| Reminds me of freshman physics... an approximate answer is easy.
Its well known that from a point outside a spherical shell of uniform
mass density that the gravitational potential is effectively due to a point
source of mass at its center; inside the shell you'll feel nothing.
On the surface, neglecting rotation of the sphere (or situated at a pole)
you will feel an acceleration of the entire mass at the center. As you move
into the sphere, you can deduct the mass of the shell outside you.
Let rho be the volume mass density, G be the gravitational constant,
g be the acceleration at the surface of the sphere, R be the radiius of the
sphere, and r be your distance from the center. The result will be simple
harmonic motion satisfying the following differential equation:
2
d r 4
--- + --- pi*rho*G*r = 0
2 3
dt
A solution satisfying the init conditions is r(t) = R*COS(k*t),
where k is the 'wave number' of the periodic motion. k = sqrt(g/R).
The period T must satisfy T = 2*pi/k. The earths R is about 6.26E6
meters, g = 9.8 meters/sec**2, so T is about 1.4 hours for a round trip.
This is reasonably less than the rotational period of the earth so this isn't
a bad approximation.
To conserve angular momentum you'd want your eastward angular velocity to
increase as you go inward and the eastern wall would hold you back on the
way down.
I'll have to think a moment about the exact effects of Coriolis forces,
its been a long time since I've done any classical mechanics.
In fact, the above may be entirely wrong!
- Jim
|
| I am going to add the assumption that your mass is insignificant compared to
the Earth, otherwise we have to worry about the Earth slowing down.
I will use these symbols:
O angular position (from the Greek letter theta),
r current distance of object from center of the Earth,
R radius of the Earth (about 6.378 * 10^6 meters),
P period of the Earth's rotation (about 86,400 seconds),
g acceleration due to gravity at the Earth's surface
(about -9.80665 meters per second per second),
g(r) acceleration due to gravity at a distance r from the center
of the Earth,
m mass of the object,
t time since the object started falling,
w a constant to be explained below (from the Greek omega),
dx differential of x, and
d^2x or dx^2 for d(dx).
Some of the constants are a little off; for example, the figure given for
gravity is actually for the Earth's surface at a latitude of 45 degrees.
At t = 0, O is 0, and r is R. Since the object is forced into the same
angular position as the tunnel, it follows the Earth's rotation. Thus,
O = 2*pi*t/P, and dO/dt = 2*pi/P.
Since the wall is frictionless, there are only two forces acting on the falling
object, in a radial direction. The first is centripetal force. Of course,
this is only an apparent force, due to the frame of reference I have chosen.
To determine the centripetal force, consider the object as if it were held in
place in the tunnel. The force required to hold it in place is
m*v^2/r, where v is the velocity perpendicular to the vector
from the center of rotation to the object.
Without such a force holding the object in place, the apparent centripetal
force is m*v^2/r, in an outward direction. F = ma, so the acceleration is
v^2/r. v is r*dO/dt. So the centripetal acceleration is
(r*dO/dt)^2/r = (r*2*pi/P)^2/r = r*4*pi^2/P^2, outward.
To find g(r), recall that gravity is proportional to the mass causing it and
inversely proportional to the distance from that mass. As stated in a
previous reply, a spherical shell farther from the center than an object can
be ignored, and a sphere can be considered as a point mass. Thus,
g(r) = r^3/R^3 * R^2/r^2 * g(R), outward,
= r/R * g, outward.
The cubes allow for the reducing mass, as an object moves toward the center,
and the squares allow for the changing distance. g(r) is directed outward
because g is negative.
The total radial acceleration of the object is:
d^2r/dt^2 = r*4*pi^2/P^2 + r/R * g, outward,
= r*(4*pi^2/P^2 + g/R), outward.
The constant multiplying r is negative (otherwise the object has reached
escape velocity and will not fall into the tunnel at all). This means that,
along the radius, there is simple harmonic motion. Digging out the physics
book, I find:
r = A cos (w*t + B), where A and B are constants, and
w = sqrt(-(4*pi^2/P^2+g/R)).
dr/dt = - w*A sin (w*t + B).
Considering dr/dt at t = 0, dr/dt = 0 (initially, the object is not moving
downward or upward),
0 = - w*A sin B.
w is non-zero, and A must be non-zero (otherwise r is always 0), so sin B must
be zero. Thus B = 0 (or a multiple of pi, which is equivalent for our
purposes). Considering r at t = 0, r = R,
R = A cos 0, so A = R.
Thus,
r = R cos (w*t).
The object reaches the center when r = 0, so w*t = pi/2. Solving for t
gives a value of about 1269 seconds. Without rotation, this would be 1267
seconds. These figures may vary slightly, depending upon what values you
chose for g, P, and R. For example, you may chose R to be the actual mean
radius of the Earth at the equator, or it might be what the radius would be
if the Earth were a sphere with the same volume as it has now.
Reaching the other side of the planet takes twice as long.
Peak velocity is - w*R, in direction pi^2/(w*P).
Does anybody want to tackle the lateral force?
-- edp
|
| Generalization: Start at any point on the Earth, and drill a hole "downward" -
NOT necessarily along a radius! - until you reach the surface again. Drop
something into the hole. Assuming, of course, no friction with the side walls,
your object falls through to the other end. How long does it take to get there?
It turns out - if I remember this right - that the transit time is independent
of the angle at which you started drilling the hole. The way you model this is
to consider the object as being a pendulum bob at the end of an infinitely long
string. The magic time value - the limiting period of a pendulum - for the
Earth works out to something like 44 minutes.
This has actually been proposed as a practical method of transportation. You
use an evacutated tube with a train kept away from the walls by magnetic
levitation. This produces extremely low, though still non-zero friction.
You compensate for that by pumping some air in behind the train to get it
started. Most of the energy to move the train comes from gravity. Some more
comes from the compressed air, but in the process of moving the train, you
compress the air in front of it, so you get some of that back, too. The end
result is, in principle, a very efficient system.
Last I heard, there was a group trying to build an evacuated-tunnel train
system (though not one with a "gravity drive", as far as I know). Tom
Stockebrand was involved - a ping-pong ball served as the train, and he
had a 50-foot (?) evacuated tube installed under his back yard.
-- Jerry
|
| In my origional reply, I hedged on the effect of the earth's rotation,
knowing the effect would be small. On thinking about it a little more
it seems that even including it the system is integrable. If we have the
angular frequency of your motion along the radius as k (bad nomenclature
since wave numbers, k, should really refer to spatial wave effects),
omega the earth's rotational rate, and (pi/2-phi) your latitude we can
make the substitution
2 2 2 2
k -> k - (sin(phi)*omega) = k1
to evaluate the effect of the earths rotation. Even at the equator, this
only has about a .1 percent effect on the round trip time. The centrifugal
force of the earth's rotation at the equator is about .33 percent of a g.
If we express your acceleration in a spherical polar frame of reference
we get
a = 2*k1*omega*sin(phi)*(R*sin(k1*t))
theta
2
a = omega * sin(phi)*cos(phi)*(R*cos(k1*t))
phi
2 2
a = (k - (omega*sin(phi)) )*(R*cos(k1*t))
r
With respect to a fixed frame of reference, your motion will be on a conic
surface with its 'waist' at the equator. Two degenerate cases occur if the
surface narrows to a line thru the poles, or flattens out to the equatorial
plane. Your motion will be multiply periodic, and will not follow a closed
path if the ratio of the earth's rotational period and your round trip
time is irrational - a very simple example of ergodic motion.
In the equatorial plane, the path is a petal shaped form - this shows why
you'll feel a force on the eastern wall - since this is the force normal
to your path that causes it to curve. It amounts to about .116 G's of
acceleration at the midpoint.
This analysis seems to agree with other replies, so probably holds good.
The brachistochrone tunnel problem in a homogenous sphere is is a classic
in the calculus of variations and evaluates to a hypocycloid, neglecting
rotation of the earth.
- Jim
|