/sci/ - Science & Math

They get burnt up by magma moving in to fill the hole. Duh.

They would bob up and down, gradually losing speed, until coming to a rest.

>>1007160
annnddd

/thread.

assuming you don't get slowed down by pressure then it would take you around 63 hours to go all the way through at terminal velocity

Assuming no friction and no burning up in the magma, you would just oscillate endlessly from one edge of earth to the other.

>>1007188
would terminal velocity differ in the hole as opposed to a skydiver?

reaching terminal velocity you wouldnt get anywhere near the other side, ignoring getting burnt by magma

It will stop in the center unless it achieved escape velocity. In that case it would break through the center and move forward through the tunnel, rapidly losing speed. It would then return to the center one last time and just stay there.

After oscillating back and forth for a while, gradually losing momentum, you would come to rest in the center and be in a state of perpetual free-fall thanks to the lack of gravity.

you'd reach top velocity at the center and then reach 0 velocity at the other edge of the planet

wouldn't you also be a pancake from the pressure?

>>1007227
Depends on if there's air in the hole or not.

If there isn't air, then there wouldn't be a terminal velocity. You'd speed up until you got to the center, then fly past it, slow down, stop, fall towards center again, repeat until you're dead center.

If there IS air, then I imagine the pressure near the center of the Earth would be REALLY REALLY high, increasing air resistance a lot and thus meaning a lower terminal velocity. That's just a guess though.

In an entirely frictionless environment, If you drill a perfect hole from one spot on the earth to any other spot, You will arrive to the other spot in 42 minutes and 30 some-odd seconds.

but doesn't time run slower at the center of huge centers of gravitation? Can we calculate how strong that force would be?

It would smack into the wall pretty hard b/c of the relative motion of the Earths rotation in the considerable time you need to reach the core (different angular velocities in inertial frame). It's a known and demonstrated effect in free fall towers I think they call it Eastern aberration.

>>1007264
The earth isn't nearly big enough to do that noticeably. It takes stars or bigger to do anything noticeable. Pic possibly related, I can't remember if it is or not, but I believe gravitational lensing is causing a few extra blue stars to appear when in reality there's only one.

>>1007270
Are you sure you're not talking about the Coriolis effect?

>>1007264
Center of Earth isn't the center of gravity. The center of the earth has no gravity.

>>1007274
Wrong picture, I realized.

Found these online, the first is Einstein's Cross. There's only one star in this picture.

The only gravity acting on the center of the earth is the gravity pulling you out. I suppose it would probably be about one half of earths normal gravity pulling away from the center in all directions.

>>1007282
maybe whatever the word is (I ahve been up for quite a while). The fact still remains it will ever so slightly drift out of the axis of your hole and further and furhter and smack. If you avoid that by feasible means it will just oscillate loosing energy and eventually stop in the middle where the theoretical center of gravity is.

>>1007303
You'd be in microgravity at Earth's center of mass, as the mass on all sides of you would be exerting forces that cancel out.

Neil Degrasse Tyson did an episode of Nova Science Now that covered this very situation. It goes something like this:

As you fall, your descent will slow. You will actually begin to fall *slower* after a certain point. Once you reach the center of the earth, you will quick literally hang in mid-air. The whole flight to the center of the earth will take nearly twenty to twenty five minutes.

Of course, this is assuming that the tectonic plates dont shift in front of the hole and kill you, that the heat doesn't melt you, that the pressure doesn't crush you, and that you dont clip any rock bits on your way down.

ASSUMING that this tube is airless and frictionless, it would take ~42.2 minutes for you to fall through the world. On the first half of your trip you would go faster and faster until you pass through the center. From then on gravity would be fighting your velocity, but would not actually overcome it and pull you back until you reach the other side.

Interestingly enough, if you moved this tube so it went from anywhere to anywhere else on earth, and didn't go through the core, the trip would STILL take ~ 42.2 minutes.

You'd oscillate back and forth every 42 minutes. It also doesn't have to be through the center of the earth, you just need to avoid all friction.

If there is no air friction, the object will never stop falling. First time, it will fall all the way through down to the diametrically opposite point on the other side, then fall again to where it was first thrown from, and so on, not slowing down because there's nothing to slow it down.
If there is air friction, it will still oscillate like that for a while, but eventually it will stop at the center.

It's a simple harmonic oscillator (ie a linear restoring force (like a spring))

(Let's assume no air)

Two things to note: 1) Spheres attract as if they are concentrated at a point (we assume earth is well modeled by a sphere 2) If you are inside a hollowed out sphere, you experience no net gravitation attraction from it.

One consequence of that second point is that dyson spheres are stable, so here: (http://www.nada.kth.se/~asa/dysonFAQ.html#STABLE)) is a decent proof.

With that out of the way, the equation of motion is simple.

<div class="math">F = ma</div>
<div class="math">-G\frac{m_{ball}\cdot m_{earth} \cdot \frac{r^3}{R^3}}{r^2} = m_{ball} \ddot{r}[/eqn}

Okay, this take some explaining. Specifically the ratio <span class="math">\frac{r^3}{R^3}[/spoiler]. This result is the reduced form of the ratio of the two volumes (volume between you and the center of the earth and the volume of the earth). It relies on assuming that the density of the earth is constant (it isn't), which lets us say that the ratio of the two volumes is equal to the ratio of masses between you and the coE and that of the whole earth. So the product <span class="math">m_{earth} \cdot \frac{r^3}{R^3}[/spoiler] gives the fraction of the earth which is actually attracting you to the center.</div>

<div class="math">-G\frac{m_{ball}\cdot m_{earth} \cdot \frac{r^3}{R^3}}{r^2} = m_{ball} \ddot{r}</div>

Okay, this take some explaining. Specifically the ratio <span class="math">\frac{r^3}{R^3}<span class="math">. This result is the reduced form of the ratio of the two volumes (volume between you and the center of the earth and the volume of the earth). It relies on assuming that the density of the earth is constant (it isn't), which lets us say that the ratio of the two volumes is equal to the ratio of masses between you and the coE and that of the whole earth. So the product m_{earth} \cdot \frac{r^3}{R^3} gives the fraction of the earth which is actually attracting you to the center.[/spoiler][/spoiler]

>>1007153

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<span class="math">\frac{r^3}{R^3}[/spoiler]. This result is the reduced form of the ratio of the two volumes (volume between you and the center of the earth and the volume of the earth). It relies on assuming that the density of the earth is constant (it isn't), which lets us say that the ratio of the two volumes is equal to the ratio of masses between you and the coE and that of the whole earth. So the product <span class="math">m_{earth} \cdot \frac{r^3}{R^3}[/spoiler] gives the fraction of the earth which is actually attracting you to the center.

Finally reducing:
<div class="math">-G\frac{M_{earth}}{R^3}=\ddot{r}</div>

This is an equation of motion (the negative sign indicating restoring force) for a simple harmonic oscillator. I'll spare you the deets but the solution to the differential equation equation is:
<div class="math">A \sin(\sqrt{\frac{GM_{earth}}{R^3}}+\phi}</div>

Where A and phi are constants we can pick.

>>1007862
so close.

<div class="math">A \sin (\sqrt{\frac{GM_{earth}}{R^3}}+\phi)</div>

I dont want to do out all of the math but if we assume that there is some simple resistive force proportional to velocity in the tunnel (this is a simplistic (an mostly wrong) model for wind resistance than the force equation looks like.

<div class="math">-G\frac{m_{ball}\cdot m_{earth} \cdot \frac{r^3}{R^3}}{r^2} - b \cdot \dot{r} = m_{ball} \ddot{r}</div>

And the solution to this differential equation is:
<div class="math">e^{\frac{-b}{2m_{ball}t}e^{i\omega t}</div>
Where <span class="math">\omega = \sqrt{\frac{GM_{earth}{R^3 m_{ball}}-\frac{b^2}{4m_{ball}}}[/spoiler]

There is no god

<span class="math">e^{\frac{-b}{2m_{ball} t}} e^{i \omega t}[/spoiler]

and

<span class="math">\omega = \sqrt{ \frac{GM_{earth}}{R^3 m_{ball}}-\frac{b^2}{4m_{ball}}}[/spoiler]