What if the hole were drilled in some other way? If it were drilled pole to pole, the rotation of the Earth wouldn't affect the fall.
There's a related problem. In Chapter 7 of Lewis Carroll's 1893 book Sylvie and Bruno. The fictional German professor, Mein Herr, proposes a way to run trains by gravity alone. Dig a straight tunnel between any two points on Earth (it need not go through the Earth's center), and run a rail track through it. With frictionless tracks the energy gained by the train in the first half of the journey is equal to that required in the second half. And also, in the absence of air resistance and friction, the time of the journey is about 42 minutes (84 for a round trip) for any such tunnel, no matter what the tunnel's length.
Martin Gardner discussed this in one of his Scientific American "Mathematical Games" columns.
The period in the frictionless hole:
Restating the question: The period for the circular "skimming" orbit (of radius equal to the Earth's radius, R) is
T = 2p (R/g)1/2
Show that this is the same as the period of motion in frictionless straight hole through the center of the stationary Earth.
Partial Answer:
First let's address the question of the force on an object within a homogenous spherical distribution of mass.
The force on a piece of mass at point P a distance r from the center is due to all mass in the sphere of radius less than r. All mass at distances greater than r contributes nothing. This can be seen by subdividing the mass in the outer shell into many shells of infinitesimal thickness. Then take infinitesimal solid angles with vertices at the point P. The two pieces of mass within the solid shell contribute oppositely directed forces at P, of size proportional to the inverse square of their distances from P. But by geometry, their volume is proportional to the square of their distance from P. Therefore, their forces are equal, and being oppositely directed, add to zero.
The mass within r is proportional to r3. Its gravitational force acts as if it were concentrated at the center of the sphere, and therefore the force it exerts at P is proportional to r-2. Therefore the gravitational force at distance r is proportional to r3r-2 = r.
Now consider the straight hole or tunnel through the Earth, but not through the center of the Earth.
The hole has frictionless walls, or tracks. The gravitational force within the Earth is f = -kd where d is the distance from the center. When d = R the force is mg. At the center of the Earth the force is zero.
At any angle ß, the component of gravitational force along the tunnel is f = F(r)sin(ß
. At distance x from the center of the hole an object is distance r from the center of the Earth. x = r[sin(ß
], and we already know that F(r) = -kr and k = mg/R so
f = -kr sin(ß
= -kx
Therefore the force is proportional to the distance, meeting the requirements for simple harmonic motion with an effective "spring constant" k.