Physics question

[StarStreak2109]

Hi there,

I have a physics question, which has bugged me for a long time.

Now imagine the following:

1. You have a planet. For the sake of simplicity, let's assume it is a perfect sphere of uniform density and composition.
2. You drill a hole exactly through the middle so that it comes out of the other end. Let's also disregard, that there is a molten core in the middle of the planet.
3. Now you throw a ball into that hole, so that it falls into it exactly through the middle of the hole.

What happens? My thoughts:

• The ball would fall into the hole. Once it passed the centre of the planet it would slow down and eventually reverse due to air friction.
• Eventually, it would stop... where? in the centre of the planet?

What do you think?

[canisin]

Hmm, I think without air friction it would go back and forth along the tunnel forever, like a pendulum or an orbit. With air friction, it would eventually stop at the center of the planet.

[Green Baron]

That's trivial.

Without friction: It falls, initially accelerating with surface g. While falling, the acceleration due to g decreases as it approaches the center, it falls through the center at max velocity, then deceleration due to g increases again and thus velocity decreases until all the gained velocity has been canceled out. Ignoring friction that would mean at the surface on the other side, exactly at an altitude where the other guy dropped it.

If you didn't catch it, it'll just keep on playing yoyo for(;;).

(If you drop it, run around quick as hell, you could catch it at exactly the same point where you dropped it. And as the sphere is featureless perfect you wouldn't even notice that you're on the other side except for the fact that you're out of breath :-))

 

Comes friction and density into play that would just limit the fall speed to an equilibrium between friction forces and weight of the body. As density and friction increases with depth and g decreases, it would arrive at the center with 0 or very little speed and just stay there. It would reach maximum velocity quickly and then slowly become slower due to lowering acceleration from g and rising friction from density, until it arrives at the center with 0 g.

 

:-)

Edit: valid for a static non-rotating massive body without a hollow inside only. If it was a a shell then > shell-theroem. If it was rotating > coriolis force, which would make your object scratch along the wall in rotation direction because of all the excess surface rotation speed it has while falling down !

Edited January 11, 2018 by Green Baron

[sevenperforce]

If the planet is rotating, you have to factor in the Coriolis effect. The tangential velocity of the surface of the spherical planet is larger than the velocity halfway down, which is larger than the velocity 2/3 of the way down, and so forth. If dropped, the ball will maintain its tangential velocity, and it will strike the side of the tunnel at a point which can be trivially calculated by the diameter of the tunnel, the diameter of the planet, and the rate of rotation.

Side note: if the Earth were suddenly to be collapsed into a black hole (which, FYI, would have an event horizon slightly smaller than a dime), all the objects on the surface would find themselves at the apoapses of highly eccentric orbits around that black hole. Don't recall exactly what the periapse would be, but I'm guessing about 50-100 miles from the center, for observers at the equator. Observers at higher and lower latitudes than the equator would be in more eccentric orbits.

But @Green Baron is right; IF the planet is not rotating, then the moment the ball drops into the shaft, it begins to accelerate at a decreasing acceleration. The further it falls, the less planet there is below it and the more there is above, so the magnitude of the acceleration vector decreases. Once it passes the center, where the forces on it are momentarily zero but its speed is at a maximum, it begins to slow down because the acceleration vector has reversed and started to go up again.

Its oscillations will continue indefinitely unless an additional force, like friction, comes into play.

[Nikolai]

39 minutes ago, Green Baron said:

Without friction: It falls, initially accelerating with surface g. While falling, the acceleration due to g decreases as it approaches the center, it falls through the center at max velocity, then deceleration due to g increases again and thus velocity decreases until all the gained velocity has been canceled out. Ignoring friction that would mean at the surface on the other side, exactly at an altitude where the other guy dropped it.

Not only that, but the time it takes to get to the other side is exactly the same as it would take for the ball to orbit to that point at that altitude.  Either way, the time to get to the other side is pi*sqrt(r/g), where r is the radius you drop it from and g is the acceleration due to gravity at that radius.  I can do the math if you like.

[Green Baron]

Interesting :-)

Well, i personally suck at math. I can use a given formula, which is enough for every day life (like throwing balls through planets :-)). But if OP likes ...

[Delay]

25 minutes ago, Nikolai said:

Not only that, but the time it takes to get to the other side is exactly the same as it would take for the ball to orbit to that point at that altitude.  Either way, the time to get to the other side is pi*sqrt(r/g), where r is the radius you drop it from and g is the acceleration due to gravity at that radius.  I can do the math if you like.

Not only that, if you imagine the sphere in two dimensions with the hole running vertically from top to bottom and you running around the circumference of the circle, your y-coordinate will always be equal to the y-coordinate of the ball. You both basically travel a r*cos(x) plot, where r is the radius of the sphere.

[wumpus]

Note that thanks to the Coriolis effect, the ball will be likely rolling down one side.  This means that in addition to the kinetic energy of the ball falling through its yo-yo sequence it will also have the kinetic energy thanks to the angular momentum.  So when it first pops out the far hole, it won't be re-entering the hole as fast as it did on the first time down.  Assuming zero-friction everywhere (and no momentum lost banging against the walls), the height it pops out on subsequent trips will all be the same.

[StarStreak2109]

Wow, thank you, more input than expected... Didn't think about the coriolis force...

[Green Baron]

Just for completeness, everyone is probably aware:

There is no coriolis when falling along the rotational axis, from pole to pole. There is max. coriolis force when falling from equator to equator, since the full speed is canceled out down and then added again on the way up. In latitudes in between an angular moment is added, depending on the radius.

Btw. that angular moment is, together with pressure gradients, one of the main driving forces for supra-regional wind in the atmosphere.

Edited January 11, 2018 by Green Baron

[PB666]

2 hours ago, sevenperforce said:

Its oscillations will continue indefinitely unless an additional force, like friction, comes into play.

It makes a number of assumptions that can be falsified. First that, the ball and the planet are inelastic, which is almost never true. I would not say indefinitely but indefinably, simply because the elasticity of neither is specified. Also this is a two-body problem, which does not really exist. If we can imagine the earth and the moon as a two body problem, both have some elasticity, and thus the Earth's rotation slows, the moons increases, and some of that energy becomes a magnetic field. In the above situation if the tube in which the ball fell was elastic, the ball would impart some of its energy on the tube (causing heat) and it would loose some of its energy, the problem is that there is no form of matter that is completely inelastic. Everything compresses if sufficient force is applied.

If a third body is placed in the system in which an object is not rotating but orbits about a star or a moon orbits about a planet, the ball would not continue to fall strait, it would eventually conduct heat to the ball in the form of friction, over time the amount of collisions would increase. A third instance is that the planet is orbiting is parent galaxy, although this is more of a hairy problem since in a galaxy the forces are spread out in many directions and one of the tensors is dark matter (an is, except for the vector product, intangible).  As a consequence a non-rotating objects orientation relative to galactic basis is changing over geological time scales.

The forth problem is gravitational waves, since two objects oscillating with respect to each other are expected to create gravitational waves at very small levels, since wave production is energy, this is loss of energy to the system. The falling object is transferring through more u/r than a satellite leaving earth orbit as it falls, this is very tiny amount of energy loss since most i trapped within the system but if the system is observed from the outside then this is not true.

So the problems apriori's are insufficient. For example, if this is the only two objects in the visible universe (which is counter to the definition of the universe because we must see it to be visible), and its light is what we observe by (including falling objects on kinetically inactive spheres), and those two objects are at 0'K (non-elastic), which by definition we cannot observe them except through gravitational observations and Heisenberg uncertainty applies. You cannot observe a system even with gravitational waves, without affecting the system. So if your system is perfectly conservative you cannot observe it, and if you observe it, then it cannot be indefinite.

 

[Nikolai]

2 hours ago, Delay said:

Not only that, if you imagine the sphere in two dimensions with the hole running vertically from top to bottom and you running around the circumference of the circle, your y-coordinate will always be equal to the y-coordinate of the ball. You both basically travel a r*cos(x) plot, where r is the radius of the sphere.

Well, right, and you can see that if you take the time to simplify the expression of force on the ball into a constant times the distance of the ball from the center -- which is a spring equation.  Springs follow sinusoidal motion.  (The "spring constant" is equal to m*g/r, where m is the mass of the ball, g is the acceleration felt by the ball at the height from which you drop it, and r is the radius at which you drop the ball.)

Edited January 11, 2018 by Nikolai
Took out an unnecessary "the"

[Steel]

1 hour ago, PB666 said:

It makes a number of assumptions that can be falsified. First that, the ball and the planet are inelastic, which is almost never true. I would not say indefinitely but indefinably, simply because the elasticity of neither is specified. Also this is a two-body problem, which does not really exist. If we can imagine the earth and the moon as a two body problem, both have some elasticity, and thus the Earth's rotation slows, the moons increases, and some of that energy becomes a magnetic field. In the above situation if the tube in which the ball fell was elastic, the ball would impart some of its energy on the tube (causing heat) and it would loose some of its energy, the problem is that there is no form of matter that is completely inelastic. Everything compresses if sufficient force is applied.

If a third body is placed in the system in which an object is not rotating but orbits about a star or a moon orbits about a planet, the ball would not continue to fall strait, it would eventually conduct heat to the ball in the form of friction, over time the amount of collisions would increase. A third instance is that the planet is orbiting is parent galaxy, although this is more of a hairy problem since in a galaxy the forces are spread out in many directions and one of the tensors is dark matter (an is, except for the vector product, intangible).  As a consequence a non-rotating objects orientation relative to galactic basis is changing over geological time scales.

The forth problem is gravitational waves, since two objects oscillating with respect to each other are expected to create gravitational waves at very small levels, since wave production is energy, this is loss of energy to the system. The falling object is transferring through more u/r than a satellite leaving earth orbit as it falls, this is very tiny amount of energy loss since most i trapped within the system but if the system is observed from the outside then this is not true.

So the problems apriori's are insufficient. For example, if this is the only two objects in the visible universe (which is counter to the definition of the universe because we must see it to be visible), and its light is what we observe by (including falling objects on kinetically inactive spheres), and those two objects are at 0'K (non-elastic), which by definition we cannot observe them except through gravitational observations and Heisenberg uncertainty applies. You cannot observe a system even with gravitational waves, without affecting the system. So if your system is perfectly conservative you cannot observe it, and if you observe it, then it cannot be indefinite.

 

Are we really going to nitpick about n-body interactions and gravitational waves in a question that involves a completely implausible shaft drilled exactly straight through the centre of the Earth?  

Has the old spherical cow in a vacuum assumption gone out of fashion these days?

[YNM]

 

[PB666]

On 1/11/2018 at 12:09 PM, Steel said:

Are we really going to nitpick about n-body interactions and gravitational waves in a question that involves a completely implausible shaft drilled exactly straight through the centre of the Earth?  

Has the old spherical cow in a vacuum assumption gone out of fashion these days?

YES! why wouldn't we.

The reason I do that is the same reason a forensic pathologist claims there are no traceless crimes. The point from a physical point of view is that every action has a reaction. So whereby we apply this to our logic. If we are detecting the motion of two neutron stars 2 million light years away detecting the gravitational effects of a local object would be all the more powerful.

Step 1 a celestial . . .Celestials are found where, in space, surrounded by other matter stars planets, other, parts, of the galaxy . . . . so are we to remove these. Most solid celestials either orbit stars or other planets.
Step 2 a dig a hole . . Most celestials of any given size do not have rigid interiors, if you dug far enough down you would start seeing the effects of pressure on its structure . . .so the celestial have to be absolutely rigid and cold at the center.
Step 3 drop and object into the hole . . . wait a second most bodies have some atmosphere, and even if a little by the time you finished digging the hole would have filled with something. Even the moon would if you dug a hole through the middle accumulate significant hydrogen over time because a hole is a lower energy state for the hydrogen. . . .so you would have drag.
Step 4 you also have friction. . .all celestials rotate even tidally locked celestials are rotating once per orbit . . . to prevent frictional encounters. . . . .there would have to be no motion.

So now we need something that does not exist. A celestial with no companions . . .anywhere.... not rotating . . . .solid as a diamond. (although how did we dig a hole through solid diamond). we drop an object into a space that has nothing but space . . .and we observe . . .but without a meter (light source and a mass that measures) how do we know.

IOW if you  remove the universe from a problem, you cannot use universe's information to deduce a solution. 

[kerbiloid]

I strongly believe that in the name of forensic we must run a practical verification test.
Any suggestions about drilling through the Earth are greatly appreciated.

[PB666]

Why would you start with earth, start with something smaller . . . . . . . .say a black hole.