Ball Dropped in Tunnel Through Earth: What Happens Next?

[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 interesting because it is proven that its period is the same as a satilite orbiting the earth!

 

[Gokul43201]

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

Yes, it would.

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

 

[ascky]

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

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) ! :grumpy:

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.. :uhh:

 

[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]

Yes, it would.

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

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

42m 10s

 

[Rogerio]

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...

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:

$$T = 2\pi\sqrt{\frac{l}{g}}$$

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

$$T = 2\pi\sqrt{\frac{6378100}{9.80665}} \approx 5067 seconds = 84 minutes, 27 seconds$$

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]

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!

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.

 

[Bartholomew]

I looked up the formula for the period of a spring and if you plug in it works out to the same answer ceptimus got.

Edit: How would you do that analytically, though, going from acceleration as a function of distance to acceleration/speed/distance as a function of time?

 

[Rogerio]

...
How would you do that analytically, though, going from acceleration as a function of distance to acceleration/speed/distance as a function of time?

Take a look at my 2 first posts, and see how to get the differential equation d2x/dt2 = -K*x , where ' x' is the distance to the center of the earth, ' t' is time, and 'K' is a positive constant.

The (trivial) solution is x= A*sin(sqrt(K)*t+B) , where A and B are constants to be determined from the contour conditions.

Here, at t=0 we have x=R_earth and dx/dt = 0 , so B=0 and A=R_earth.

Using the known values, you'll get something like 42m 10s.
:-)

 

[ceptimus]

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.

No it isn't. Try reading it again.

Why do you think the pendulum method comes up with exactly the same answer that you did (except you have slight errors in your constants)? I used the officially accepted values for Earth radius and little g.

I admit that the pendulum method does take advantage of the assumption that the time taken to traverse a straight tunnel is independent of the length of the tunnel. It is by no means obvious that this should be the case. But if we accept that this has already been proved, then the pendulum method is sound.

Remember the formula for the period of a simple pendulum also makes assumptions, (small angles of swing, sin(angle in radians) = angle for small angles, etc.

Also, the Earth isn't actually made of uniformly dense material. The core is denser than the mantle. If you take this into account, you'll find that the time taken to traverse a deep straight tunnel (in a vacuum, in the absence of planet spin) is actually less than that taken to traverse a short tunnel.

Yeah - I've studied this problem before.

 

[BobG]

There's still a problem with this example. All three, Hooke's Law, the formula for a pendulum, and the formula for the period of an orbiting object should give the same answer if the problem's set up right.

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

$$T = 2\pi\sqrt{\frac{l}{g}}$$

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

$$T = 2\pi\sqrt{\frac{6378100}{9.80665}} \approx 5067 seconds = 84 minutes, 27 seconds$$

This is the time taken for the return trip.

The acceleration due to gravity is:

$$g=\frac{GM}{r^2}$$

Substituting this into the formula for a pendulum and you get the formula for computing the period of an orbit:

$$T=2\pi\sqrt{\frac{r^3}{GM}}$$

As long as 'r' stays constant, everything works out. If 'r' varies, it falls apart. In an elliptical orbit, it takes less time to go from one covertex to perigee to the opposite covertex than it does to go from one covertex to apogee to the opposite covertex.

In fact, to account for elliptical orbits, the average radius (the semi-major axis) is substituted for r.

For it to work for the tunnels (with a varying radius), you have to assume the highest point is directly above the fulcrum and the lowest point directly below the fulcrum. In other words, your pendulum length is one half the radius of the Earth with a period of 59.75 minutes.

Actually, I just rounded off the WGS-84 (1996) value for the equatorial radius of the Earth (6378.137 km), since WGS-84 is the datum most commonly used on maps (thanks to GPS).

 

[edwinwinlim]

just wondering, i know nothing about pendulums and orbits or epogee and apogees, but I am just wondering, if your calculations involved the acceleration of an object due to gravity, (which is essentially due to mass/weight, whatever you want to call it) wouldn't you need to compensate for the weight loss in the making of the tunnel, and as such change your values, so that the world in theory is a little lighter?

 

[Rogerio]

just wondering, i know nothing about pendulums and orbits or epogee and apogees, but I am just wondering, if your calculations involved the acceleration of an object due to gravity, (which is essentially due to mass/weight, whatever you want to call it) wouldn't you need to compensate for the weight loss in the making of the tunnel, and as such change your values, so that the world in theory is a little lighter?

Of course not!

For a 100cm diameter tunnel, the relative difference is less than 1/200,000,000,000,000 .

 

[Bartholomew]

Okay, Rogerio, I think I get it--thanks.

 

[regor60]

pendulum

Doesn't the fact that gravity acts on the pendulum in the direction of its motion only in two places on the arc and is diminished elsewhere, in contrast with the tunnel, where gravity always points in the direction of motion, suggest that the pendulum model will overstate the time taken ?

 

[BobG]

Doesn't the fact that gravity acts on the pendulum in the direction of its motion only in two places on the arc and is diminished elsewhere, in contrast with the tunnel, where gravity always points in the direction of motion, suggest that the pendulum model will overstate the time taken ?

Nope.

All three methods work (spring, pendulum, Kepler's third law).

But you have to set the equilibrium point for the spring at the correct location and the length of the pendulum to the correct value. You can't get distracted by the opposite side of the Earth. It's a one hemisphere problem - starting from 0, you want to reach the center of the Earth, then return to the surface of the Earth reaching a velocity of 0 just as you reach the surface of the Earth.

If you use the radius of the Earth for the equilibrium point, the pendulum length, or the semi-major axis, you've basically moved all the Earth's mass to one opening of the tunnel.

 

[ceptimus]

Doesn't the fact that gravity acts on the pendulum in the direction of its motion only in two places on the arc and is diminished elsewhere, in contrast with the tunnel, where gravity always points in the direction of motion, suggest that the pendulum model will overstate the time taken ?

Gravity always attracts the tunnel object directly towards the centre of the Earth. The force can be resolved into two components: the force normal to the tunnel (that holds the object down) and the force parallel to the tunnel axis (that accelerates or decelerates it). See diagram below.

http://img6.exs.cx/img6/7773/tunnel4yn.gif

Turn the diagram upside-down and you have the classic simple pendulum diagram.

Keep the tunnel very short (say a few metres) and any effects due to part of the Earth mass being above the tunnel become negligible.

This isn't the same as the O.P's original tunnel that went right through the centre of the Earth - like I said, I'm relying on the fact (proven by others) that ANY straight tunnel joining two equal height points on the Earth's surface has the same gravity-powered transit time.

 

[Bartholomew]

The orbit doesn't work because the net gravity within the Earth does not follow the inverse-square law.

 

[BobG]

42 ... obviously!

But the answer does rely on the gravitational physics that exist inside a sphere (you can't use the 'simple' version).

Kind like Rogerio said. Per the shell theorem, you can disregard all the mass further from the center of the Earth than the falling object. If the object is closer to one side, there is more mass between it and the opposite side, but it's closeness to the near side winds up a wash with no net acceleration.

The amount of mass 'inside' the object's radius decreases as the object gets closer to the Earth. But all of the inside mass can be treated as existing at the center of the Earth. The mass is decreasing by the cube of the radius while the force of gravity is decreasing by the square of the radius. The net affect is a nice linear reduction in gravitational acceleration. (This is why ceptimus's pendulum formula works with no adjustments - the numerator and denominator are being reduced by the same rate).

42 winds up being the answer to everything.

Using ceptimus's gravity tunnel trains:
How long from New York to LA? 42.
How long form New York to London? 42.
How long from Sydney to Beijing? 42

By the way, Lewis Carroll was fascinated by the idea of gravity tunnel trains and the number 42 permeates his Alice in Wonderland Books.

I think Douglas Adams must have have been a Lewis Carroll fan, especially given the hint he left in "The Restaraunt at the End of the Universe". How many cards in a deck of cards?

 

[Bartholomew]

I've wondered before, is there an intuitive reason why you can disregard the mass outside your radius? Or do you just have to get that result by calculus?

 

[DaveC426913]

"is there an intuitive reason why you can disregard the mass outside your radius"

Yes. Because the net gravity inside a hollow sphere is zero.

It can be shown mathematically that, at any point inside a hollow sphere, the gravitational pull of the near side of the sphere is exactly canceled out by the pull of the far side of the sphere, leaving you at rest at any point inside.

That means that when you are standing 100 km below the surface of the Earth, a shell of the entire Earth 100 km thick all around has no net effect on your weight. You are effectively standing on an Earth that is (E(r) - 100km) in radius.


P.S. This is why Larry Niven's Ringworld was- and in the more general sense, Dyson Spheres are- unstable. The Sun at the centre has no reason to stay put.

 

[Bartholomew]

Well, yes, but my question is, is there a way to get that result intuitively without using math? Like is there some intuitive property of the interior of a hollow sphere?

 

[Rogerio]

Well, yes, but my question is, is there a way to get that result intuitively without using math? Like is there some intuitive property of the interior of a hollow sphere?

No, you need calculus to get the result. And it has to do with the uniform density external shell.

Same way, for external objects , a body acts like all its mass was at its center, but you need math to say that.

 

[BobG]

Well, yes, but my question is, is there a way to get that result intuitively without using math? Like is there some intuitive property of the interior of a hollow sphere?

It has to be figured out using math, but you could figure it out with a spreadsheet if you didn't know calculus (calculus is definitely the easiest manual method). If you break your sphere into a lot of equal size sections (actually, you could get a feel for the problem just using a slice of the sphere - a circle) and calculate the gravitational attraction of each section in a spreadsheet, you can then sum up the sections to left of your object, sum the sections to the right of your object and compare them. The hardest part is setting the location of each section in your spreadsheet. You'd like to avoid having to manually enter each location (i.e. - you usually want to figure out a formula to calculate the next location based off the previous cell). In this case, the center of the circle makes a good reference point. The location of the object is measured against the center, giving you one side of a triangle. If you move around the circle in small angles, you can get side-angle-side of a triangle with the third calculated side being the line between your object and a section of the circle. In this case, you'd want that line in vector format (x-component, y-component) with your gravitational force in vector format. That way you could sum the entire spreadsheet and let the x's and y's cancel themselves out instead of deciding which sections to sum yourself. If you get everything down to formulas, by copying and pasting, you can brute force the problem, doing hundreds of calculations at once.

 

[Bartholomew]

Well, I'm not going to believe so easily that there is no way to get an intuitive answer. Using a spreadsheet is less intuitive than calculus, not more. Is a hollow sphere the only shape for which net gravity is zero at any point in its interior? If it is there must be something special about that property which can be understood--perhaps first with math but then with intuition. I'm not talking about just plugging in the formulas and getting the result of zero gravity--that's not really understanding it.