Ball Dropped in Tunnel Through Earth: What Happens Next?

[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.

 

[hitssquad]

Radial transport

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

Yes, it would.

Wrong, Gokul. HowStuffWorks made the same error:
http://www.howstuffworks.com/question373.htm

• If you could do this on earth, one amazing effect would be the ease of travel. The diameter of the Earth is about 12,700 kilometers (7,800 miles). If you drilled the tunnel straight through the center and could create a vacuum inside, anything you dropped into the tunnel would reach the other side of the planet in just 42 minutes!

It would not, unless the tunnel went from pole to pole. Jim Hoerner of Know Nukes fame has pointed out essentially the same problem in regards to the space elevator.

 

[Rogerio]

...
If you drilled the tunnel straight through the center and could create a vacuum inside, anything you dropped into the tunnel would reach the other side of the planet in just 42 minutes![/list]

Yes, that's right!

It would not, unless the tunnel went from pole to pole. Jim Hoerner of Know Nukes fame has pointed out essentially the same problem in regards to the space elevator.

Well, if the tunnel were at equator, the trip would take just 4 or 5 seconds more, due the centrifugal force.

In other words, basically the same 42 minutes Gokul had said.

 

[hitssquad]

Equatorial freefall tunnels

if the tunnel were at equator, the trip would take just 4 or 5 seconds more

Then why might Ceptimus have mentioned the Coriolis effect?

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.

A related pair of questions might be:

1. What shape might take an evacuated free-fall tunnel spanning two equatorial points of a) a uniformly massive spinning body; and b) the earth?
   
   and,
   
   
2. Would there be any excess speed in an object arriving at the far end of this equator-point-to-equator-point free-fall tunnel (i.e., might it launch out the other end of this tunnel into space)?

 

[Bartholomew]

I'm going to post my question about the hollow sphere in the appropriate forum.

Edit: (the Classical Physics forum)

 

[ceptimus]

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 formulae and getting the result of zero gravity--that's not really understanding it.

One way of getting a handle on this is to consider a large gaseous body - a gas giant planet, or a star. Let's say a non-rotating one to keep things simple.

Consider an atom of the star at some random position inside it. The gravitational effects on this atom can be divided into the part from all the material closer to the centre of the star than it (a sphere) and the remainder (a hollow shell).

If there were a net force inside a hollow spherical shell that attracted a body towards the 'closest wall', then in the star case, this force would be tending to pull the star apart. Clearly any such net force would have to be less than the attracting sphere 'beneath' the atom (if it were not the star would indeed fly apart) and this must be true for all radii from zero out to the star radius.

So we are left with the options that any attracting force towards the shell wall must be fairly weak, or that there must be a net force tending to push an object away from the wall.

It's not an intuitive proof as it stands, but it points you in the direction of thinking that the net gravitational force inside a hollow shell might be zero.

 

[BobG]

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.

Unfortunately, there's just not too much about it that's intuitive, even when considering objects outside the sphere. Kepler came up with his first two laws of panetary motion based strictly on organizing his observations vs. any 'physics'. It took him 10 more years to come up with the third (the observations didn't match what he 'intuitively' expected, that the period should be proportional to the square of the radius).

 

[T@P]

another ineterstuing phenomenon, is if you were to take a satelite which orbits the Earth *directly on the surface of the earth*, that is ignoring the terrain, wind, etc. but hypothetically, if you find the time it takes for that satellite to go around the earth, it is exctly the same if it were to go right through the Earth (!). Moreover, if you dig *any* tunnel through the earth, the oscillating ball will have a period of 42 min. (the same as the one right through the center). ofcourse i am assuming many things, such as the possibility of doing it, the fact that the Earth is not a perfect uniform sphere, etc.

I don't know about you, but as a student taking physics for the first time i find that very exciting/interesting.

 

[Bartholomew]

In the Classical Physics forum Krab and Kuiper explained the zero-gravity-in-a-hollow-sphere effect. See Kuiper's link, http://ebtx.com/ntx/Newton.htm

 

[Rogerio]

if the tunnel were at equator, the trip would take just 4 or 5 seconds more

Then why might Ceptimus have mentioned the Coriolis effect?

I don't know what he had in mind. But in order to avoid extra considerations about the interaction among the ball and the tunnel, we may consider there is no friction force between the ball and the tunnel wall.

And since Coriolis doesn't contribute anything to the radial acceleration (center of the Earth direction), it doesn't change the calculations.

So, it would be the same 42 minutes.

 

[Rogerio]

In the Classical Physics forum Krab and Kuiper explained the zero-gravity-in-a-hollow-sphere effect. See Kuiper's link, http://ebtx.com/ntx/Newton.htm

Well Bartholomew, maybe after take a look at those drawings, you can easily understand how to divide the big problem into small ones, but I don't think the final effect (zero gravity) is intuitive.

They use geometry in the same way I would use calculus to get the result.

 

[ceptimus]

I don't know what he had in his mind. But in order to avoid extra considerations about the interaction among the ball and the tunnel, we may consider there is no friction force between the ball and the tunnel wall.

And since Coriolis doesn't contribute anything to the radial acceleration (center of the Earth direction), it doesn't change the calculations.

So, it would be the same 42 minutes.

Are you sure?

Let's consider a tunnel starting from the equator, and initially heading north east, that emerges at a point 60 degrees north. The eastward velocity at the equator due to Earth rotation is 1674.4 km/hr. At 60 degrees north the eastward velocity is exactly half as much. So the vehicle that traverses the tunnel must lose 837.2 km/hr of eastwards velocity on its journey if it is to come to a standstill at the tunnel exit. Now it must do this by applying an eastwards force against the tunnel wall.

If the tunnel walls are frictionless, then they can only impart a reacting force normal to the tunnel walls, and this will be in a roughly north west direction. Won't this cause the vehicle to accelerate northwards?

I'm not saying I think you are wrong. I'm not sure. I'd appreciate a better explanation.

 

[Bartholomew]

Rogerio, geometry is more intuitive than calculus. Calculus is just a numerical way to represent geometry; geometry deals with more real, understandable, distinct ideas. With calculus, everything is just a function, which (once you've set up, and the setting-up _is_ an intuitional activity) you crunch through without really thinking about the ultimate meaning of each step, until you arrive at an answer.

It is like the difference between using calculus to determine that the integral from a negative value to the corresponding positive value of sin x will be 0, and looking at the actual graph of the function to see that it is so.

 

[Rogerio]

Are you sure?

Let's consider a tunnel starting from the equator, and initially heading north east, that emerges at a point 60 degrees north. The eastward velocity at the equator due to Earth rotation is 1674.4 km/hr. At 60 degrees north the eastward velocity is exactly half as much. So the vehicle that traverses the tunnel must lose 837.2 km/hr of eastwards velocity on its journey if it is to come to a standstill at the tunnel exit. Now it must do this by applying an eastwards force against the tunnel wall.

If the tunnel walls are frictionless, then they can only impart a reacting force normal to the tunnel walls, and this will be in a roughly north west direction. Won't this cause the vehicle to accelerate northwards?

In this case? Yes.

But it is not the kind of tunnels we were talking about.

Just to refresh:

...
If you drilled the tunnel straight through the center and could create a vacuum inside, anything you dropped into the tunnel would reach the other side of the planet in just 42 minutes!
...
It would not, unless the tunnel went from pole to pole.

Well, if the tunnel were at equator, the trip would take just 4 or 5 seconds more, due the centrifugal force.
In other words, basically the same 42 minutes Gokul had said.

As you can see, we're discussing tunnels that go straight through the center of the earth.
And yes, I'm sure the trip would take about 42 minutes (but of course I can be wrong!).

I'm not saying I think you are wrong. I'm not sure. I'd appreciate a better explanation.

Take a look at my calculations below:

The differential equation, without considering Earth's rotation would be
d2x/dt2 = -K*x , so,

x= R_earth * cos( sqrt(K)*t ) R=6366000 m , K= 9.81/(6366000 s)^2

Now, taking into account the Earth's rotation:

d2x/dt2 = -K*x + (w*cosA)^2*x

where 'w' is the angular speed ( 2*pi/86400 s) , and 'A' is the angle between the tunnel and the equatorial plane.

So, x= 6366000 * cos( sqrt(9.81/6366000 - (2*pi*cos(A)/86400)^2) * t) , or

x=6366000 * cos( sqrt(291.39 - cos(A)^2) * t / 13750.99 )

Using these values, the time diff between the trips using a polar tunnel and a equatorial tunnel would be 4.35 seconds.

 

[Rogerio]

Rogerio, geometry is more intuitive than calculus. Calculus is just a numerical way to represent geometry; geometry deals with more real, understandable, distinct ideas. With calculus, everything is just a function, which (once you've set up, and the setting-up _is_ an intuitional activity) you crunch through without really thinking about the ultimate meaning of each step, until you arrive at an answer.

It is like the difference between using calculus to determine that the integral from a negative value to the corresponding positive value of sin x will be 0, and looking at the actual graph of the function to see that it is so.

Bartholomew, I agree that geometry is not so abstract as calculus,
but the Euler triangle, for instance, is pure geometry. However, it is not intuitive.

In the same way, I don't think the zero gravity in hollow spheres is an intuitive result, even using only geometry to get it.

 

[ceptimus]

The north-east (south west when making the return journey) tunnel that I described in my earlier post seems to have paradoxical properties:

When the vehicle makes the journey away from the equator, it has to lose its excess eastward velocity, and this would seem to accelerate it northwards (due to the reaction against the south east wall).

When making the return journey the vehicle has to gain eastward velocity, this requires a reaction against the north west wall, which would accelerate it southwards.

So when going in either direction, the vehicle would gain speed, and rather than coming to a stop at the end of the tunnel, it would arrive there with excess speed.

Is this really what would happen? If not, where is the flaw in my description?

 

[Bartholomew]

The geometrical approach is far more intuitive than the calculus approach. This could turn into a lengthy debate so I'll say right now that you are not going to change my mind. You keep your opinion and I'll keep mine. Unless that really bothers you, in wihch case let's use the general philosophy forum.

 

[jlsoto]

Yes, that's right!



Well, if the tunnel were at equator, the trip would take just 4 or 5 seconds more, due the centrifugal force.

In other words, basically the same 42 minutes Gokul had said.

yes, but what if the tunnel is neither at the equator nor from north to south pole? what if it was let's say from San Francsico to Los Angeles? or from the north pole to Los Angeles, would it still be the same time?

 

[Doc Al]

yes, but what if the tunnel is neither at the equator nor from north to south pole? what if it was let's say from San Francsico to Los Angeles? or from the north pole to Los Angeles, would it still be the same time?

Set up the problem and see if that matters.

 

[jlsoto]

i'm just a simple layman that got interested in the subject. i have no idea of formulas or anything, you would have to explain it to me in terms that i would understand, that's why i asked.

 

[K.J.Healey]

I only get about 14.9 minutes: Plug into Mathematica:
F = -G m M /r[t]^2;
U = Integrate[F, r[t]];
\[Rho] = ME/((4/3)*Pi*RE^3);
M = \[Rho]*(4/3)*Pi*r[t]^3;
T = (1/2) m (r'[t])^2;
L = T - U;
a = D[L, r[t]];
b = D[D[L, r'[t]], t];
d = Solve[a - b == 0, r''[t]][[1]][[1]][[2]]
f = DSolve[{r''[t] == d, r[0] == RE, r'[0] == 0}, r[t],
t][[1]][[1]][[2]];
Time = -Solve[f == 0, t][[1]][[1]][[2]]
ME = 5.9742*10^24 _kg;
RE1 = 6356.75 *1000 _m;
RE2 = 6378.135 *1000 _m;
G = 6.667428*10^-11 (_m^3) _kg^-1 _s^-2;
RE = RE1;
Sqrt[Simplify[Time^2]*(1 _min^2/(60 _s)^2)]
RE = RE2;
Sqrt[Simplify[Time^2]*(1 _min^2/(60 _s)^2)]OUTPUT:
14.8658 minutes
14.9408 minutes

For the longest and shortest radii.

 

[Doc Al]

I only get about 14.9 minutes: Plug into Mathematica:
F = -G m M /r[t]^2;
U = Integrate[F, r[t]];
\[Rho] = ME/((4/3)*Pi*RE^3);
M = \[Rho]*(4/3)*Pi*r[t]^3;
T = (1/2) m (r'[t])^2;
L = T - U;
a = D[L, r[t]];
b = D[D[L, r'[t]], t];
d = Solve[a - b == 0, r''[t]][[1]][[1]][[2]]
f = DSolve[{r''[t] == d, r[0] == RE, r'[0] == 0}, r[t],
t][[1]][[1]][[2]];
Time = -Solve[f == 0, t][[1]][[1]][[2]]
ME = 5.9742*10^24 _kg;
RE1 = 6356.75 *1000 _m;
RE2 = 6378.135 *1000 _m;
G = 6.667428*10^-11 (_m^3) _kg^-1 _s^-2;
RE = RE1;
Sqrt[Simplify[Time^2]*(1 _min^2/(60 _s)^2)]
RE = RE2;
Sqrt[Simplify[Time^2]*(1 _min^2/(60 _s)^2)]


OUTPUT:
14.8658 minutes
14.9408 minutes

For the longest and shortest radii.

Not quite sure what you did here. Making the usual assumptions (uniform density, spherical, non-rotating Earth & frictionless/airless tunnel), the transit time from one end to the other is about 42 minutes. It's easy to show that the time does not depend on whether the tunnel goes through the center of the Earth or not. (Sorry for the delayed response; been away for a few days.)

 

[K.J.Healey]

I just used the Force/Potential of gravity to get equations of motion.
I get :
R'' = -((2 G ME r(t))/RE^3)
Where G is G
ME is mass of Earth (total)
and RE is surface radius of the earth.

I Made my "mass" of the Earth in:
F=-G m M[r] /r^2 a function of the distance from the center, that is calculated using the density of the Earth (total mass/total vol) times the current radius.

so as the object falls it feels less and less gravity.
The acceleration I got was above, which is linear in "r"

Solving this equation (simple ODE) you get a sinusoid that has:
r''=-K*r
so its standard cos+sin, but at t=0, r'[0] == 0 and r[0] == Radius of Earth
initial conditions.

My "K" is : 3.07034*10^-6 s^-2

And I solve t differential in the DSolve line. Plugging in a bunch of constants I get:

R[t]== RE cos((Sqrt[2] Sqrt[G] Sqrt[ME] t)/RE^(3/2))

which simplifies to:
R[t]==(6.37814*10^6)*Cos[0.00175224 t]
in meters, and seconds.
First 0 point is at about 900 seconds ~15 minutes.

 

[Doc Al]

I just used the Force/Potential of gravity to get equations of motion.
I get :
R'' = -((2 G ME r(t))/RE^3)
Where G is G
ME is mass of Earth (total)
and RE is surface radius of the earth.

Where did the 2 come from?

 

[K.J.Healey]

Where did the 2 come from?

When you take the potential
U = G m M[r] / r
And you calculate the mass of everything inside a circle of radius r:
M = density * (4/3) Pi r^3
so the potential goes like r^2
When you apply Euler-Lagrange the 2 comes down in the d/dr

 

[K.J.Healey]

Heres how i did it in mathematica:
<br /> F= -G m M /r[t]{}^{\wedge}2<br />
<br /> -\frac{G m M}{r(t)^2}<br />

<br /> U=\text{Integrate}[F,r[t]]<br />
<br /> \frac{G m M}{r(t)}<br />

<br /> \rho = \text{ME}/((4/3)*\text{Pi}*\text{RE}{}^{\wedge}3)<br />
<br /> \frac{3 \text{ME}}{4 \pi \text{RE}^3}<br />

M=\rho *(4/3)*\text{Pi}*r[t]{}^{\wedge}3<br />
<br /> \frac{\text{ME} r(t)^3}{\text{RE}^3}<br />

T=(1/2)m (r&#039;[t]){}^{\wedge}2<br />
<br /> \frac{1}{2} m r&#039;(t)^2<br />

<br /> L=T-U<br />
<br /> \frac{1}{2} m r&#039;(t)^2-\frac{G m \text{ME} r(t)^2}{\text{RE}^3}<br />

\text{c1}=D[L,r[t]]<br />
<br /> -\frac{2 G m \text{ME} r(t)}{\text{RE}^3}<br />
\text{c2}=D[D[L,r&#039;[t]],t]<br />

<br /> m r&#039;&#039;(t)<br />
\text{accel}=\text{Solve}[\text{c1}-\text{c2}==0,r\text{&#039;&#039;}[t]][[1]][[1]][[2]]<br />
<br /> -\frac{2 G \text{ME} r(t)}{\text{RE}^3}<br />

\text{motion}=\text{DSolve}[\{r\text{&#039;&#039;}[t]==\text{accel},r[0]==\text{RE},r&#039;[0]==0\},r[t],t][[1]][[1]][[2]]<br />
<br /> \text{RE} \cos \left(\frac{\sqrt{2} \sqrt{G} \sqrt{\text{ME}} t}{\text{RE}^{3/2}}\right)<br />
\text{Time}=-\text{Solve}[\text{motion}==0,t][[1]][[1]][[2]]<br />

<br /> \text{Solve}\text{::}\text{ifun}: \text{Inverse functions are being used by $\backslash $!$\backslash $(TraditionalForm$\backslash \grave{ }$Solve$\backslash <br /> <br /> $), so some solutions may not be found; use Reduce for complete solution information. $\backslash $!$\backslash $($\backslash $*ButtonBox[$\texttt{&quot;}<br /> <br /> \rangle\rangle \texttt{&quot;}$, ButtonStyle-$&gt;\texttt{&quot;}$Link$\texttt{&quot;}$, ButtonFrame-$&gt;$None, ButtonData:$&gt;\texttt{&quot;}$paclet:ref/message/Solve/ifun$\texttt{&quot;}<br /> <br /> $, ButtonNote -$&gt;$ $\texttt{&quot;}$Solve::ifun$\texttt{&quot;}$]$\backslash $)}<br />

<br /> \frac{\pi \text{RE}^{3/2}}{2 \sqrt{2} \sqrt{G} \sqrt{\text{ME}}}<br />

\text{ME}=5.9742*10{}^{\wedge}24 \text{$\_$kg};<br />

<br /> \text{RE1} = 6356.75 *1000 \text{$\_$m};<br />

<br /> \text{RE2} = 6378.135 *1000 \text{$\_$m};<br />

<br /> G = 6.667428*10{}^{\wedge}-11 \text{$\_$m}{}^{\wedge}3 \text{$\_$kg}{}^{\wedge}-1 \text{$\_$s}{}^{\wedge}-2<br />

<br /> \text{RE}=\text{RE1};<br />

<br /> \text{Sqrt}[\text{Simplify}[\text{Time}{}^{\wedge}2]*(1\text{$\_$min}{}^{\wedge}2/(60 \text{$\_$s}){}^{\wedge}2)]<br />

<br /> \text{RE}=\text{RE2};<br />

<br /> \text{Sqrt}[\text{Simplify}[\text{Time}{}^{\wedge}2]*(1\text{$\_$min}{}^{\wedge}2/(60 \text{$\_$s}){}^{\wedge}2)]<br />

14.8658 \sqrt{\text{$\_$min}{}^2}<br />

14.9408 \sqrt{\text{$\_$min}{}^2}<br />