Hypothetically, suppose that there was a straight, air-free bore hole that ran from sea level at the equator to the center of the Earth. If I were to jump in, for how long would I be in free fall? How would you go about calculating this?
Cecil has covered this. It’s in a Straight Dope book I have. I’m not sure if the column is online.
The basis of what would happen is that gravity would decrease as you got nearer the centre of the Earth, and you would be totally weightless at the very centre, then gravity would start increasing again on the other side.
You would continue accelerating in this airless, frictionless environment until you got the to centre of the Earth, whereupon you would start decelerating until you came to a complete stop at the surface at the antipodes. You would then start falling back in the opposite direction, and, because it’s a frictionless environment, you would bounce back and forth in this manner between the two opposite sides of the planet forever.
Fill the tube with air, however, and things are different. You’d accelerate a relatively short distance until you reached terminal velocity, then you’d fall at that steady rate until a point just past the centre of the Earth, where you would start fallingback to a point just this side of the centre. You would bounce back and forth a few times until you came to a complete stop at the centre of the Earth.
To more directly answer your question, if the tube stops at the centre, you’d wind up a pancake, in an airless or air-filled tube. You’d continue accelerating until impact if airless, and would accelerate only a short way if air-filled. I’m not sure what you mean by ‘free-fall’; that word, as I understand it, refers to falling without artifical retardation regardless of whether you are still accelerating or not.
Actually, given the question as stated, he would continue accelerating until he reached the center of the earth, then there would be a rather rapid deceleration, seeing as how he didn’t specify that the shaft continued past the center. So, he’d be in free fall for however long it took to reach the bottom of the shaft, about 4,000 miles, starting out accelerating at 1 g, with the rate of acceleration gradually tapering off until the sudden stop at the end.
Thanks for the responses – I should have been clearer in my OP. I understand the principle that I would fall starting at 1 g and then a decreasing percentage of g until I slammed into the core. What I’m trying to figure out here is the specific math to get the amount of time from jump to splat.
Thanks!
Right, you can’t use 3.8 m/sec^2 because you’d not be accelerating at that speed the whole time. Initially you would, but you’d reach points where your acceleration would slow. Sounds like a problem for the super-duper calculus guys, who should be along shortly.
The gravitational pull of the Earth on you as you fall through the hole will, as you say, decrease as you fall because the mass interior to your radius will be decreasing. The formula for your acceleration is:
a®=GM®/r[sup]2[/sup].
The tricky thing is that M® isn’t going to be a smooth function because the density of the Earth varies with depth. The rocky mantle has a lower density than the metallic core, and in both zones the density increases with depth due to gravitational compression.
If you assume that the Earth is a uniform sphere of density D, then M®=4/3 pi r[sup]3[/sup] D,
so
a®=G 4/3 pi r D
Then since the acceration is the second time derivative of your position, you just have to solve the 2nd order ordinary differential equation:
d[sup]2[/sup]r/dt[sup]2[/sup] = G 4/3 pi r D
which I will leave as an exercise for the reader, 'cause I’m supposed to be writing my thesis.
There are some simple physical effects to deal with in this hypothetical story problem. Are we to ignore them, for the sake of the puzzle, or find ways to deal with them?
You say your tunnel is air-free, so I’m presuming an airlock at the surface to keep it air-free.
You’d need some sort of space suit to keep you from going splat the instant you open the airlock to the air-free tunnel.
At sea level, I suppose your airlock would keep the sea from pouring in at high tide.
On the other hand, if the end of the tunnel were open, the equatorial winds passing over the opening would make a musical note, as if Og were playing in a jug band.
I guess we’ll just gloss over the fact that nobody has ever drilled a hole even a tenth of that depth.
You’re right, you wouldn’t. But, that’s because the Earth’s surface gravitational acceleration is 9.8 m/s[sup]2[/sup] (~32 ft/s[sup]2[/sup]. At some point in your descent, it would be 3.8 m/s[sup]2[/sup], but I’m not certain how to calculate how far down that might be.
I hope someone will solve the differential equations above, but as a shortcut, this is mathematically equivalent to a spacecraft in low-earth orbit.
Imagine a perfectly spherical earth of uniform density, with no atmosphere, and a hollow shaft from the North Pole to the center. Imagine a spacecraft whizzing along in ultra-low-earth orbit, from pole to pole, along one of the meridians. Imagine a solid plane, extending through the equator, and out to infinity. Call it the “Plane o’ Death”.
Now imagine Santa Claus standing at the North Pole, watching for the spacecraft to pass over, and at the moment it passes (a few inches) overhead, dropping the rock down the shaft. At that moment the rock and the spacecraft both have the full force of gravity pulling them toward the Plane o’ Death. The only difference is that the spacecraft also has a horizontal motion vector.
As time goes by, the rock continues to accelerate, but by an ever-decreasing amount, because more and more of the earth is in its rear. The spacecraft continues to accelerate toward the PoD, also by an ever-decreasing amount, because more and more of its acceleration vector is in a different plane. The two continue to move toward the PoD, with increasing velocity but decreasing acceleration, until both go splat 21 minutes later.
Which is one-fourth of the time necessary to complete a low-earth orbit. (The astronauts take about 90 minutes, but they have to clear the atmosphere.)
Airless or not, it doesn’t matter, you would be a smear along the side long before you reached the core. Angular momentum has your ass.
That’s hot!
That’s because on MY keyboard the 3 and the 9 are right next to each other. Been meaning to get that fixed. 
Not if the shaft was drilled at the pole.
Podkayne, step away from the thesis. If you’re so zonked that you can’t recognize a Simple Harmonic Oscillator, you really need a break.
As far as the density profile, RM Mentock could tell us more, but I seem to recall that it’s such that the local value of g is approximately constant until you reach the core, at which point the density is more or less constant (so g decreases linearly in the core).
I’m thinking terminal velocity (in the “air filled tube ending at the center” scenario) could make for an interesting twist.
V[sub]term[/sub] is reached when the force of the air pushing up on you (generated by your fall speed) is equal to the force of gravity pulling down. This varies based on things like atmospheric pressure, shape and density of the falling object, etc. However, it also depends on the gravitational force (since that’s what it’s balancing against). And as you fall down the tube, the force of gravity is decreasing. So if you start out at terminal velocity of (let’s say) 55 m / s, after a bit the force of the air at that speed is going to be larger than the (now lower) gravitation pull, so you have a net accelleration up, slowing you down to a new V[sub]term[/sub]. This process continues as you fall towards the center of the earth, with gradually lower terminal velocity.
Exactly at the center of the earth, terminal velocity would have to be zero (since gravity is also zero). The question becomes: are you slowed down enough (essentially by your own body surface acting as a parachute in a low-g environment) to survive “landing” on the bottom of the tube?
Yes, but you’ll be going faster than terminal velocity, because you were accelerated earlier. It’s like being shot out of a cannon toward the Earth–you don’t slow down to terminal velocity if you’re already going faster.
Nope. It’s a big center-of-the-Earth splat!
As a matter of fact, you would slow down, if fired at greater than terminal velocity, and you would approach terminal velocity from above. The question is just whether you would slow down soon enough, which is a much more complicated question. Off the top of my head, I suspect you would: In addition to the lower gravity, the air is also denser at depth, which would also decrease your terminal velocity. At a few miles, say, from the center of the Earth, the terminal velocity should be almost zero, and a few miles should be plenty long enough to slow down.
I tried throwing a feather - hard - at the ground. It slowed down.
Sure you do. Not instantaneously, but if you’re dropping faster than terminal velocity, the net forces on you act to reduce your speed. The terminal velocity of a freefalling skydiver is lowered when they deploy their parachute (due to a higher cross section area), so the skydiver feels a force upward as they slow to the new terminal velocity.
So as you fall through the tube, once you accelerate up to terminal velocity (at whatever depth you reach that), after that (as the gravitational force decreases) you’ll always have a net upward force as the air resistance tries to slow you down towards whatever terminal velocity is for the value of g at your given position.
It may not be the case that this braking is enough to avoid the “pancake” end result, but who knows.
{goes off to hack out a perl script…}
Ok, I did an iterative simulation, with some interesting results.
Without drag: you hit bottom at roughly 20 minutes with a final velocity of 7600 m/s.
With drag, the trip took almost 88 hours, the maximum velocity was 40 m/s (which was hit after 20 seconds, gradually slowing after that), and the final velocity was 0.148 m/s. That definitely looks survivable. Anyone know what a deadly fall velocity would be? 5 m/s? 10?
Here’s the code. For the drag constants, I used the default values provided in the calculator here, converted into metric. Radius of the earth obtained from wikipedia.
#!/usr/bin/perl
# Falling through tube with air resistence
# Constants
# Radius of earth: 6356750 m
# Drag coefficient: 0.7
# Air density: 1.22 kg/m^3
# Cross section: 0.5 m^2
# mass: 70 kg
$radius_earth=6356750;
$drag_coef=0.7;
#$drag_coef=0;
$density_air=1.22;
$area=0.5;
$mass=70;
$g_surface=9.81;
# Initial conditions
$distance=0;
$velocity=0;
$t=0;
$max_velocity=0;
print "Time Distance Velocity Fgravity Fair Net Accel
";
while ( $distance < $radius_earth )
{
$t++;
$f_gravity=$g_surface * ( 1 - $distance / $radius_earth ) * $mass;
$f_air=0.5*$density_air*$velocity*$velocity*$drag_coef;
$accel=($f_gravity - $f_air) / $mass;
$distance+=$velocity+0.5*$accel;
$velocity+=$accel;
if ( $velocity > $max_velocity )
{
$max_velocity = $velocity;
$max_velocity_time = $t;
}
if ( $velocity < 0 )
{
print "Something's gone horribly wrong...";
print "$t $distance $velocity $f_gravity $f_air $accel
";
last;
}
if ( $t % 3600 eq 0 )
{
print "$t $distance $velocity $f_gravity $f_air $accel
";
}
}
print "$t $distance $velocity $f_gravity $f_air $accel
";
print "
Max velocity = $max_velocity at time tick $max_velocity_time
";
Feel free to nitpick my distance calculation, I’m not sure I got the iterative calculation right. Is there supposed to be a 1/2 a * t * t in there or not?