If a tunnel is drilled through the center of the Earth and elevator conveyance is installed?

Question

it will take 42 minutes to travel to the other side,
if the conveyance is allowed to free fall.

This energy-effective method of transportation
has the disadvantage that passengers will remain
in weightlessness for prolonged period of time,
and may feel uncomfortable.

If we want the passengers to feel normal weight
during the travel, and minimize travel time,

how much time will the trip take in that case?

Our teacher gave up on us already,
all he wants is a just number.

28.15 minutes is very close to
exact answer (in case of uniform density).

Answer 1

Air resistance would make this trip impossible. However, if you can evacuate the tunnel, the time to make the complete trip is twice the time to make it to the center.

Inside the Earth, the gravitation strength is proportional to the distance from the center (compared with outside the Earth, where it's inversely proportional to the square of the distance from the center).

I did the problem numerically in MS Excel, using a time step of 1 second. The acceleration in each step size is 9.81 m/s^2 multiplied by x/(6378 km), the fraction of the distance from the center, with 1 being the surface and 0 being the center. The velocity in each step is simply the acceleration of the last step, multiplied by the time step, then added to the velocity of the previous step. The position in each step is just the position of the last step, subtracting the product of the time step and the velocity.

I got 1266 seconds, or 21.10 minutes to make it to the center of the Earth, so 42.20 minutes to make the entire trip.

If we want the passengers to experience 9.81 m/s^2 the entire trip, we have two options. They can stay perfectly still and experience this due to gravity, in which case they never make the trip at all, or they can accelerate downward, matching gravity PLUS 9.81 m/s^2. So, their acceleration would be g + 9.81 m/s^2, where g varies with distance.

a = (9.81 m/s^2) * (x / (6378 km)) + (9.81 m/s^2)

Altering the acceleration formula in Excel, I find that the trip time to the center is 844.4 seconds, so the total trip time is twice this:

1688.8 seconds = 28.15 minutes.

Note that to use this method, once you reach the center, you have to start accelerating in the opposite direction. So, while it's true that the passengers always perceive normal acceleration, the direction of the acceleration switches abruptly halfway through the trip, tossing the passengers on their heads if they're not strapped in (or ready to perform an acrobatic maneuver).

For tricky differential equations like these, I prefer numerical integration, such as Excel does.

Answer 2

Retardis is right. Gravity changes continuously as you go through the earth, and the math to calculate the magnitude of the field at any point is complicated. The elevator would have to constantly change speed through the journey to maintain a constant weight for the passengers. I think you got the problem statement wrong, or the teacher doesn't know what he is talking about.

Your weight would be normal at the surface, before the elevator started moving. As soon as the elevator starts to move downward your weight decreases, so how do you maintain normal weight? Also, as you move toward the center of earth, gravity decreases to zero at the center and then starts to increase in the other direction. Therefore, to experience normal gravity you would have to rotate the elevator 180 degrees at some point and accelerate continuously toward the center, and then rotate again at some point and decelerate to the surface. Then rotate again at the surface. There is no way you could rotate the elevator while maintaining "normal" weight for the passengers.

Let's say you strap the passengers into seats at the start, then flip the elevator around and start down. This would get the discomfort of the rotation out of the way quickly. You could accelerate at 2g (64.4 ft/s/s) to give them back their normal weight. Then you would have to gradually slow the acceleration to 1 g at the center, and further reduce to 0g (constant speed) just before reaching the other side. Since you have been accelerating for 8,000 miles you would be travelling at a tremendous speed. You would have to decelerate for some time before stopping at the surface. The passengers would experience some decreased weight during the deceleration, but would arrive in the right-side- up position. The speed could be calculated, but it would vary continuously through the journey, and in a complicated nonlinear way. You would need a lot of calculus to determine the time of the trip. The time would also depend on how long you take to decelerate at the end. Could be a fun problem if you have nothing else to do with your summer.

Addendum: After a little more thought, it appears that an estimate could be made assuming the average acceleration would be 1g, or 32.2 ft/s/s. Over 8000 miles this would give a transit time (d=1/2 at^2) of about 27 minutes, which is faster than the free fall trip. However, the elevator would be travelling around 35,000mph at the other side, so there would have to be a significant period of deceleration, during which the passengers would not experience "normal weight".

Answer 3

Wow, that's a tricky one.

Since gravity decreases as you go down, the only way you could manage it would be to launch the elevator down FASTER than gravity. This would make the passengers feel their "normal weight" all right, except that they'd be sticking to the elevator's roof! So let's assume you've installed some comfy chairs on the roof, making it into a nice "floor".

Now, as they descend toward the center, there are two forces acting on the passengers:

1) The force of gravity, F_g. If we pretend that the earth's mass is distributed evenly, then F_g decreases linearly from mg to zero as we approach the earth's center. That is:

F_g = mg·(r/R), where r is our current distance from the center, and R is the earth's radius.

2) The normal force F_n acting on the passenger's butts (pushing down on them from the "roof"). We want to arrange things so that F_n is always equal to a constant mg; this will make the passengers "feel" their normal weight.

So the net force is F_g + F_n. By Newton's 2nd Law:

a = (F_g + F_n)/m
= (mg·(r/R) + mg)/m
= g(1 + r/R)

Expressing this in terms of the distance x=(R-r) which we've descended so far:

a = g(1 + r/R)
= g(1 + (R-x)/R)
= g(2 - x/R)

Now, since "a" is the 2nd derivative of "x", we have this differential equation:

x′′ = g(2 – x/R)

I'm rusty with differential equations, but somebody else can take it from there if they want to give it a shot. After that, you should have x as a function of t, and you can then determine the value of t that would cause x to equal 2R.

I don't know what grade level you're at, but unless your teacher expects you to be able to solve differential equations, I think he's asking you too hard a question.

Answer 4

You and your girlfriend are taking GR together? How romantic.

Assume spherical, even density earth (a little stretch, but go with it)

a = G(mass inside)/r^2 + g

massinside = M * r^3 / R^3

a = GM/R^3 r + g

So it's like a spring still, but the equilibrium point isn't the middle, which complicates the calculation a bit because you can't just use the natural period of the motion to get your answer.

Divide the journey into two pieces, the journey to the middle and the journey out. You switch the polarity of the g-force halfway through. (Apologize to the passengers for this inconvenience).

Equilibrium position:
requilibrium= gR^3 / GM (on the backside).
Magnitude of oscillation = R + requilibrium

position = (R + rgR^3/GM)cos (omegat)

omega = sqrt (GM/R^3) = 2pi / T
They give you T/2 = 42 minutes

Solve for the t where position = rgR^3/GM

Double that. You're done.

Answer 5

Neither of the above answers are correct, the second is a rough approximation but the gravity of earth does not originate from the exact center of earth. Gravity is a function of the presence of mass, therefore as you travel closer to the center of the earth you will be feeling less and less gravity. Once you are a few miles from the center you will feel hardly any pull in any direction. The math required to produce an even approximately close answer is a little involved but would have to use the formula's for gravity's dependance on mass and distances - which is not linear. Depending on what grade you are in I would say your teacher should check first to see if they could even reasonably approximate an answer before giving it to students.

Cheers

Answer 6

Sorry I think your teacher is tugging on your toga. That question is like "how far can you walk into a woods?" the answer is Halfway after that you are walking out. If you free fall through the center of the earth, after you hit the center you would have to fall up. Not a likely scenerio. Tell your teacher to teach you not BS you. That or you didn't fully explain the question.

Answer 7

if allowed free fall, it will move with simple harmonic motion