Can anyone show that the period for dropping a rock through a hole in the earth is just like low earth orbit?

Well, we could invoke Kepler's 3rd law to determine the orbital period of the rock, but in the interests of using more elementary physics, I'll derive it from what we know about centripetal force; namely, equating the expression for the magnitude of the centripetal force, mv²/R, with the magnitude of the gravitational force, GMm/R², that provides it:

mv²/R = GMm/R²

v² = GM/R

v = (GM/R)^(1/2)

This gives the linear instantaneous speed of the orbiting rock, and we know in one orbital period it covers once circumference of the Earth, of length 2πR:

2πR = vT

T = 2πR/v = 2πR^(3/2)/(GM)^(1/2) (*)

To compute the period of the falling rock, we note that, when the rock is at distance r from the center, it lies above a sphere of volume (4/3)πr³, with mass M_r = (4/3)πr³ρ. Using the assumption of constant density, we may also write M = (4/3)πR³ρ to determine:

M_r/M = r³/R³

M_r = Mr³/R³

Going back to Newton's law of gravitation, the gravitational force acting on the falling rock at depth r is equal to:

F = G(M_r)m/r²

= G(Mr³/R³)m/r²

= GMmr/R³

Using Newton's second law F = ma = m(dv/dt), we have:

GMmr/R³ = m (dv/dt)

GMr/R³ = dv/dt = v dv/dr (using the chain rule, dv/dt = dv/dr * dr/dt = v dv/dr)

We can use this differential equation to relate the distance r(t) at time t to the velocity v(t) at time t by separating variables and integrating:

∫[R, r(t)] GMr/R³ dr = ∫[0, v(t)] v dv (using initial conditions r(0) = R and v(0) = 0)

GM/(2R³) r² | [R, r(t)] = v²/2 | [0, v(t)]

-GM/(R³)(R² - r(t)²) = v(t)²

This next part...I sincerely hope someone submits a more elementary way to derive. We can now, at least for the first fourth of the descent from R to the center, determine that:

v(t) = dr/dt = [-GM/R³(R² - r²)]^(1/2)

Another separation of variables gives;

-dr/(R² - r²)^(1/2) = (GM/R³)^(1/2) dt

Integrating over the first fourth of the period yields:

∫ [R, 0] -dr/(R² - r²)^(1/2) = ∫ [0, T/4] (GM/R³)^(1/2) dt

I'll spare the trig substitution to evaluate the first integral and simply say it is equal to π/2 [1]. The right integral is clearly equal to (GM/R³)^(1/2)*T/4, and we obtain:

π/2 = (GM/R³)^(1/2) * T/4

T = 2πR^(3/2)/(GM)^(1/2)

as we needed.

Does the diameter of the hole count?

If so, then how big is it?

One would assume that it would get "stuck" in the center, no?

_________________________

Edit: Ok, so for the rock in orbit it is T^2=4*pi^2* R^2/V^2

R being radius of orbit to center

For the falling rock, it is not clear, since it has a linear velocity down the rabbit hole, and you mention that the only force acting on the rock is gravity.

Perhaps using something like ke equation but ignoring mass would help, since r of the earth is 1/2 diameter, or lenght of the tunnel, 2 r*g=V^2

And since V ^2 is 4*pi^2*r^2*f^2

Then period is

T^2=4*pi^2*r^2/2*r*g

T^2=4*pi^2*r/2g

_________________

I'll leave it to the nerds to calculate the minimum orbit height, time it takes (extracted from g) and the lenght of the tunnel and all that kind of stuff.

Goodby!

if the hollow exceeded throught the middle of mass of the earth and ignoring air friction - it may oscillate to and fro. With air friction, it may finally stop on the middle of mass, at this element there is 0 internet gravity. If teh hollow did not bypass throughout the middle of mass, the rock could hit the wall of the hollow in some unspecified time sooner or later, and are available to a stop. Edit - clarification: without air friction, there is not any rigidity to gradual the fee of fall of the rock because it fell in direction of the middle - the rock will enhance up all teh thank you to the middle of gravity. The momentum could then carry it previous teh middle, and it may behave as though it have been "thrown" the different direction, faraway from teh middle of mass because it travels. because of conservation of momentum, the rock could proceed against the pull of gravity to a element precisely opposite from the place it became into initially dropped, the place it could have 0 momentum, and as quickly as back fall back in direction of the middle. truly, that's an "orbit" with the only coordinate axis of 0 length. if unacted on by using the different exterior rigidity, the rock will in basic terms oscillate to and fro between the two factors at opposite ends of the hollow. If air friction is taken under consideration, the rock will attain a terminal velocity previously achieving teh middle of the earth, at which it could have a definite momentum that must be appreciably decrease than if there became into no air. The rock could nevertheless bypass by using teh middle, yet does not commute as some distance in teh different direction previously it falls back. this could repeat with the gap getting smaller and smaller, the place the rock will finally come to chill out on the middle of mass the place there could be 0 gravity.