thought experiment

[oe39]

what is this thought experiment supposed to illustrate?

[Ohgod]

Alternatively - lined with what metal? One with a melting point > 8000K?

[PairTheBoard]

I don't see the point to the complication of dropping it from one meter, unless you're looking for the answer that it would pop out the other side, before falling back.

Without that complication I think it's easy to see that it ocillates from one side of the earth to the other. The Net Force of gravity acting on it from all sides on its trip will always be in the direction of the Earth's center, with equal and opposite Net Force at respective points on both sides. So it will accelerate all the way toward the center and decelerate in equal fashion on the other side.

The only wrinkle is in calculating the magnitude of the Net Force Vector at points along it's trip. Even after making simplifying assumptions about uniform density for the Earth, you have the problem of computing how much of the Earth is pulling upwards and how much is pulling downwards at any point along the trip. But you don't need to exactly compute that to know that there will be more Earth Left pulling down toward the Center than up toward the Surface. Of course at the Center the Net Force will be zero. But that doesn't matter because the ball will quickly pass that point on its way to the other side.

PairTheBoard

[Ohgod]

If we assume uniform density, the hole remains intact throughout etc., we don't need to worry about the Earth 'pulling upwards' - the net gravitational field inside a uniform spherical shell due to that shell is always zero - Gauss FTW.

The gravitational field declines linearly from its value at the surface to zero at the centre as you move down through a uniform sphere:

Mass inside radius r (where r < Re the radius of the Earth)

= (r^3 / Re^3) * M

g at radius r / g at surface

= [GM(r^3/Re^3)/r^2] / [GM/Re^2] = r / Re

[PairTheBoard]

[ QUOTE ]
If we assume uniform density, the hole remains intact throughout etc., we don't need to worry about the Earth 'pulling upwards' - the net gravitational field inside a uniform spherical shell due to that shell is always zero - Gauss FTW.

[/ QUOTE ]

Right. Although you should give a clear definition of "Net Gravitational Field" as it applies here if you want people to know what you're talking about. Also, if you wanted to calculate how long it would take for the ball to reach the other side, the complications apply.

PairTheBoard

[Ohgod]

You're right of course - the gravitational field at any point inside the sphere is determined only by the mass 'inside' any given point. What I was really objecting to was describing bits of the Earth as 'pulling up'. Although it's true, of course, I've found in the past that talking about the shell as a single entity which produces no field inside itself helps people understand it most easily.

And from that, since the restoring force is linear with distance from the centre, won't we get simple harmonic motion in the idealised case?

[PairTheBoard]

[ QUOTE ]
You're right of course - the gravitational field at any point inside the sphere is determined only by the mass 'inside' any given point. What I was really objecting to was describing bits of the Earth as 'pulling up'. Although it's true, of course, I've found in the past that talking about the shell as a single entity which produces no field inside itself helps people understand it most easily.

And from that, since the restoring force is linear with distance from the centre, won't we get simple harmonic motion in the idealised case?

[/ QUOTE ]

Oh! I was thinking of what you said about the Net Gravitational Field the wrong way. I thought it was some complicated concept involving energy required to move from one point to another - or something like that. You're talking about a "Shell". A picture would have helped me.

So if the radius of the Earth was say 4000 miles, and the ball was 3500 miles from the center, then the "Shell" you are talking about is the 500 mile thick shell consisting of that part of the Earth between 3500 and 4000 miles from the center. And the Theorem you're referring to says that for the ball located on the inside surface of that shell, 3500 miles from the Earths center, the Net Gravitational force on the ball from that Shell is zero. That's Gauss for you. I don't think that's at all obvious. But Gauss proved it. And it solves the complications I was worried about.

You can therefore calculate the Net Force of Gravity acting on the ball at that location as if the only thing there was a smaller Earth with radius 3500 miles and with the ball on its surface. As the ball falls the Force of Gravity increases in proportion to the square of the distance to the center of gravity. But the Mass of Earth decreases in proportion to the cube of its remaining radius. So I can see how it could work out that the Force on the ball varies Linearly as it travels. I'll take your word on it that it does anyway.

Yea. You're right. The Gauss Theorem really simplifies things.

Suppose the vacuum tube extended 1 meter above the earth to accomodate the OP. Would that affect the "harmonic" part of the oscillation? Would the Force still vary linearly over that one meter?

PairTheBoard

[Ohgod]

True - I wasn't as clear as I could have been, and a picture would definitely have helped - but I'm lazy [img]/images/graemlins/smile.gif[/img]

The Gauss result is definitely not obvious, at least at first. I was hoping to just quote it and that would do...

Regarding the section of tube above the surface of the earth - the force would start to decrease again in an inverse square fashion, so the motion would now no longer be harmonic. However, since 1m <<<< 6000km it would only produce a tiny perturbation from harmonic behaviour, I think.

[BigSoonerFan]

[ QUOTE ]
what is this thought experiment supposed to illustrate?

[/ QUOTE ]

That someone needs help with their homework?

[HP]

To actually add something cool to this thread, imagine you dropped a second steel ball 1 metre above the first. You release them at the same time

while the balls are hurtling towards the centre of the Earth, the gravitational force will be slightly greater on the second ball, with is further from the centre of the Earth. This means the distance between the two balls is decreasing, and not at a constant rate. They will actually appear to be accelerating towards each other

Einstein says the balls actually have no forces acting upon them. It's more correct to say the surrounding earth is accelerating upwards, instead of saying the balls are accelerating downwards

So if no forces are acting on the balls, how are they getting closer? The curvature of space-time!

And as the balls pass the through the centre of the Earth and go through to the other side, on the way up, the opposite effect will occour. The distance between them will increase.

btw some one needs to confirm what I said makes any sense, it's been a while since I took gen rel