How long would it take?

Question

say you could tunnel through the earth north to south pole and be able to drop a object through the centre of the earth . how long would it take for it to travel through. no saying it would melt at the middle, as you have dropped it through a tunnel thanks

Answer 1

Under the conditions you describe, no object will ever even get close to the center of the Earth.

There's air in the tunnel. Let's say that the object is you. Terminal velocity for a human in Earth freefall is in the range of about 100 (if your fall flat, arms and legs extended) to 200 miles per hour (if you "dive", body and limbs straight). If it weren't for the fact that the density of the air wouldn't be constant with depth, you'd maintain close to this speed for the first few hours (remember, it's 4000 miles to the center from here), until you were deep enough that a significant percentage of the mass of Earth was behind you. Then you'd gradually begin slowing down due to air friction versus the decreasing net gravitational force on you. After a couple of days of this, you'd be approaching the center of Earth at about the same speed as a feather falling through the air. BUT you haven't yet accounted for the change in air density, so what was just described is not what will happen.


Factor into your well the fact that air pressure will increase by a factor of about 10 for every 10 miles down you go, you'll figure that you're going to have real trouble even get close to the center. The reasoning is this: at sea level, the volume of air equal to the volume of a person (about 3 cubic feet maybe) weighs about 0.006 pounds. Every 10 miles down you go, that same volume will weigh 10 times more. If I have it figured right (anyone out there???), somewhere between about 40 and 50 miles down the air density will equal that of water. You will float at that point just as if you were in water. Swimming down much further will be increasingly difficult since unlike water, the density of the air will still keep increasing with depth, making your bouyancy progressively more difficult to overcome.

The only way to be able to fall all the way to the center would be to seal and apply vacuum to your well. Then when you jumped in with your spacesuit on, no air or air friction will slow you down. If the vacuum was close to perfect, you'd accelerate at a 1G rate to thousands of miles per hour within minutes (100 seconds at 1G = 32ft/sec^2 =3200ft/sec~= 0.6mps=2182mph), Gradually, this rate of acceleration will decrease as more of the mass of Earth becomes behind you, becoming 0 as you pass through the center, then gradually tending to reverse your acceleration to the point where your speed would be zero just about time you hit surface level on the opposite side (you did dig ALL the way through, right?). Then you would start falling back again, and if your vacuum were perfect, continue this cycle indefinitely.

Each lap, under such conditions, would take only a few hours.

Answer 2

Morningsomething has it right, although it wouldn't quite reach the opposite surface thanks to air resistance. It would oscillate a few times and finally end up at the center of the earth.

The falling distance for the object is the radius of the earth, about 6,400 km. You can't use the simplified distance formula of s=1/2 g t^2, however, because g will be decreasing all the way down, then increasing on the way up to the other pole.

Yet another complication is that the earth's density increases to a maximum at or near the center, so the change in g would not be constant as the object falls. This is turning out to be a very interesting problem indeed...

By the way, all you calculator jockeys giving us long strings of digits as answers - only the first 2 or 3 of those numbers have any meaning; the rest are the result of the calculator assuming that all the inputs are exact to the 10th digit or so, which of course is not true (gravitational acceleration itself varies by several hundredths around 9.81 m/s^2 depending on where you are on earth, so no calculation using this number can ever have more than 2 siginificant digits in the result [maybe 3 since 9.8 is so close to 10.0]). The rule of thumb is to keep the extra digits in intermediate calculations but to round off the answer at the end.

Long-distance underground subways have been proposed using this "gravity assist" principle. It works even when the tunnel does not go thru the center but is drilled as a straight line (or chord) from one station to the next. At first the train would roll downhill into the earth, and its momentum would bring it close enough to the destination such that little fuel would be needed to complete the trip. The problem is the huge expense of drilling and maintaining such long tunnels.

Answer 3

Previous answers have assumed that the acceleration due to gravity would be constant. That is a big assumption. And I guess another responder pointed out about the presence of air -so we need to do this in a vacuum.

Just as a thought experiment, put the object near the centre of the earth, is it going to feel 1g acceleration to move the last little bit toward the exact centre? the object is being pulled at by the mass of the planet all around it in every solid radial direction.

Okay so maybe we can agree that the acceleration is 1 g near the surface. Now as you fall down the hole the acceleration may increase a little because the distance between the object and the centre of gravity of the earth is decreasing. However in this same thought experiment, some of the mass of the earth is now on the oppposit side of the object.

I think it may take a long longer than 30 minutes to fall through the earth therefore. The trip may actually begin at zero velocity, speed toward the centre but then as the acceleration decreases, the object will approach some limiting velocity and drift along at some constant velocity and then it will be decellerated as it falls up toward the surface on the other side.

Answer 4

RL612 is half right. The accelleration towards the centre in these cases is dependant on the total amount of mass between the current position and the centre.

This means that gravity starts to lessen as soon as you descend below the surface, and gradually reduces to zero when you reach the exact centre. The effect on velocity is to increase it at a slower and slower rate as you get deeper under the surface, until the object is not actually accellerating at all at the centre.

At this point momentum will carry the object towards the surface on the other side, slowed by increasing gravity, and assuming zero air resistance an oscillation will ensue. I'm not sure if this is simple harmonic motion, for which the force would have to be proportional to the distance from the centre, and not the volume.

A satellite in close Earth orbit has a period of about 97 minutes, but this is determined by full Earth gravity for the whole of the cycle, and I would expect the trans-centric period to be somewhat longer as a result.

Answer 5

About 90 minutes for a two-way trip -- that about the same time as a satellite going around the Earth in low orbit. That's assuming that somehow you got rid of the air friction.

And no, it would not stop at the center of the Earth. It would accelerate while on the way down, reach the fastest speed at the center, then slow down on the way up. It would slow down to zero speed just at the top of its travel. That is, at the surface of the Earth; on the other side of the tunnel, and back where you dropped it.

Answer 6

About a half hour.

Ignoring air resistance, the object would start to bounce up and down, in simple harmonic motion.

Imagine a mass hanging from a spring, going up and down, up and down...

Let's look at it from an energy point of view.

Initially, the object has mgh for it's energy (h = radius of earth).

When it hits the center of the earth, its maximum velocity, v = h * w. w is related to the period (I'll tell how later). Conservation of energy tells us

mgh = ½m v²

mgh = ½m (h * w)²

gh = ½ h² w²

w = sqrt(2 g / h)

For SHM, T = 2 * pi / w (where T = the period)

T = 2 * pi * sqrt(h / 2*g)

T = 2 * pi * sqrt(6378.137 km / 2*g)

T = 3583.04 s

That's how long it will take to get there and back, so cut it in half.

t = 1791.52s = 29.8587 min, or about a half-hour.

Please check and see if I'm right!


Another way to look at it.

The total energy in the SHM system is

E = ½ k h²

where k = spring constant

the object started the system with E = m g h

So

mgh = ½ k h²

k/m = 2g / h

In SHM, w = sqrt(k/m) = sqrt(2g/h), which is was we came up with earlier.

Answer 7

Wouldnt the object just float at the centre of the earth?

Gravity pushing on it equally from all directions?

I may be awfully wrong so i look forward to other peoples answers

Although i also think ive heard of the object rebounding back and forth across the centre of gravity

Come on science geeks, put us out of our misery!

Answer 8

Eternity. Gravity pulls objects towards the center of the earth, so when the object got half way between the two poles it would stop.

Answer 9

It would actually start to execute simple harmonic motion around a point at the centre of the earth.

Answer 10

i would think that it wouldn't go to the other side cause of gravity stoping it in the middle or maybe cause the earth is spinning it will keep going up and down. but without those thinks i would guess a good couple of days.