Are there any physicists around Y!A? I am truly desperate?

Question

A hole is drilled through the center of Earth and elevator car is installed inside?
The car departs from the surface (1km above sea level), and after quick initial acceleration continues to move towards the center of the earth at constant velocity 100 m/s.

Few seconds into constant velocity phase of the trip, when elevator just crosses sea level, a passneger drops elactic ball on the floor, which continues to bounce elastically from the floor with initial period 1s.

As the car approaches the center of the earth, period of bouncing gradually increases.

What is the period of bouncing half-way towards the center of earth?
[in assumption of uniformly dense planet, gravity at this point is g/2]

Jack:
nobody ever saw the inner structure of the Eaarth, there are only theories and some limited experimental data.

That some goverment agency posts thier version of physical truth does not make it the true truth.

Dear Dr. Bekki:
this problem belongs to the class of so-called 'adiabatic invariant' problems.

In 8 month around Y!A I saw only
Prof. ***shimrod***
answering such problem correctly.

I invite you to join his company.

"Kinetic energy is conserved so v(bounce) = constant"

With all due respect, sir, no.
Total mechanical energy E = KE+PE is conserved only when external potential does not explicitly depend on time.

In this case (non)conservation of energy
is written as
dE/dt = ∂PE/∂t =
= mh(t) ∂g(r)/∂t = -PE(t) v / Rearth

Ok.
Lets denote local gravity as g(t).
dE/dt = ∂PE/∂t = mh(t) dg/dt
dE/dt = mgh(t) 1/g dg(t)/dt
dE/dt = PE(t) 1/g dg/dt

Averged over period:
dE/dt = (t) 1/g dg/t

In iniform field parabolic motion = 2/3 E
= 1/3 E

dE/dt = 2/3 E 1/g dg/t
dE/E = 2/3 dg/g
E/Eo = (g/go)^(2/3)

Period:
T/To = (E/Eo)^1/2 (go/g)^1
= (g/go)^-2/3 =
³√4 = 1.5874 s

Answer 1

Assuming constant acceleration for a moment:

Time to fall a distance d is given by:
d = 1/2 at^2
so t = sqrt (2d/a)
so total period = 2t = sqrt (8d/a)

It seems to me that the height of the bounces shouldn't change if gravity is changing slowly. I have to think about that assumption for a bit.

If the gravity drops by a factor of 2, the period goes up by a factor of sqrt(2).

Answer 2

Let's ignore the rω^2 component of the radial acceleration, or just lump it in there with g.

Let's assume a varies linearly with position, r, such that
a = g*r/R
g = constant = 9.81 m/s^2
r = R - 100t

So relative to the floor, which is moving at -100 m/s
the ball has equations of motion
dv/dt = -a = -gr/R
= -g*(R-100t)/R
= -g + (100g/R)*t
v = -gt + (50g/R)*t^2
Initially the ball has a velocity of -100 m/s so relative to the floor, it has zero initial velocity

x = -(g/2)*t^2 + (50g/3R)*t^3 + h
The initial height h determines the period

Now the 50g/3R term is negligible since R is very large.
So x ~= -(g/2)*t^2 + h
which is precisely the equation of motion we expect to have if there is no change in gravity.

So I say the period does not change significantly from 1 sec.

Answer 3

The period doubles.
Kinetic energy is conserved so v(bounce) = constant.
Period t = 2v/g and g is reduced to half its initial value, so t doubles.

Answer 4

Here's the dimensions:

http://www.nasa.gov/images/content/103949main_earth10.jpg

If you already knew the answer then why are you so desperate for a physicist?