Why wouldn't hole through earth work?

[YT2095]

yup, magnetic flux is subject to same rule as for distance and strength, something like if a magnet has 9 tesla at 1 inch then 2 inches it`ll be like 3 and not 4.5 as you`de assume by double the distance halve the power.

it drops off exponentialy rather than linear .

 

something like that anyway )

[swansont]

i_a is right, dipoles drop off as r³

[YT2095]

so magnets will do it, but gravity won`t.

is that correct?

 

and yes I see the ^2 or ^3 sqr/cubed thingy now

[Sisyphus]

 

by this same ilustration' date=' 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).[/quote']

 

No! The size doesn't matter. The walls could be lightyear thick lead and the cavity ten lightyears across. There would still be no gravity anywhere within the cavity.

[YT2095]

so the ground Directly under your feet is worth the same as that above your head even though it`s much further away?

[Sisyphus]

Yes, because there's much more of it. In a sphere, in fact, the amount overhead increases with the square of your distance from it, just as the gravity decreases with the square. So the pull is the same from all directions all the time, and it all cancels out.

[insane_alien]

YT, yeah. you got to remember that there is a LOT more above you than below you.

enough so it will equal whats below you even though its further away.

[Klaynos]

so magnets will do it' date=' but gravity won`t.

is that correct?

 

and yes I see the ^2 or ^3 sqr/cubed thingy now [/quote']

 

 

There are no dipoles for gravity as tehre is no negative mass...

[woelen]

Janus' date=' 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.

 

<math snipped for brevity>

 

never quoted my self before.

 

what CPL came up with is orbital velocity at the surface of the earth.[/quote']

The potential energy formula you use is only useful for objects outside the planet. If all mass were concentrated in a point, then you could use that formula and at infinite distance from the point, where all mass is concentrated the potential energy equals 0.

 

Inside the earth (again, assuming uniform distribution of mass), only the part of the mass of the sphere with size r', with r' the radius of the remaining radius between you and the center of the earth counts for potential energy. The outer mantle cancels precisely. That already is stated before, if you are in a hollow globe, with mass uniformly distributed over its mantle, then inside you feel no gravity, regardless of where you are, not even near the mantle.

[woelen]

There are no dipoles for gravity as tehre is no negative mass...

The essential difference between magnetic fields on one side, and gravitational and electrical fields on the other side, is that in magnetism there is no analogue for the mass (or charge in the case of electrical fields).

 

For gravity and electricity, a closed surface (e.g. a sphere) can be chosen, such that the total incoming or outcoming flux of field lines can be non-zero. This is ONLY the case if there is a non-zero mass (for electricity: non-zero charge) inside the closed surface. For a magnetic field, the total incoming or outgoing flux of field lines for ANY closed surface equals 0. At a local point, there can be magnetic flux (e.g. going from outside to inside), but then there also is another place on the surface, where there is flux going in the opposite direction, from the point of view of the inside/outside relation.

 

For gravity, mass can be regarded as gravitational monopole, and for electricity, charge can be regarded as electrical monopole. In magnetism, there are no monopoles (as far as we now know). This has a large consequence. Field lines are always closed loops for magnetism. When there are monopoles, then field lines can start (or end) at the monopoles. The important difference between gravity and electricity, is that gravity only knows one type of monopoles (positive mass), while electricity knows two types of monopoles (positive and negative charge). The existence of monopoles also makes the concept of potential a sensible one. Both in gravity and in electricity we know the concept of potential (and it changes proportional to 1/r as function of distance r from the monopole). The E-field and G-field can be derived as gradients of potential.

[Mokele]

Proving that a hole through the earth would not produce perpetual motion, or much of anything of interest, is simple: the mantle (the layer between the crust and the core) is fluid, albeit a very viscous one. It'd be like trying to drill a hole through oatmeal.

 

Mokele

[DaveC426913]

By the way' date=' 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

 

 

I'm sorry, I don't believe you. If this were true:

1] an everyday object will have accelerated to 1/10th of the speed of light in less than 4,000 miles, merely due to the pull of gravity.

2] at that speed, (if it were a constant speed) it could traverse 8000 miles less than 1/2 second.

 

Step back and use your common sense. Is that really how fast you think it's going?

 

 

 

I'm going to bet dollars to doughnuts that this is where you made your mistake:

 

You assumed that it would continue to accelerate at 30ft/s^2 all the way along its path to the centre.

 

But that is not a simulation of freely falling in Earth's gravity, that is a simulation of freely falling toward a point-sized black hole that has 1 Earth mass.

 

Inside a real, non-zero-sized body, the gravitational attraction begins decreasing immediately, as soon as you are below the Earth's surface. eg. At a 1000 mile depth, you are only being pulled by the gravity of a sphere that is 4000-1000=3000 miles in radius (which is only about half the gravity at the surface). By the time you are halfway to the centre, gravity is only 1/4 as strong.

 

I think you'll find the passage through the core is a much more leisurely pace than you reported.

[timo]

Inside a real, non-zero-sized body, the gravitational attraction begins decreasing immediately, as soon as you are below the Earth's surface. eg. At a 1000 mile depth, you are only being pulled by the gravity of a sphere that is 4000-1000=3000 miles in radius (which is only about half [/i']the gravity at the surface). By the time you are halfway to the centre, gravity is only 1/4 as strong.

Gravitational force is proportional to the distance from the center, not ~r² as you seem to imply. Reason: Assuming a homogeneous mass-distribution for earth and m® being the mass that effectively pulls (the mass being in a sphere around the center of earth with a radius equal the object´s distance to the center), then F ~ m®/r², m® ~ r³ => F ~ r.

 

Ok, that already was my point, but since I´m at it I´ll take the fun provide just another guess of the velocity at the center:

Knowing the dependence of the force on the distance from the center and ignoring friction, you can easily compute the difference in potential energy between the center and the surface: [math]E= \int_0^R m r g / R \ dr = 0.5 mgR [/math]. Equating this with the kinetic energy (potential energy difference is converted to kinetic energy), you end up with [math] 0.5 mv^2 = 0.5 mgR \Rightarrow v = \sqrt{gR} \sim \sqrt{10 \cdot 6\cdot10^6 \frac{m^2}{s^2}} \sim 8000 m/s [/math]

 

One thing: Please use SI units. Not everyone wants to look up the definition of feet and miles and SI is common knowledge.

[DaveC426913]

Perhaps I'm not explaining myself adequately.

 

If you stand at the bottom of a tunnel 1000 miles deep, you will not experience a stronger force of gravity than at the surface, you will experience a weaker force - you will weigh significantly less.

 

 

Why? Becasue there is a large mass of Earth overhead pulling you up that cancels out much of the downward pull. In fact, it's not very difficult to calculate how much. For the purposes of calculating gravity, you can divide the Earth into two concentric spheres:

1] below your feet, a solid sphere, 3000 miles in radius

2] above your head, a hollow sphere, inner radius 3000 miles, outer radius 4000 miles, thickness 1000 miles

 

The gravity anywhere inside a hollow sphere is 0. The effect this has is that everything above your head (the entire 1000 mile thick hollow sphere) cancels out - you experience no net gravitational pull from it at all.

 

Thus, the only gravity that you experience is from the 3000 mile sphere under your feet. This sphere, having a mass that is much less than that of Earth, gives you a much smaller g, which means you don't accelerate as fast.

 

At 3,999 miles down, ALL but a 1 mile radius of Earth is cancelled out, and you only feel the pull of a sphere 1 mile in radius. So, ultimately as you fall toward the centre of the Earth, your acceleration drops steadily (thouigh not linearly) from 30ft/s^2 to 0.

 

 

 

 

Don't take my word for any of this, this is all perfectly well-known.

[swansont]

Atheist is quite right. If you assume constant density the force, and thus acceleration, is linear in r, and you will end up at around 8 km/s absent other forces.

[DaveC426913]

Elsewhere I got an answer the same as Atheist: 8km/s. That's a little more reasonable than 31,000km/s, yes?

[timo]

Elsewhere I got an answer the same as Atheist: 8km/s. That's a little more reasonable than 31,000km/s, yes?

Yes. And Woelen got that result before, also. I was not saying that you´re wrong by saying that gravitational force weakens when you come closer to the center. In fact, I explicitely said it did by stating F~r (r=0 should be the center of earth, r=6000 km the surface). However, what your statement I replied to implied (or at least how I read it) was that at half the radius you´d have a quarter the force as on the surface. It´s half the force, not a quarter.

Your argumentation about when you are at 1000 km depth you are only pulled by a sphere of a radius of 5000 km is absolutely correct. That is exactly what goes into my m®. What is incorrect is your claim about the force not dropping linearly. I have an idea of why you think so (namely because the amount of mass effectively pulling does not go linearly) but you seem to forget that gravitational acceleration is not only a function of mass but also of distance. The two effects combined give you a linear dependency. I´d normally say "do the math to see it yourself" but in this case I have already done the math, so perhaps take another look at my calculation sketch in my previous post to see why the force drops linearly.

[woelen]

Yes, it is nice to see that this kind of problems can be solved in so many different ways. Atheist uses the energy approach, I used the approach of deriving dynamic equations of motion. But of course, the results are the same and that nicely demonstrates the close relation between the two different concepts used. Frequently, there are multiple ways to solve mechanical problems, and sometimes method A is easier, and other times method B is easier. Best insight is obtained if one finds the theoretical underlying connection between energy and dynamics. That however, requires the use of Hamiltonian functions, describing total energy of the system in terms of generalized velocities, which is beyond the head of high school students and first year university students. That kind of math, however, allows one to solve similar questions for much more complicated systems than the hollow tube through a single ball of mass.

 

For the interested reader: http://www.answers.com/topic/hamiltonian-mechanics

[Janus]

Just one more point. It turns out that the speed at the center (for uniform density) equals

 

[math]\sqrt{\frac{GM}{R}}[/math]

 

Where M is the mass of the Earth and R its radius.

 

This is also equal to the orbital velocity for an object at the surface of the Earth (ignoring air resistance)

[J.C.MacSwell]

Just one more point. It turns out that the speed at the center (for uniform density) equals

 

[math]\sqrt{\frac{GM}{R}}[/math]

 

Where M is the mass of the Earth and R its radius.

 

This is also equal to the orbital velocity for an object at the surface of the Earth (ignoring air resistance)

 

Interesting how that works out.

[Janus]

Interesting how that works out.

 

Actually, you would expect it to work out that way.

 

As has been pointed out, the force acting on the falling body decreases porportionally with the decrease in distance from the center. This implies harmonic motion on the part of the object.

 

Harmonic motion can also be described as a projection of circular motion on the diameter of the circle. If you start with an object moving in a circle with a centripetal force of

[math]f= \frac{mv^2}{R}[/math]

 

You will find that the component of that force acting along the line of the diameter varies linearly. (as per harmonic motion).

 

Since the centripetal force acting on a surface orbiting object and the intial restoring force of our falling object are the same, if you start the two objects at the same point at the same time, you will find that a line drawn through either object, and perpendicular to the path of the our falling object will always pass through both objects.

 

Thus when the orbiting object reaches a point 90° from its starting point, the falling object has reached the center and both objects at that moment are heading in the same direction at the same speed.