Why wouldn't hole through earth work?

[YT2095]

ok lets simply further, there are 2 equal mass planets side by side, in the dead center between the cog of each planet will be an equal pull either side between A and B.

you stray a little towards A (further from B) and A will have the greater effect on you and slowly you`ll start to move towards A.

until you hit the surface of it.

(you can`t discount B, same as you can`t discount the Moon or the Sun on the Tides).

 

so if you have 2000Kms of land under your feet, and have a 4000Km gap of air above with a roof 2000kms thick, you`ll be able to walk on that ground!

just the same as we can walk on the moon with the Earth directly above us, we Don`t fly off into space.

[CanadaAotS]

I'm pretty sure you wouldn't be able to walk on the cylinder walls for the same reason that gravity wouldn't let you walk on the walls of a sphere from the inside.

 

Try looking at how the forces work from birds eye view (only looking laterally).

 

What you get is the object inside a ring. This is the 2D analogy of charges canceling out in a sphere. This means that laterally, no forces are effecting the object. When you look at it from a front view, the earth is still pulling down to the center so all you have is a vertical force.

[YT2095]

ok, I`ll simplify my thinking further.

 

imagine 2 bar magnets, and to keep things even we`ll only use the north poles of each.

both mags have the same gauss and equal in all aspects.

One magnet has 5mm of cardboard on the end though.

the other has none and left as plain.

in the middle is a steel ball bearing, bring the 2 magnets together so they touch the ball, one each side.

which magnet will the ball stick to?

 

the one With the cardboard spacer or the one without?

 

gravitation has a function of distance, as do magnets.

[Severian]

ok lets simply further' date=' there are 2 equal mass planets side by side, in the dead center between the cog of each planet will be an equal pull either side between A and B.

you stray a little towards A (further from B) and A will have the greater effect on you and slowly you`ll start to move towards A.

until you hit the surface of it.

(you can`t discount B, same as you can`t discount the Moon or the Sun on the Tides).

[/quote']

Yes. In fact, there is a technical name for this point: the Lagrange point.

[YT2095]

Thank you )

[swansont]

One disctinction that needs to be made, I think, is that Lagrange points (be they stable or unstable equilibrium) are still points. If you move away, you feel a force. The situation inside of a gaussian surface is that the force is zero at all of the points; no restoring nor disrupting force is present.

[CanadaAotS]

exactly... thats it, I'm drawing a diagram! lol

 

 

EDIT: damn photobucket... it made my diagram alot smaller I dont know if you are able to read it.

Anyways the small circle with the black dot is the birds-eye view of the tunnel. It forms a 'ring' and inside the ring charges cancel each other out (2 dimensional version of a sphere). This means that the object falling down the tunnel isnt effected by the sides at all. The bigger circle shows the front view where the only force acting on the object is the vertical pull of gravity.

 

By the way, the object would (apparently) be going at 3 m/s when it reached the center. I think this is wrong though... check out my thread where I try to figure it out here

[Sisyphus]

Now' date=' you can add all the assumptions back if you like. If you add air resistance, the air resistance will become greater and greater the closer to the center you get, hence your terminal velocity will increase. Your terminal velocity will NOT hit 0 however. The only time terminal velocity goes to 0 is when you hit a brick wall haha

 

[/quote']

 

This is incorrect. First of all the denser the air, the more air resistance there is, and therefore the lower the terminal velocity. Secondly, irregardless of what it does before that, it will hit zero at the exact center. There is no force of gravity at the center, therefore any air resistance at all will be greater than it (greater than zero, that is), and will cause a negative acceleration. The only velocity at which there is equilibrium is zero.

 

Now, this doesn't mean you will actually stop. It just means the terminal velocity is zero, which is different. Terminal velocity is just the velocity at which there is equilibrium between forces and therefore towards which you tend.

[CanadaAotS]

Ohhhhh

 

Yes that makes sense then.

[CanadaAotS]

By the way, thanks to Sisyphus, if you dropped some object down the tunnel (with all the previous assumptions like no air friction etc.) the object will be going at a wopping

 

31,375,360 meters per second

 

which is more then 10% the speed of light!

 

EDIT: 400th Post w00t to moleculism

[Janus]

ok lets simply further' date=' there are 2 equal mass planets side by side, in the dead center between the cog of each planet will be an equal pull either side between A and B.

you stray a little towards A (further from B) and A will have the greater effect on you and slowly you`ll start to move towards A.

until you hit the surface of it.

(you can`t discount B, same as you can`t discount the Moon or the Sun on the Tides).

 

so if you have 2000Kms of land under your feet, and have a 4000Km gap of air above with a roof 2000kms thick, you`ll be able to walk on that ground!

just the same as we can walk on the moon with the Earth directly above us, we Don`t fly off into space.[/quote']

 

Not the same thing. In your two planet scenerio the mass to either side of you does not change as you move toward one planet or the other.

 

With the hollow Earth, it does. Check out the following diagram. The line represents your position in the sphere.

 

 

Notice, that as you move "down" towards the bottom of the sphere, less and less of the Earth's mass is "below" you pulling you "down" and more and more is "above" you pulling you up. The two pulls will exactly cancel each other out.

 

Newton proved this mathematically long ago.

[CanadaAotS]

I thought that was Guass' law...

[CPL.Luke]

gauss's law also applies to gravity in the newtonian sense. Gauss's law affects all fields that radiate out radially from a point.

[5614]

Gauss's law affects all fields that radiate out radially from a point.

Gauss's law can be used wherever the inverse-square law is true.

[Klaynos]

Gauss's law can be used wherever the inverse-square law is true.

 

Guasses law an be used (classically) for any system inwhich a guassian surface can be created, encloseing some charge (taken to mean both electrical and mass). the inverse-square law is the result of solving guases law for a spherically symetric charge structure, which is easiestly done using a spherical guassian surface...

 

Note when reading this, I'm not really keeping up with reading this thread...

 

The "no force inside a hollow sphere due to the sphere" is trivial to show with Guasses law...

[CPL.Luke]

also the velocity at the center is wrong, I posted acomment in the other thread that the math behind that is wrong.

 

you'd be going 7905.96 m/s

 

which makes sense as the escape velocity for the earth is 11 km/s so at 8km/s you would have 0 chance of getting out of the earth's gravitational field, wheras at .1C you would rocket away from the earth.

[J.C.MacSwell]

Ok...

 

 

A few people dont seem to understand what terminal velocity is... Its when Force of friction and the Force of Gravity equalize and cancel each other out (exactly why you stay at the same velocity' date=' no acceleration is happening).

 

[/quote']

 

Drag or air resistance (as you said elsewhere) is much more accurate.

[insane_alien]

CPL, you would be going at around 11 km/s at the centre but you would slow don to 0 by he surface. the escape velocity of 11 km/s is for going 11km/s at the surface. not the centre

[Janus]

CPL, you would be going at around 11 km/s at the centre but you would slow don to 0 by he surface. the escape velocity of 11 km/s is for going 11km/s at the surface. not the centre

 

No, CPL's figure for the speed is correct. It's what you get when you take the potential energy difference between the center of the Earth and the surface, and then figure out the speed needed to equal this in kinetic energy for a given mass.

[woelen]

The speed at the center of the earth, when you start off with zero velocity cannot be computed easily, because we do not know the mass distribution of earth precisely.

 

But for simplicity, assume that the mass is evenly distributed and that density is all over the same inside the earth.

 

Then the depth x as function of time t can be written in differential form as

 

d²x/dt² = g * (1 - x/r), here r is the earth radius, and g is the acceleration at the surface.

 

Solving this differential equation for x(0) = 0 and xdot(0) = 0 yields:

 

x(t) = r - r * cos(wt), with w = √(g/r).

 

The velocity, xdot(t) can be written as

 

xdot(t) = w*r*sin(wt).

 

At the center, the velocity is maximum, so sin(wt) equals 1 over there. The velocity in the center of the earth will be

 

w*r = √(g*r).

 

With g equal to 10 m/s², and r approximately 6,300,000 meters, this is approximately 8000 m/s.

 

All calculations are very rough, so this number only is an indication of the velocity at the center. In all these computations, the air friction is not taken into account.

[insane_alien]

Janus, thats what i did. i did the math in the other thread. i'll post it here too and you can tell me where its wrong.

 

lets look at this from the energy perspective.

 

at the centre you will have 0J of gravitational potential energy

at the surface you will have

(-6.6742*10-11*5.9742*10^24*70)/6278100

=-4376084406 J of potential energy

KE = 0.5*m*v^2

v^2=2*KE/m

v=sqrt(2*KE/m)

v=11181.7 m s^-1

 

which just so happens to be escape velocity as well.

 

never quoted my self before.

 

what CPL came up with is orbital velocity at the surface of the earth.

[CPL.Luke]

insane alien your misusing newtons law of gravitation, the formula changes once you go inside the radius of the earth

[swansont]

at the centre you will have 0J of gravitational potential energy

 

No, using the formula PE=-GMm/r the potential energy is zero at infinite separation, which is why that calculation yields the escape velocity from the surface.

[YT2095]

Inverse sqr law, THAT is the wording I was looking for rather than "A Function of distance"

 

so if you have a ring of bar magnets arranged in a circle, and a ball bearing (steel) inside this circle, the ball will roll to the magnet Closest to it, this Will howver be a point DEAD CENTER where the ball is equidistant from each and so it will remain.

 

by this same ilustration, if the cavity inside the planet was large enough, you would be able to walk on its inside walls (providing you made no sudden movements as the grav would be quite low).

[insane_alien]

don't magnets work on a inverse cube law because they have poles?

 

ok, i was wrong about the speeds and gravity and all that.