How long would it take for a ball dropped in a vacume tube to reach the center of the earth.?

Answer 1

Assuming a spherical Earth of uniform density, the answer is: about 21 minutes and 5 seconds.

Due to the mathematics of the inverse square law (with a spherical distribution of mass) the centripetal attraction at a distance z from the center is exactly the attraction which would be caused with all the mass located at a LESSER DISTANCE from the center if it was concentrated at the center. (Curiously enough, there's a zero net gravitation from a uniform sperical hollow shell if you're inside it.)

The attraction is thus proportional to the cube of z (that's what the total mass of the inner sphere is like) divided by the square of z (the inverse square law of gravitation). All told, it's simply proportional to z.

A motion in which the acceleration is proportional to the displacement is called an "harmonic" motion. This is exactly what we have here: The amplitude (z) is a sine function of time and the period of such a motion is like the period of a pendulum; it does NOT depend on the maximum amplitude (if you drop the ball from the halfway point to the center, it will also take 21 minutes and 5 seconds to reach the center; as it just travel shorter distances more slowly).

Curiously, the result remains the same for any straight frictionless vacuum tunnel which does NOT go through the center of the Earth. This is to say that such a tunnel allows gravity to carry a ball from one point of the Earth to any other in just 42 minutes and 10 seconds, starting at sero speed and arriving at zero speed).

Nice piece of "fantasy engineering". See link:

Answer 2

Since the force of gravitational attraction increases nonlinearly as the distance between the two objects decreases, the ball would find itself accelerating MUCH quicker once it got very close to the center of the Earth. However, at that point it would have so much acceleration and velocity it would overshoot the center and from the potential energy that it gained would go pretty much to the other end of the surface (other side of the earth, opposite of the starting end) and at this point the amount of potential energy gained is dispensed and must respond to the gravitational pull of the earth - possibly this oscillation of the ball going in and out of alternating sides of the earth would go indefinately when ignoring friction. If we take Air resistence into account, then the ball would slowly lose some of the "potential energy" of the "fall" towards the center of the earth as heat and other forms of energy and so it would NOT make it quite to the end of the other surface and slowly, but surely it would oscillate in a "damped" motion where it would finally come to rest after some time in the center of the earth (center of the tube). This is of course ignoring any and all circular/corriolis effects which could affect its path and contribute to other forms of loss. Very interesting question - wonder where you came up with the idea. Thanks for asking - hope this suffices.

Answer 3

Be careful! The acceleration due to gravity decreases from approx. 9.8 m/s/s at the surface to zero at the center of the Earth. At a distance R from the center of the Earth and BELOW THE SURFACE, only the mass within a sphere of radius R contributes to the acceleration. All of the other mass outside of the sphere cancels to zero! Assumptions are uniform mass and a spherical Earth.

Answer 4

Scott H may be incorrect.
Sir Isaac Newton demonstrated in the form of a proof, that the gravity remains constant. He demonstrated that the gravity of the particles behind it as it travels exactly equalizes the increasing attraction as the particle grows closer to the center. He did this proof using infinitely thin shells, overlaying each other to combine to make the Earth.
I am sure that many things in physics have changed since his time, but I had not heard of this change.

Answer 5

About 21.1 minutes. I did this calculation for someone else who asked the same question.

When you reached the center, you would be traveling at about 7800 meters per second.

Answer 6

LiD is correct.

Exact fomula in assumption of uniform spherical Earth is
T = πi/2 √(Re/g)

Answer 7

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