This is an excellent question.  In fact, I would say it is an advanced mechanics question worthy of a late undergraduate or early graduate school class.  Typically in the basic gravity tunnel question, rotation is ignored or removed by rails or other guidance system.

Solving this problem and presenting a solution at a 10^th to 12^th grade level is a bit tricky.

 

Here are few points to consider that can give clues on how to solve this problem.

 

The total energy (kinetic and gravitational potential) is conserved.

 

The gravitational potential is proportional to the mass included inside a shell of radius equal to the distance from the center of the earth.

 

Solving the non-rotational problem is probably a good start to understanding the problem.

 

If we set up the problem so that the object is held over the hole at the surface of the earth then it will have an initial angular speed relative to a frame that is stationary with the center of the earth.  Since, gravity is a central force there will be no torque induced.  This means angular momentum will be conserved like in a planetary orbit problem.  This is true in the non-rotating case but is not as interesting because the angular momentum is zero.

 

This problem may be similar to a Foucault pendulum.   Basically, a Foucault pendulum is a huge pendulum that oscillates side to side but as another oscillation as well due to the rotational motion of the earth.  Example and explanations can be found online.

 

http://www.calacade my.org/products/pendulum.html

 

 

 

Part 1: Solving for the potential. 

 

These sorts of problems are common in E&M courses but can be extended to gravitation.  The technique is referred to as Gauss’ Divergence Theorem.   In words it can be expressed as: The force at any given point comes from the mass contained within a sphere centered on the mass distribution with a radius equal to the distance between the object and the center of the mass distribution.  The equation looks like:

 

g * dArea = -4*PI * G * Mass_enclosed.

 

Where g is the gravitation field (an acceleration), dArea is the area element (technically there is a dot product of the force and the normal vector to the enclosed surface).  There is a negative to indicate the field is directed toward the mass.  The 4 * PI is in there to get the form to work out.  Consider the case of a point mass:

 

g * 4* PI * r² = - 4 * PI * G * Mass

 

g = - G * Mass / r²

 

 

If we want the force, we multiple the g-field by mass_object.

 

From this we can get the potential, which is what we want for setting up the conservation of energy equation (Field Potential = - Del (field))

 

Field Potential = -G * Mass /r.

 

Note that the potential energy is the field potential multiplied by the mass of the object, mass_object.

 

One simplifying assumption I will make is that the earth has uniform density.  This is not true, but it makes the problem tractable.   The earth is also assumed to be a sphere.

In the following Re is the radius of the earth, G is the universal gravitation constant, m is the mass of the object, r is the distance between the object and the center of the earth. 

 

 

 

g-field outside the earth (r > Re):

 

g * 4 * PI * r² = -4 * PI * G * Integral^0->Re{* density_earth * 4 * PI * r² *dr}

 

The term Integral^0->Re{* density_earth * 4 * PI * r² *dr} = Mass_earth, so if the radius is bigger than the radius of the earth we get the expected result of

 

g = -G * Mass_earth/r².

 

 

g-field inside the earth (0< r < Re):

 

g * 4 * PI * r² = -4 * PI * G * Integral^0->r{* density_earth * 4 * PI * r² *dr}

 

The term Integral^0->r{* density_earth * 4 * PI * r² *dr} = (4 * PI * density_earth /3) * r³

 

or is equal to Mass_earth * (r/Re) ³.

 

So

 

g = - G * Mass_earth * r /Re³.

 

 

 

 

The potential energy outside the earth (r > Re):

 

U₁ = -G * Mass_earth * mass_object/r

 

The potential energy inside the earth (0 < r < Re):

 

U₂ = ˝ * (G * Mass_earth * mass_object/ Re³) * r² + constant to be worked out for continuity.

 

Notice the g-field is continuous.  In other words, both equations give the same value at

r = Re.  However, the potential is not.  So I am leading you a bit by stating that a constant will be worked out.  A constant potential creates no force because the derivative of a constant is zero.  Usually for gravitational potential we like to say that at r = infinity the potential approaches zero, so we will not add a constant to that part.

 

We need to calculate this constant to make our potential continuous at r = Re. 

 

U₁(r = Re) = U₂(r = Re).

 

Constant = - 3/2 * G * Mass_earth * mass_object / Re.

 

So the final form of the potential energy U:

 

r > Re:  U = -G * Mass_earth * mass_object/r

 

r < Re:  U = ˝ * (G * Mass_earth * mass_object/ Re³) * r² – 3/2 * G * Mass_earth * mass_object / Re

 

 

 

Part 2: What we know from the non-rotating problem

 

When we multiple the g-field by mass_object to get force this looks just like Hooke’s Law for restoring forces (like springs), Force = - k * displacement.   This where the concept of a gravity tunnel train with the train oscillating between end points evolved.

 

The energy equation is:

 

Energy of object(r, dr/dt) =

   ˝ * mass_object* (dr/dt) ²

+ ˝ * (G * Mass_earth * mass_object/Re³) * r² 

- 3/2 * G * Mass_earth * mass_object / Re

 

This equation is just like a bouncing spring or an oscillator.  If the object is initially at rest when at position r = Re, then the initial energy is just potential energy:

 

E (start) = - G * Mass_earth * mass_object / Re.

 

Energy < 0 indicates the system is bound.  In other words, the object is stuck in our system and is not going to go flying off into space unless it receives extra energy from somewhere that gives the whole system positive energy.

 

The force equation is:

 

mass_object * d²r/dt² – (G * Mass_earth * mass_object/Re³) * r = 0 

 

From which the frequency of oscillation can be found (d²r/dt² + (2*PI * Freq)²r = 0)

 

Frequency = sqrt (G * Mass_object / Re³)/ 2 * PI.

 

You could also figure out the speed as a function  distance r from the center of the earth.

 

 

Part 3: Apply the rotating effects.

 

We assume that we will only be working inside the earth, since we expect some symmetry and conservation of energy (a mass bobbing on a spring) to indicate that the object will not drop from one side of the tunnel and fly out above the ground on the other side.  This is the case of the non-rotating problem.  It seems valid here as well.

 

Since angular momentum is conserved, our energy equation has a term that comes from the angular part:

 

E_angular = L²/ (2 * mass_object* r²),

 

where L is the angular momentum.

 

The rotation can be treated as a Coriolis force so, the g-field is written as

 

g = g(r ) – w X (w X r).

 

Where w is the angular frequency and X represents the vector cross product.  This force is directed away (outward) from the axis of rotation.   For example, if we were doing this at the equator then the Coriolis force lifts away from the center of the earth.  At the rotational poles the Coriolis force disappears and the object is going to spin relative to the earth as it drops.  The strength of Coriolis force will go as

 

Magnitude Coriolis = w² * r * sin (theta),

 

where theta is the angle between the rotational pole and the position of the object.

 

Suppose the path in the non-rotating case goes right through the center of the earth.

To what I think is a pretty good approximation, this extra force will bow the path of the object away from straight through the center and then back again on the other side.  The exact shape, I think, is difficult to calculate.  There are all sorts of possibilities if the line of the path does not go through the center of earth or is angled relative to the axis of rotation. 

 

In this way, this problem also feels a bit like a Foucault pendulum.

 

 

The energy equation can be written as:

 

Energy of object(r, dr/dt) =

   ˝ * mass_object* (dr/dt) ²

+ L²/ (2 * mass_object* r²)

+ ˝ * (G * Mass_earth * mass_object/Re³) * r² 

- 3/2 * G * Mass_earth * mass_object / Re.

 

 

 

Now, this problem is a bit trickier.

 

Let’s try to just get some of the more obvious behavior, including an approximate idea of the path out of this equation.

 

First of all, because of the 1/r² term the object will never pass through the point r = 0.

Secondly, this is like an elliptical orbit except that the mass of the central body is changing.

 

At the start the object is rotating with the earth with an angular frequency of 2 * PI / (seconds in a day), or = 2 * PI / 86400 radians/second. 

 

The angular momentum is

 

L = mass_object * Re * (2 * PI * F_earth)². 

 

The initial energy is

 

E(start) = L²/ (2 * mass_object* r²) -  (G * Mass_earth * mass_object/Re)

 

 

So the speed squared in the radial direction (after some algebra) is given by:

 

(dr/dt)² = L²/ (mass²_object* Re²) * (1 – (Re/r)²) + (G * Mass_earth/Re) * (1 – (r/Re)²).

 

 

From this equation, we see that the radial speed dr/dt = 0 when r = Re.  Also because (dr/dt)² must be greater than 0, we see there must be a minimum r.

 

I hope this explanation helps at least to indicate the difficultly of solving the seemingly simple question you asked.

 

 

Sincerely,

 

Tom “Foucault” Cull

[note added by MadSci Admin: There are some previous answers in our archives dealing with gravitation and the Earth, particularly this one, which deals with the detailed derivation of the gravity inside a sphere.]