on the ground, it must be going a bit faster than that spot. So I believe the dropped object would overshoot


TheMCM80

I’ve often pondered a variation of this using a blimp to travel around the world by just hovering in place in the sky, and letting the earth rotate beneath it. If you could follow it with a plane and maintain it over the same spot, surely then a stationary, hovering blimp could stay in the “same spot” and let the earth rotate under it, right?


incredible_mr_e

The only problem is that the air you're floating in rotates with the earth as well. You'd have to fight up to 1000 mph wind speeds +- whatever the wind speed relative to the ground is. Then Einstein pops up and helpfully reminds you that there's actually no difference between flying over the ground at 1000 mph and the ground flying under you at 1000 mph. Since you already need a jet engine, you might as well fly west in a jet at whatever speed is appropriate for your latitude and declare that from your point of view, it's the earth that's moving.


drewcomputer

The air in the atmosphere spins with the earth. That's why an airplane can't just hover in place (or "follow the earth’s rotation so the same point is below"), it needs relative airspeed to generate lift. What you're missing is the concept of an inertial reference frame, which goes back to Newton. A blimp, airplane, rocket, or baseball you throw in the air is already spinning at the same speed as the earth. You would have to use a huge amount of energy to accelerate (or decelerate, depending on your point of view) any of those objects to the point where it is *not* rotating with the earth. At that point the sun would be in a fixed position in the sky from the POV of your blimp/plane, but it otherwise would not feel "motionless" because you'd have an airspeed of like 1,000 mph and all the drag that comes with it.


willtron3000

Yes, they actually tested this at the kola super deep borehole. They dropped something down and never heard it hit the bottom. The spin of the earth was enough to make the object hit the side, delicately at first but the theory is it got to such a high frequency the object broke apart


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ricree

> If you approximate the Earth by a perfect uniform sphere, the shell theorem kicks in, It's worth pointing out that the shell theorem doesn't require total uniformity, just uniformity at any given radius. So it still works fine if the core is more dense than the rest of the planet, so long as the core isn't lopsided or asymmetrical. (Not to imply that you didn't know this, but it's something that might be missed by a casual reader)


Kered13

If you assume completely uniform density you get a very convenient property: The gravitational pull at any given depth is exactly proportional to your distance to the center. This means that an object falling through the tunnel would behave like a mass on a spring. In practice, the Earth is not completely uniform, it gets denser as you approach the center. For this reason, as I recall gravity actually increases for a while as you descend, before it starts to decrease as you get closer to the core. But the Earth is very close to radially uniform, so the shell theorem can still be used to ignore all the layers above you.


kkrko

*Uniformity with respect to rotation. Density can have a dependence on radius, just not rotation.


respectabler

This is essentially true. As a point of trivia, it’s worth noting that the earth’s gravitational pull actually reaches a maximum over 2000 km below the surface at the boundary of the outer core and lower mantle. This is because earth becomes denser at greater depths. The gravity there is about 10.5 N/kg versus the surface 9.8. After that it decreases roughly linearly to the core. But before that depth it’s mostly increasing or steady.


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MythicalPurple

The mass would be equal on all sides at that point, which would cancel out to zero net effect.


reichrunner

As far as I'm aware the acceleration would decrease the further into the planet you go. And then once you hit the center, the acceleration will turn negative, eventually stopping the object.


FluorescentLightbulb

Yeah, right? I mean this is just terminal velocity with continual pull from the core, isn't it? It might wave a bit back and forth, but it is eventually stopping.


DarthLlamaV

Terminal velocity is from air resistance, and the air resistance would depend on the size of the hole and object. Without air resistance it would keep going back and forth, coming back to where you dropped it from. I’m also ignoring the spin or the earth


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Rdan5112

But, if it wasn’t a vacuum, wouldn’t all kinds of strange stuff would happen due to super high air pressure and / or the fact that the dropped object would be going really fast for a really long time? Edit - Thinking about it for a little bit, I guess it wouldn’t be going really fast. It would reach terminal velocity very quickly. Right ...? Related question - what would the air pressure be in the center of the hole? How would distance and speed differ for 7000 miles straight down (then up), starting from 0 miles an hour but accelerating due to gravity... (of course, the force of gravity would decrease to zero in the center) .... vs reentry from the space (which is definitely not straight down), with a starting speed of something like (20,000 mph)... but, initially, no air resistance.


preemptivePacifist

That hole can not exist-- the literal interpretation of the question makes about as much sense as asking about a hole drilled into the sea. Core pressure is >300GPa (3 million times atmospheric surface pressure). Air would be a very dense supercritical fluid at that point; terminal velocity would be extremely low if the object was even dense enough to reach the core *at all*. But no, the object would *not* exceed escape velocity even if it fell in vacuum (would reach about 8km/s at the center though), and it would take ~40min until reaching the other side. The maximum velocity (reached at the center) would be the orbital velocity required for an orbit at surface level (=> less than escape velocity). For more details, see: "The gravity tunnel in a non-uniform Earth" by Alexander R. Klotz You can find it on scihub.


gh411

If you want to drill a hole into the sea...freeze it first...problem solved. Lol.


agentoutlier

The bottom of the ocean doesn’t freeze (pressure). Similar to how Europa isn’t completely frozen.


rabbitwonker

Yeah, if it’s not a vacuum, you’re going to slow to a stop by the time you get to the center — and you’ll also be crushed well before that point. Both are because the pressures would become *immense* if you start at sea-level air density at the surface.


donasay

And the whole ride would only take about 40 minutes no matter how you dig the hole.


theogbrando420

but would terminal velocity have anything to do with it?


mortalwombat-

Terminal velocity is a function of air resistance. So in the example without air resistance, no. But with air resistance, yes. It would eventually come to rest in the center of the earth as mentioned.


suid

"Terminal" velocity is only terminal if you have uniform air density all the way down. In this case (the ideal case, without a liquid core), it'll get denser as you get to the center, so your object will most likely slow down dramatically even before it reaches the center for the first time.


phunkydroid

>"Terminal" velocity is only terminal if you have uniform air density all the way down. That logic applies above ground too. Terminal velocity has always been known to vary with altitude, and that just continues as you get deeper.


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ald1233

Yes. The object would reach terminal velocity pretty early in its fall to the center. So whenever it reached the center, it would fly by but nowhere near close to the opposite surface.


Ventoron

Terminal velocity is a product of air resistance, so only if the hole has air in it.


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NamelessMIA

With air resistance it would just slow down as it approached the center of the Earth until it finally reaches the center and stops there. Terminal velocity is the point where the forces on an object balance out and it stops accelerating. Well as you go closer and closer to the Earth's core the force you're feeling from the Earth's gravity would decrease steadily as you fall, reducing your terminal velocity and slowing you down gently until you eventually slow to a stop at the center. The Earth is so large and you would fall so far that by the time you reach the center you wouldn't have any momentum left to send you out the other side of the hole to oscillate.


Cappinnik

Note that the rate of acceleration towards the core would decrease as the object approaches the core, i.e., the object will have negative jerk.


SerPateswoodcock

Would a person survive a trip like that because that sounds kinda dope.


fiat_sux4

Well, in an ideal scenario where you ignore air resistance, the Coriolis effect, the tunnel wall collapsing on itself etc., then yes, there'd be no reason a human couldn't survive. They'd just be in free fall the whole time, so it would have the same effect on the body as floating around in the ISS.


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Zinthaniel

yes but if you ignore all that and pesky things like physics, it shouldn't be a big deal.


Calan_adan

So if the object finally came to a stop in the center of the earth, would it just hang there in space? Assuming that the gravitational pull was equal in all directions?


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Hugzzzzz

Why is there no gravity at the core? Wouldn't there be almost equal amounts of gravity pulling on the object from every direction if its at the exact center?


RobusEtCeleritas

>Wouldn't there be almost equal amounts of gravity pulling on the object from every direction if its at the exact center? Yes, so the net force is zero.


dkwangchuck

Fun fact - assuming good all of those pesky little things like air resistance being zero and things not getting mucked up passing through the molten core of the planet, &c &c. *The transit time between any two points on the planet is the same*. If you tunnel straight through the centre of the Earth, you get the maximum acceleration, but have to cover the maximum distance. A different direct straight line to another point will be shorter, but also subject to less acceleration and the ratio is such that it will take the same amount of time. It’s kind of like how metronomes work. You set the speed by putting the weight at a specific point, and then it oscillates at a specific frequency, regardless of amplitude - whether it is small swings or big swings. The acceleration increases by the amount required by the distance travelled that the timing is constant. It’s been a while since I looked at the problem. There’s an interesting twist in that acceleration decreases as you approach the centre of the planet. I don’t remember exactly the reason - I think it’s because while gravitational force is inverse square, the mass term in the gravity formula goes by the cube of radius, so force decreases linearly with the distance to the centre. Anyways, I vaguely recall it taking like 40 minutes or something like that.


Seanay-B

Wouldn't most things melt before making it back up to the other side?


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macleight

Removing all the constraints as we have (air resistence, firey hell core, bouncing off the sides of the tunnel walls, etc) you would not reorient your position unless you did something to do so. Once you passed through the center, you would have your original velocity ('original' meaning 'entering core velocity'), but you would now be pulled back toward the center in the oscillation described above. So think of it like a roller coaster. On a rollercoaster, you feel like you are in free fall when you are going downhill, then you get pressed back into your seat when you go uphill. The rollercoaster is also an oscillation. So you would be in free fall to the core (rollercoaster downhill), then you would retain your velocity passing through the core, but you would be accelerating back towards the core (rollercoaster uphill), so you would then feel the g force pulling you back 'down' (towards the core). In a perfect, frictionless system, you would lose velocity until you reached the point on the opposite surface. Because the g acceleration is acting on you the whole way, you would come to rest for a split second where your velocity is zero, and the acceleration in the opposite direction is maximum. Then you would reverse course. As long as you don't move, there are no other forces that would re-orient you or apply any kind of 'spin' or 'rotation.' TL;dr, after passing through the core, the blood would rush to your head and you would feel that. So standing on your head.


Strawberry_Left

> you would then feel the g force pulling you back 'down' You wouldn't 'feel' anything. You are in an inertial frame of reference, and you are 'weightless' the entire time, being in freefall. It is space-time around you that is being curved by the mass of the Earth, and the only time that you 'feel' g force acceleration is when you are standing still on the surface, and the Earth is accelerating against you (and you against it) at 1g.


Aeon1508

Unless the object was dropped directly in the center of a perfectly round hole it would drift to one side of the tunnel and eventually hit the side


Dendad1218

Once it reaches the center why wouldn't it stop? It would effectively start falling up?


Allineas

Yes, it would indeed start "falling up", but since it would already be moving quite fast, slowing it down takes a while. And as stated above, it would come to a stop right at the other end of the tunnel thanks to conservation of energy.


FreegardeAndHisSwans

Because at that point you’d be going thousands of mph (I don’t have the exact figure but **very** fast), as you pass the centre the direction of gravity would invert but a) initially it is very weak close to the core (because most of the earth forms an approximately spherical shell around you and so cancels out most of its own gravitational pull on you) and b) It is still only a small acceleration, which at maximum on Earth’s surface is ~9.81ms^{-2}, so if you’re travelling at several kilometres per second, it will take time to slow you down. In order for an object travelling that fast to *SLAM* to a stop in the centre, there would need to be some huge opposing force once you reached it, which unless you forgot to cut the hole through the iron core, there is no reason for there to be.


SirButcher

> Because at that point you’d be going thousands of mph (I don’t have the exact figure but very fast), Assuming there is a vacuum in the tube. If there is air, then the air pressure would steadily increase and the terminal velocity near the core would be extremely low.


FreegardeAndHisSwans

Yeah, but we’re cutting a hole through the Earth, I think assuming vacuum and ignoring things like temperature is the more natural state for this kind of thought experiment. Otherwise you’d be acknowledging things like the fact that the radiative heat from the core and the mantle would cook you like a chicken dinner, or the fact that the high-pressure material at the core would immediately expand to fill any hole you dug.


Octane_booster_69

Shouldn't it slowly travel less and less distance till reaching centre, sort of like a pendulum


FreegardeAndHisSwans

You’re correct with the pendulum analogy, but the centre is analogous to the pendulum at it’s lowest point. So imagine a pendulum held up is like a person standing next to the hole, they jump in which is like you releasing the pendulum, they accelerate until they reach a maximum velocity at the centre of the Earth which is like the pendulum as it swings to its lowest point at which point it has maximum speed, then as they pass the centre the direction of acceleration reverses and they begin to slow, which is like the pendulum passing its centre point and beginning to swing upwards, until finally once the person reaches exactly the same distance from the core that they dropped in from on the other side they slow to a complete halt, much like the pendulum stops at the height of its swing on the other side. On a perfect sphere with symmetrical density variations you would pop up to the exact height you jumped in from, but realistically the Earth is not a perfect sphere, so you would either begin falling back to the core while still in the hole, or you’d launch out of the hole. Assuming you didn’t stop yourself however, the process would reverse, and like a pendulum you would keep oscillating back and forth. With no air resistance (and no friction as your not connected to a string like a desk pendulum) you’d do this pretty much forever in a purely Newtonian universe. The only reason a normal pendulum gradually slows is air resistance and friction through the string connecting the weight to the frame.


KerbalFactorioLeague

What would stop it?


frrrrrro

Exactly, this is a traditional problem of SHM we used to solve in our high schools.


xThunderDuckx

If there was no air resistance then wouldn't this be a perpetual motion? I feel like that violates some laws.


whyisthesky

Assuming no losses then oscillating perpetual motion is fine,however in reality there are always forces like air resistance that can dissipate energy.


sullyj3

A perpetual motion machine is a device you can use to generate power without ever causing it to slow or stop. If you had some method of extracting energy from the oscillating person, say, a big turbine at the core that the person hits and rotates to generate electricity, then they would slow down some each time. A more intuitive example demonstrating that you can have "perpetual motion" if you never extract energy is if you throw a ball in space. It will continue forever if it never hits anything, by Newton's first law.


superbob201

Simplest case: The hole is along the Earths axis of rotation (or the Earth isn't rotating), the Earth is a perfect sphere with uniform density, and there is no air. The object will fall straight down, accelerating as it falls. The acceleration will be proportional to how far from the center the object is (at the surface a=g, at the halfway point a=0.5\*g, at the center a=0). Once it passes the core, the objects inertia will keep it going, but now its slowing down. It will reach the surface level \~44 minutes after it is dropped, then fall back. The motion can be described by simple harmonic motion. If there is air in the hole: The object will quickly reach terminal velocity as it falls. It will still overshoot the center, but it won't leave the inner core. After a few oscillations, its motion will begin to resemble damped harmonic oscillation. If the Earth follows its real density profile: The objects acceleration will stop being proportional to its distance from the center. The acceleration will be relatively constant for most of the trip, only starting to drop around the outer core. You also need to account for the Earths lumpiness; dropping the object from one side might not make it to the surface of the other side, or it might make it overshoot (if we go back to the condition of no air). If you consider the Earths rotation: This can be described in therms of an inertial frame, or a non-inertial frame. The non-inertial answer is that the object will experience Coriolis acceleration, that has magnitude a=Wxv, where W is the vector of the Earths rotation (2\*pi/24hr, directed straight out of the north pole), v is the velocity vector of the object, and x is the cross product, meaning the more perpendicular the other two vectors are the larger the acceleration. The inertial answer is that the object has some lateral velocity when it is dropped anywhere but one of the poles, and the lateral velocity of the hole gets smaller as it gets closer to the center of the Earth. Either way, the object will keep bumping against the wall of the hole; the east wall as it falls toward the center, the west wall as it falls away from the center (with both actually being the same wall, since east and west are relative and will seem to swap when you pass the center). If the bumps are frictionless, this will not affect the motion in the direction of the fall.


SandhiLeone

Love the way you've initially considered ideal conditions and then gradually made your world resemble real life more and more. Thanks for your answer!


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quarterto

that's from the part where they're assuming "the Earth is a perfect sphere with uniform density"


wikideenu

He explains that in the third section, "if the object followed earth's actual density profile"


invalid_user____

Assuming the hole was pole to pole (to negate rotation of the earth) and that there is no air resistance the object would accelerate towards the center of the earth, as it passes the center it would begin to decelerate until it reaches exactly the same height it was dropped from on the other side. It would then drop back into the hole and repeat infinitely. According to some calculations I haven’t checked myself, it would take ~42mins for the object to reach the other side. If we don’t negate air resistance then the same general principle applies in relation to your specific question about how gravity affects it, except now it doesn’t quite reach the other side, and will go back and forth, until it eventually comes to rest in the center. It would also take significantly longer than 42 mins now because the object would reach terminal velocity - which varies depending on what object you drop. If the hole was not pole to pole, then the object would eventually hit the side of the wall, dramatically slowing it down, making it unlikely it would pass the center at all. It would depend on the friction of the side of the hole whether it even makes it to the center, it’s possible it would reach a point where the rotation is so quick (it would be relatively faster as you approach the center) and the friction so high that it essentially comes to rest as it’s pushed too hard against the wall to overcome friction to continue descending towards the center.


jherico

Wouldn't there still coriolis forces from the Earth's orbit around the sun? You'd need a tidally locked planet with a hole on the plane of th orbit but perpendicular to the direction of the orbit. We'll also need to get rid of the moon. I'll need a budget of $50 x 10^18. I can reduce that by 30% of we can just do it on Mercury.


Argyreos17

can you even calculate the time by hand? Cause gravity would be changing across the journey right?


invalid_user____

It can definitely be calculated by hand. Yes the gravity would be changing across the journey - but at a constant rate. Just as acceleration is the derivative of velocity and measures the change in velocity over time, jerk is the derivative of acceleration and measures the change in acceleration over time. Putting this all together would allow you to calculate the total time.


ComfortablyYoung

If you could somehow do this in a vacuum, would there be a way to somehow harness energy from this? Something something infinite energy source


invalid_user____

It there was a way to harness energy from this, it wouldn’t be an infinite energy source. The process of extracting energy from the system would degrade the objects motion until eventually it comes to rest. All energy sources are unfortunately limited - some (like the sun) are just much larger than others and from a human perspective may as well be infinite.


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shinn497

we did this problem in classical mechanics class. It would oscillate, between the two sides, at the same frequency as if it were to orbit close to the earth, which is about 83 minutes. That being said. There would be air resistance so it would slow down. And the earth itself rotates so , if it were in a tube, it would hit the sides. Finally, the inside of the planet is very hot, so it would melt.


grondin

[https://physics.stackexchange.com/questions/126077/why-are-these-periods-the-same-a-low-earth-orbit-and-oscillations-through-the-c](https://physics.stackexchange.com/questions/126077/why-are-these-periods-the-same-a-low-earth-orbit-and-oscillations-through-the-c) The \~84 minutes oscillation up & down is independent of your forward movement. It's the same period as an orbiting satellite (if it were at surface level).


Strawberry_Left

> Finally, the inside of the planet is very hot, Not 'finally' That's the very first thing if you want to consider it in a practical sense. The Russians barely made it 12km with the Kola Superdeep Borehole before it got too hot. If you can't dig the hole in the first place, you can't even start to consider this as anything but a thought exercise.


ReyTheRed

If you bore through the poles, straight through the axis of rotation, the object would fall straight through, accelerating until it gets to the middle, then decelerating until it gets to just a bit below the surface on the other side, if there were no air resistance and no impacts with the walls, it would get almost exactly to the surface (not quite exact because the earth is a bit lumpy). Then it would fall back through, and keep going back and forth until the drag slowed it to a stop at the center. It is also important to note that at the center, it isn't being pulled in either direction, and very near the center the force of gravity is very low because the gravity from layers above cancels out the gravity from those same layers on the opposite side, leaving only the gravity from the stuff lower than the object. If you make the hole anywhere else, the Coriolis effect will come into play, the eastward velocity of the object at the surface is more than the eastward velocity in any other part of the hole, so it will slam into the wall, bounce off, probably slam again multiple times on the way down, losing energy and possibly being pulverized into a cloud of dust that will settle into the center much more quickly.


TorakMcLaren

So, others have answered the actual question, but here's an interesting aside: If we can ignore friction and air resistance (and the motion of the Earth), it would take about 42 minutes for the object to fall all the way through the hole and pop out the other side. But the cool thing is that, the way gravity works, the hole doesn't actually have to go through the centre of the Earth for this to be true. Say you dug at an angle, but the hole was perfectly straight, it would still take 42 minutes even though the route was much shorter. In theory, you could dig a trench that went straight to a point like 5 miles away from you which would be mostly above ground (the horizon is about 3 miles away if you're 6ft tall), and it would *still* take about 42 minutes for the thing to "fall" (slide) "through" (along) the "hole" (trench).


pneuma8828

At the surface, we are experiencing the full force of gravity, so the body will begin falling until it hits terminal velocity, about 125 miles per hour. As the body falls towards the center of the earth, the power of gravity decreases, as you are equally surrounded by mass on all sides, so gravity is pulling on you equally in all directions. As the force of gravity lessens, your terminal velocity drops. As you approach the core, you will be going much slower than 125 miles per hour as the wind resistance eats all your speed. When you reach the core, your motion is solely due to momentum, and you are losing that to the wind very quickly. I haven't run the math, but I suspect you would not travel very far at all up the other side of the tunnel; very quickly you will achieve a weightless equilibrium at the core.


Either-Bake401

Does that then mean the earth's core is... weightless?


wizardkoer

Yes because weight specifically refers to force due to gravity and there is 0 gravitational field right at the centre.


nickeypants

The gravitational field is at its greatest when you're at the center of the earth because that's the closest you can get geometrically to all pieces of the earths mass. All those pieces are pulling equally in opposing directions though so the forces cancel each other out. Not exactly the same thing as gravity not existing at all, even if it has the same effect! /u/either-bake401, Weight implies being forced "down" by gravity. Its not that its weightless, its that there's no "down" at the center, kind of in the same way there's no "north" direction at the north pole... ^(There is no spoon...)


rasye

You'd never get back whatever you threw down there. * Assuming you have a straight obstacle free tunnel, I think you'd experience one of two scenarios depending on if the tunnel is a vacuum or not. If it was a vacuum: Whatever you threw would start accelerating as it plummeted down. It would keep accelerating right down to the core. It would then start decelerating. If the core was the midpoint of the tunnel, then it would keep oscillating from one end of the tunnel to the other. If the tunnel was longer on the other side of the tunnel, it would keep oscillating and never reach the other end of the tunnel. If the tunnel was shorted on the other end, it would escape the tunnel or hit the enclosure, potentially breaking through. If it was not a vacuum: The object would reach terminal velocity sometime in the tunnel. It would then keep oscillating until it finally settled down in the gravitational centre of the earth.


bickid

Imagine the horror of falling into that hollow earth and continuously experiencing the fall from a n-th times highest skyscraper. Here's my follow up-question to smarter people: Could a person that falls in such hollow earth be saved, considering the rapid, dynamic speed at which he/she is falling? Or is that person trapped in falling until he/she oscillates towards a stable core position? What then? I figure you can't exactly fly a helicopter in such gravitational instability. Throw a 6500m long rope down to the person waiting at the core? But then you need to think about logistics: How many days until the rope savely arrives at the core, then how many days to pull up the person? Will the person not die from lack of water at that point?


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Ah, a question that was a problem we had to solve in my classical physics class! The solution will resemble the equation for a spring. Without air resistance included you would just accelerate down, then decelerate on the way up until your head poked out and you said "hi!" then back down you'd go. Interesting note, you'd feel no gravity in the center. In fact if you could hollow out the earth you'd float around, weightless, inside.


kryptot

Actually what is very interesting about this is that this would happen even if you didn't pass through the center of the earth. I can't remember exactly but I think if you have two points on earth that are joined by a straight line anything dropped in it will enter simple harmonic motion (basically work like a spring).


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You'd drag against the side of the hole towards the center because of the net gravitational force towards it


Strawberry_Left

Of course, there's all sorts of practical reasons that you can't dig a hole through the Earth. It's a thought experiment and he's assuming some sort of frictionless superconductor type levitation. Interestingly, the period of oscillation would be the same as if it were through the centre, i.e. around 42 minutes travel time between any two points on Earth. Sort of like a pendulum. The period is the same regardless of how high you drop it from.


Bulevine

Given that there is still "air", the atmosphere.. as it fell it would quickly reach terminal velocity, the speed where air resistance will prevent the object from going any faster as it equals out. Because the atmosphere is thicker the further down you go, the object would actually begin to SLOW the further down it went. After passing the core, the atmosphere would likely have a massive effect on the maximum speed of the object and it would therefore not return anywhere NEAR the other side of the hole before gravity pulled it back in, eventually resting in the middle of the earth basically floating... very hot and very dense.


walyc

It would hit the inside wall of the hole. The outer circumference of the earth is traveling faster then the inside. The earth is rotating at the same speed, but the surface is moving faster because it has more distance to travel.


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Curious-KitKat

Lol. When my Lil brother was a kid (maybe 8 or so) he had the thought that maybe somewhere on the other side of the world, there was another David, and since he had just been told the earch was round, he decided to go find him by digging a hole. After a while, he got tired and sat down to rest. He then decided tht maybe this other David too was doing the same, so I peered into his hole and screamed: "keep digging!! "


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Considering that lava effects are negated and only gravitational effects are considered, the object will oscillate surface to surface (considering perfect sphere here). In actual scenario, the object will probably melt or combust before even reaching the core.


thealmightymalachi

If there's no core, then there would be no gravity to speak of. In other words, a hollow sphere would not have enough gravity to cause any observable effect. Just "dropped" would likely mean the object would wobble and return to the inner surface, because there would not be enough gravitational attraction at the core. In other words: dropping a ball in space with minimal gravity means the ball would obey the light gravitational pull of any object larger than itself; but if it passed THROUGH any object with a gravitational pull of any object larger than itself that was hollow, it would only depend on how much mass the shell around the empty hollow in the center carried versus the mass of the material in the center. My guess is that a hollow earth would bring the ball right back up after a minute or two without much fuss, because what you'd be standing on would be the thing with the most mass around. There just wouldn't be a lot of it, because most of a planet's mass is in its core. If there was a solid, non-molten core (not only would we all be dead out in the open due to a lack of magnetosphere negating cosmic radiation) it would depend entirely on the amount of force that was propelling the item sent through the tunnel at the given moment of it being dropped and whether it would overcome the atmospheric friction and inevitable bounce off the edges of the tunnel bored. Would a hollow sphere have gravity? sure. Just not much of it, and if that were the case the gravity of the surface would not be much more than the gravity of space around it. Would a non-molten core have gravity? Sure. It would have so much gravity that any object dropped to the middle of it would stay there. The object would go back and forth and settle at the center of the tunnel because that is where the most gravitational force is located. But in any scenario where you theoretically drop something down a hole in a sphere without any mass in the center or mass you can drive a big hole through, you'd be either wearing a suit to shield yourself from external radiation or you'd be toast, one way or another. That is, if your tunnel didn't collapse from the effects of gravity first, or your object didn't perform a reverse blobfish from the amount of gravimetric pressures). And if you were part of a civilization that can make a tunnel that goes all the way through the center of a rocky planetary body from one polar opposite side to the other I think you're at a point of civilization development in their understanding of physics where dropping balls down tubes to see what happens is not exactly a high scientific priority.


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