tunnel through the centre

vikasj007

this is supposed to be a very easy one, but i am still putting it up for fun.







suppose a tunnel is dug right through the centre of the earth, from one end to another,( say from north pole to south pole), and then from one end of the tunnel a ball is dropped in it.


what will finally happen to the ball????




keep in mind this is a hypothetical situation, so please post sensibaly.

matthyaouw

it will combust or melt.
Assuming we ignore the role of heat, it will hover almost stationary half way down/up/however you care to phrase it.

ascky

Doesn't it just oscillate forever? (with no air resistance)

derekmohammed

IF there is no air resistance it will ossilate forever. It is intresting because it is proven that its period is the same as a satilite orbiting the earth!

Gokul43201

Yes, it would.

First extra question : How long would it take to travel from one end to the other ?

ascky

I get around 42 minutes

BobG

59.75 minutes.

The radius of the Earth isn't the semi-major axis.

Look at it this way.

You start wtih a circular orbit with a semi-major axis that matches the center of the Earth.

Start making the orbit more elliptical, but keep apogee at the surface of the Earth. For example. With the center of the Earth still the focus, and an eccentricity of .1, an object would be at 5218.48 at its closest point to the center of the Earth - 6378.14 at its furthest (still at the surface of the Earth). The semi-major axis is reduced to 5798.

At .2 - perigee at 4252, apogee at 6378, semi-major axis at 4252

Etc.etc. to eccentricity of .9999 - apogee 6378, perigee 0.03, semi-major axis 3189. The satellite reaches perigee and heads right back to the same point on the surface of the Earth that it started on.

Continue increasing eccentricity until, finally, eccentricity is undefined (object headed straight at the center of the Earth). You can use the the extremely high eccentricity of .9999 something and the semi-major axis of 3189 as a pretty good esitmate (the limit).

Apogee is the surface of the Earth, perigee is the center of the Earth. The big difference is that with no tangential velocity, the flip side of the orbit (perigee to apogee) is in the same direction as apogee to perigee side of the orbit.

In other words, reaching the other side of the Earth (apogee) is the equivalent of completing one 'orbit'. It takes two 'orbits' to reach apogee on the same side of the Earth.

There is no way you can have two apogees and no perigee!

That also means its period is not the same as an orbiting satellite.

Gokul43201

Ooh Bob-gee, you lost me there (actually I didn't read the post carefully, after finding your anwer disagree with mine) !

I first got about 85 minutes, then realized that was twice the time...so my number is pretty close to what ascky gets.

Now I guess I'll really have to read your post

cronxeh

BoBG..

Bartholomew

If you assume the earth to be of the same density throughout, the gravity _decreases_, linearly, as you go from the surface to the center. I do not know the proof of this result but I read it somewhere... so you can't treat it as just an ordinary orbit. I'm not sure how to solve it, though, without just using a computer program for an approximation.

Rogerio

Solving d2x/dt2 = -k*x , with g=9.81m/s^2 at R_earth=6366km , I've got

42m 10s

Rogerio

Bartholomew is absolutely right!

Considering g(x) the gravity acceleration at distance 'x' from the center:
g(x) = -G * M_earth/x^2

But , at x < r_earth , the mass outside (x>r) doesn't affect the object, so the "new" mass to be considered is:
M_earth * x^3/r_earth^3

So, the "new" g , if x<r_earth, is:
g(x) = -G * M_earth * x / r_earth^3 , or

finally g(x) = -K*x ,

where K = G*M_earth/r_earth^3 = 9.81/(6366000*s^2)

ceptimus

Easiest solution is to use the formula for the period of a simple pendulum:

[tex] T = 2\pi\sqrt{\frac{l}{g}}[/tex]

And put l (pendulum length) equal to Earth radius:

[tex] T = 2\pi\sqrt{\frac{6378100}{9.80665}} \approx 5067 seconds = 84 minutes, 27 seconds[/tex]

This is the time taken for the return trip.

Hurkyl

I don't see any similarity between this problem and that of a pendulum...

OTOH, it is like a spring (compare the force of gravity to Hooke's Law)

ceptimus

The pendulum analogy is perfect given the way this question is normally phrased:

"A straight tunnel is driven between two points on the Earth's surface. How long would a train powered by gravity alone, take to traverse the tunnel? Ignore friction and air resistance."

Now the reasoning goes like this:

1. The question wouldn't have been asked like that unless the time taken is independent of tunnel length, so if we solve it for one case, we have automatically solved it for all.

2. So that we don't have to worry about decreasing gravity with depth we will consider a very short tunnel, so that it only penetrates a negligible distance below the Earth's surface.

3. So we draw a 'tunnel' one metre long, and work out the accelerating force on the 'train'. This involves a bit of pythagoras, and we see that the force pushing the train towards the centre of the tunnel is proportional to the distance from the tunnel centre (for the very small angles involved).

5. Now, we recognise that this is exactly the same situation as a pendulum bob, with the length of the pendulum equal to the Earth's radius. Bingo! we already know the formula for that!

6. 30 seconds on a calculator later, and we're done!

ceptimus

Most such tunnels would actually be affected by all sorts of coriolis effects, but I notice that the O.P. specifically used an example that isn't - the tunnel that connects the two poles.

BobG

Aside from the fact that Bartholomew's point about the acceleration due to gravity decreasing due to the distribution of mass both above and below, there's a problem with your solution even in the simplified example (all the Earth's mass exists at the center).

A pendulum assumes 'l' stays constant. If the two points are more than 12,756 km apart, the pendulum is so long that it swings past the center of the Earth. That definitely causes a problem. At the 'bottom' of the pendulum swing, the pendulum is 3640 km past the center of Earth, meaning your motion is going to be a little more complicated than a simple pendulum swing.

Edit: Oh, I read your original solution. Except your original solution is for a curved tunnel with the openings 12,756 km apart, not a tunnel straight through the Earth. And I don't see how a pendulum applies to a straight tunnel in any event. The pendulum length will be constantly changing, reaching a minimum length at the 'bottom' of its swing and then expanding back to its original length.