question revisited

[DrPizza]

I thought I'd revisit this topic...

If you were able to drill a hole through the center of the earth (from N pole to S pole to avoid coreolis effect) and jumped in, *NOT* ignoring air resistance (and wearing a magic pressurized suit to avoid being crushed), would you make it to the center of the earth?
Let's even assume at first that the matter on the earth is a uniform sphere and get rid of the moon and any effects it would have.
Also, I suppose, make the assumption that friction (air resistance) is proportional to the square of the velocity.

It would, of course, be a damped harmonic oscillation... but how damped - that's the question not answered last time... would you never quite make it? Make it and never quite make it back to the center? Or oscillate indefinitely?

Also, the air pressure would be tremendous at the center of the earth (I think) but gravity would = 0.. does that make sense? Somehow I can't get those two concepts to come into agreement in my mind.

 

[S0Y73NTGR33N]

Gravity would not be 0... gravity is never 0... it would just be pulling in all directions equally at once.

-green

 

[blahblah99]

It would be a damped oscillation, with wind resistance acting as your source of friction. Eventually you'll just be floating in the center of the Earth. Not sure about the air pressure part though.

 

[CycloWizard]

Haven't thought about it much (about ten seconds really ), but the damping would likely be proportional to the aspect ratio of the earth in the direction of the hole that you drilled. Since the earth isn't a sphere, the location of the hole would affect the body forces at a given location. I would imagine, therefore, that the damping would be greater if you drilled from the north pole to the south than if you drilled from one spot on the equator to the other, since the earth has a larger diameter at the equator than around the prime meridian. I know you said assume it's a sphere, but I'm guessing that this effect would be somewhat important.

 

[Armitage]

The air pressure would certainly be high ... the weight of a column of air about 6,600 Km high! But the weight of a given volume of air would change through the distance ... increase due to pressure as you go down, but decrease due to the gravity gradient inside the earth. Sounds like a fun problem in integral calculus! Matbe I'll fire up Mathematica over lunch and have a wack at it.

As to whether you get there or not ... unless the pressure gets so high that the density of the air becomes greater then the density of whatever is falling such that boyancy becomes an issue, I don't see how you wouldn't. Drag alone wouldn't prevent it because drag is proportional to velocity, and we're talking about aerodynamic drag, not static friction ... as long as there is a force on the body, there will be motion, and as long as you aren't exactly at the center of the earth there will be a force due to gravity.

 

[DrPizza]

Originally posted by: Armitage
The air pressure would certainly be high ... the weight of a column of air about 6,600 Km high! But the weight of a given volume of air would change through the distance ... increase due to pressure as you go down, but decrease due to the gravity gradient inside the earth. Sounds like a fun problem in integral calculus! Matbe I'll fire up Mathematica over lunch and have a wack at it.

As to whether you get there or not ... unless the pressure gets so high that the density of the air becomes greater then the density of whatever is falling such that boyancy becomes an issue, I don't see how you wouldn't. Drag alone wouldn't prevent it because drag is proportional to velocity, and we're talking about aerodynamic drag, not static friction ... as long as there is a force on the body, there will be motion, and as long as you aren't exactly at the center of the earth there will be a force due to gravity.

It's not just a foce, but a net force on a body which will produce acceleration and motion. At super slow speeds, is the force of friction of the air equal to the gravitational force of attraction? If so, it may stop before it gets to the center.

Here's a program to describe this motion better... is the air resistance enough to cause critical damping?
here
(think of the 0 on the y-axis as being the center of the earth)

 

[CycloWizard]

Originally posted by: Armitage
The air pressure would certainly be high ... the weight of a column of air about 6,600 Km high! But the weight of a given volume of air would change through the distance ... increase due to pressure as you go down, but decrease due to the gravity gradient inside the earth. Sounds like a fun problem in integral calculus! Matbe I'll fire up Mathematica over lunch and have a wack at it.

As to whether you get there or not ... unless the pressure gets so high that the density of the air becomes greater then the density of whatever is falling such that boyancy becomes an issue, I don't see how you wouldn't. Drag alone wouldn't prevent it because drag is proportional to velocity, and we're talking about aerodynamic drag, not static friction ... as long as there is a force on the body, there will be motion, and as long as you aren't exactly at the center of the earth there will be a force due to gravity.

The 'weight' is a function of pressure, which is a linear function of normal distance from the source of gravity (pressure = density*gravity*height). In this case, gravity at the very center of the earth should be zero, since you would have equal and opposite gravitational forces acting in all directions (except along the tunnel, obviously, since there is much less mass density there than in the earth). I'll have to think about what the pressure distribution actually would be, but I don't think it would be terribly high, certainly much less than 5 atmospheres. How high does our atmosphere extend?

 

[Armitage]

Originally posted by: DrPizza

Originally posted by: Armitage
The air pressure would certainly be high ... the weight of a column of air about 6,600 Km high! But the weight of a given volume of air would change through the distance ... increase due to pressure as you go down, but decrease due to the gravity gradient inside the earth. Sounds like a fun problem in integral calculus! Matbe I'll fire up Mathematica over lunch and have a wack at it.

As to whether you get there or not ... unless the pressure gets so high that the density of the air becomes greater then the density of whatever is falling such that boyancy becomes an issue, I don't see how you wouldn't. Drag alone wouldn't prevent it because drag is proportional to velocity, and we're talking about aerodynamic drag, not static friction ... as long as there is a force on the body, there will be motion, and as long as you aren't exactly at the center of the earth there will be a force due to gravity.

It's not just a foce, but a net force on a body which will produce acceleration and motion. At super slow speeds, is the force of friction of the air equal to the gravitational force of attraction? If so, it may stop before it gets to the center.

Fluid drag is proportional to velocity, so as velocity approaches 0, so will the force due to the drag. So, if there is a gravitational force, no matter how small, then the velocity has to be non-zero, right? It will be at critical velocity ... Fdrag = Fgravity, and for there to be a drag force, there must be a velocity. No velocity = no drag = unbalanced force

Here's a program to describe this motion better... is the air resistance enough to cause critical damping?
here
(think of the 0 on the y-axis as being the center of the earth)

Meh ... need some plugin that I don't have installed. Here's a static graph of what I think it likely shows: http://hyperphysics.phy-astr.gsu.edu/hbase/oscda2.html

Maybe I see what you're getting at ... if the system is overdamped, it will approach, but never quite reach 0. In most systems where there is a component of mechanical friction involved as well, the velocity will go to 0 and you won't get any closer, but in this case, since we only have fluid friction, I think that your velocity will never go precisely to 0 ... it will continue to approach the 0 position indefinitely.

Note that an overdamped oscilator with a high enough initial velocity can overshoot 0 once, and then will approach from the other side.

 

[Armitage]

Originally posted by: CycloWizard

Originally posted by: Armitage
The air pressure would certainly be high ... the weight of a column of air about 6,600 Km high! But the weight of a given volume of air would change through the distance ... increase due to pressure as you go down, but decrease due to the gravity gradient inside the earth. Sounds like a fun problem in integral calculus! Matbe I'll fire up Mathematica over lunch and have a wack at it.

As to whether you get there or not ... unless the pressure gets so high that the density of the air becomes greater then the density of whatever is falling such that boyancy becomes an issue, I don't see how you wouldn't. Drag alone wouldn't prevent it because drag is proportional to velocity, and we're talking about aerodynamic drag, not static friction ... as long as there is a force on the body, there will be motion, and as long as you aren't exactly at the center of the earth there will be a force due to gravity.

The 'weight' is a function of pressure, which is a linear function of normal distance from the source of gravity (pressure = density*gravity*height). In this case, gravity at the very center of the earth should be zero, since you would have equal and opposite gravitational forces acting in all directions (except along the tunnel, obviously, since there is much less mass density there than in the earth). I'll have to think about what the pressure distribution actually would be, but I don't think it would be terribly high, certainly much less than 5 atmospheres. How high does our atmosphere extend?

For most purposes, the atmosphere extends to about 100 Km.
Even though the force due to gravity is 0 at the center, the volume of air there is still supporting the entire column of air above it. So take your equation ... pressure = density*gravity*height, plug in the equation for gravity as a function of distance from the center of the earth, and the relation between pressure and density and integrate.

 

[CycloWizard]

Originally posted by: Armitage
For most purposes, the atmosphere extends to about 100 Km.
Even though the force due to gravity is 0 at the center, the volume of air there is still supporting the entire column of air above it. So take your equation ... pressure = density*gravity*height, plug in the equation for gravity as a function of distance from the center of the earth, and the relation between pressure and density and integrate.

It's only 'supporting' it relative to the source of gravity. The 'bottom' of the air column will be the location with the highest gravity field.

 

[Armitage]

Originally posted by: CycloWizard

Originally posted by: Armitage
For most purposes, the atmosphere extends to about 100 Km.
Even though the force due to gravity is 0 at the center, the volume of air there is still supporting the entire column of air above it. So take your equation ... pressure = density*gravity*height, plug in the equation for gravity as a function of distance from the center of the earth, and the relation between pressure and density and integrate.

It's only 'supporting' it relative to the source of gravity. The 'bottom' of the air column will be the location with the highest gravity field.

Not following you there? What do you mean by "highest gravity field"? That is at the surface of the earth of course, but that isn't the bottom of the air column in any way that I can see.

 

[DrPizza]

Originally posted by: Armitage

Originally posted by: DrPizza

Originally posted by: Armitage
The air pressure would certainly be high ... the weight of a column of air about 6,600 Km high! But the weight of a given volume of air would change through the distance ... increase due to pressure as you go down, but decrease due to the gravity gradient inside the earth. Sounds like a fun problem in integral calculus! Matbe I'll fire up Mathematica over lunch and have a wack at it.

As to whether you get there or not ... unless the pressure gets so high that the density of the air becomes greater then the density of whatever is falling such that boyancy becomes an issue, I don't see how you wouldn't. Drag alone wouldn't prevent it because drag is proportional to velocity, and we're talking about aerodynamic drag, not static friction ... as long as there is a force on the body, there will be motion, and as long as you aren't exactly at the center of the earth there will be a force due to gravity.

It's not just a foce, but a net force on a body which will produce acceleration and motion. At super slow speeds, is the force of friction of the air equal to the gravitational force of attraction? If so, it may stop before it gets to the center.

Fluid drag is proportional to velocity, so as velocity approaches 0, so will the force due to the drag. So, if there is a gravitational force, no matter how small, then the velocity has to be non-zero, right? It will be at critical velocity ... Fdrag = Fgravity, and for there to be a drag force, there must be a velocity. No velocity = no drag = unbalanced force

Here's a program to describe this motion better... is the air resistance enough to cause critical damping?
here
(think of the 0 on the y-axis as being the center of the earth)

Meh ... need some plugin that I don't have installed. Here's a static graph of what I think it likely shows: http://hyperphysics.phy-astr.gsu.edu/hbase/oscda2.html

Maybe I see what you're getting at ... if the system is overdamped, it will approach, but never quite reach 0. In most systems where there is a component of mechanical friction involved as well, the velocity will go to 0 and you won't get any closer, but in this case, since we only have fluid friction, I think that your velocity will never go precisely to 0 ... it will continue to approach the 0 position indefinitely.

Note that an overdamped oscilator with a high enough initial velocity can overshoot 0 once, and then will approach from the other side.

Bingo! Now you see what I'm talking about.
I've seen the trivia problem multiple times, the answer of which is "you'd come out at the other side of the earth, then fall back through again back and forth indefinitely" But, that's only if you could ignore air resistance. I've just always wanted to have an accurate answer for that question

It's been a while since I did any actual calculations, but I remember thinking "wow, it'd take a long time before you were even close to the center" If anyone wants to have fun, just try deriving an equation for the acceleration of gravity (or the force) as a function of distance from the center of the earth. I'm sure there's got to be an easier way than the way I started. I started with a circle, and as a point on the surface moves inward, I subtracted the area "behind it" from the area "in front of it", creating a crescent. Then, center of mass of the crescent, distance to the center.... it got messy quickly enough that I quit, waiting for a more elegant and simple solution.

 

[Armitage]

Originally posted by: DrPizza

Originally posted by: Armitage

Originally posted by: DrPizza

Originally posted by: Armitage
The air pressure would certainly be high ... the weight of a column of air about 6,600 Km high! But the weight of a given volume of air would change through the distance ... increase due to pressure as you go down, but decrease due to the gravity gradient inside the earth. Sounds like a fun problem in integral calculus! Matbe I'll fire up Mathematica over lunch and have a wack at it.

As to whether you get there or not ... unless the pressure gets so high that the density of the air becomes greater then the density of whatever is falling such that boyancy becomes an issue, I don't see how you wouldn't. Drag alone wouldn't prevent it because drag is proportional to velocity, and we're talking about aerodynamic drag, not static friction ... as long as there is a force on the body, there will be motion, and as long as you aren't exactly at the center of the earth there will be a force due to gravity.

It's not just a foce, but a net force on a body which will produce acceleration and motion. At super slow speeds, is the force of friction of the air equal to the gravitational force of attraction? If so, it may stop before it gets to the center.

Fluid drag is proportional to velocity, so as velocity approaches 0, so will the force due to the drag. So, if there is a gravitational force, no matter how small, then the velocity has to be non-zero, right? It will be at critical velocity ... Fdrag = Fgravity, and for there to be a drag force, there must be a velocity. No velocity = no drag = unbalanced force

Here's a program to describe this motion better... is the air resistance enough to cause critical damping?
here
(think of the 0 on the y-axis as being the center of the earth)

Meh ... need some plugin that I don't have installed. Here's a static graph of what I think it likely shows: http://hyperphysics.phy-astr.gsu.edu/hbase/oscda2.html

Maybe I see what you're getting at ... if the system is overdamped, it will approach, but never quite reach 0. In most systems where there is a component of mechanical friction involved as well, the velocity will go to 0 and you won't get any closer, but in this case, since we only have fluid friction, I think that your velocity will never go precisely to 0 ... it will continue to approach the 0 position indefinitely.

Note that an overdamped oscilator with a high enough initial velocity can overshoot 0 once, and then will approach from the other side.

Bingo! Now you see what I'm talking about.
I've seen the trivia problem multiple times, the answer of which is "you'd come out at the other side of the earth, then fall back through again back and forth indefinitely" But, that's only if you could ignore air resistance. I've just always wanted to have an accurate answer for that question

It's been a while since I did any actual calculations, but I remember thinking "wow, it'd take a long time before you were even close to the center" If anyone wants to have fun, just try deriving an equation for the acceleration of gravity (or the force) as a function of distance from the center of the earth. I'm sure there's got to be an easier way than the way I started. I started with a circle, and as a point on the surface moves inward, I subtracted the area "behind it" from the area "in front of it", creating a crescent. Then, center of mass of the crescent, distance to the center.... it got messy quickly enough that I quit, waiting for a more elegant and simple solution.

Don't you simply neglect the mass of the spherical shell above you, since the force due to gravity inside a spherical shell is 0?
So in the gravitational equation F = GMm/r^2, the primary mass M is now a function of r, specifically M = 4/3 pi r ^3 d where d is the density.

So now we have F = G 4/3 pi r d m

 

[sao123]

Don't you simply neglect the mass of the spherical shell above you, since the force due to gravity inside a spherical shell is 0?

I dont think so. This would only apply is the earths core was hollow.

In the basic principal, you have to account for the gravitational pull of every atom of the earth on the falling object.
the gravitational constant is not constant throughout the earths core. It is (local maximum) 9.8 m/(s^2) at the surface and dropping at some function f(x) until it is 0 at the exact center of the earth, and then raising again until it is -9.8m/(s^2) at the other surface point. you could probably approximate a x^3 order equation which would model this behavior. But this is no small task and would get very ugly quickly.

If you could graph a similar equation in x^3 order to represent the air friction with respect to column density (pressure), you might be able to use that force sum, to predict an actual velocity & position at some time (t) into the motion.

Very ugly.

 

[VTHodge]

As far as air pressure is concerned, I believe there would be no pressure at the center of the earth. Fluid pressure acts on the surface between the fluid and the center of gravity.

Now, as far as gravity collapsing your hole... :Q! Imagine cutting a thin strip out of the middle of a strong magnet.

 

[Armitage]

Originally posted by: sao123

Don't you simply neglect the mass of the spherical shell above you, since the force due to gravity inside a spherical shell is 0?

I dont think so. This would only apply is the earths core was hollow.

It doesn't matter if the earth's core is hollow or not. If you're inside a spherical shell, the effect of that shell cancels itself out ... if you do the integrals it comes out to zero. (http://www.madsci.org/posts/ar...0/962116209.Ph.r.html)

You can seperate this from the force due to the sphere beneath you, so you only need to consider the mass below you.

In the basic principal, you have to account for the gravitational pull of every atom of the earth on the falling object.
the gravitational constant is not constant throughout the earths core. It is (local maximum) 9.8 m/(s^2) at the surface and dropping at some function f(x) until it is 0 at the exact center of the earth, and then raising again until it is -9.8m/(s^2) at the other surface point.

Your f(x) is GM/r^2
where M is the mass of the earth, and r is the radius from the center of the earth. Or in the case, the mass of the sphere below you at a radius of r. It's just the universal gravitation law with a substitution so you don't have to muck around remembering the mass & radius of the earth for basic calculations.

you could probably approximate a x^3 order equation which would model this behavior. But this is no small task and would get very ugly quickly.

Nope

If you could graph a similar equation in x^3 order to represent the air friction with respect to column density (pressure), you might be able to use that force sum, to predict an actual velocity & position at some time (t) into the motion.

Very ugly.

 

[Armitage]

Originally posted by: VTHodge
As far as air pressure is concerned, I believe there would be no pressure at the center of the earth. Fluid pressure acts on the surface between the fluid and the center of gravity.

eh? What's holding up all that air then? At the surface of the earth, the ground holds up about 14 lbs/in^2
If you have a hole all the way through, what's holding up that 14 psi? The 14PSI on the other side. Draw an FBD of the air in that hole ... there's nothing else to take that load, and the load will only increase the deeper you go because there is more air above you, even though the weight/mass of the air decreases the lower you go.

 

[DrPizza]

Originally posted by: Armitage

Originally posted by: DrPizza

Originally posted by: Armitage

Originally posted by: DrPizza

Originally posted by: Armitage
The air pressure would certainly be high ... the weight of a column of air about 6,600 Km high! But the weight of a given volume of air would change through the distance ... increase due to pressure as you go down, but decrease due to the gravity gradient inside the earth. Sounds like a fun problem in integral calculus! Matbe I'll fire up Mathematica over lunch and have a wack at it.

As to whether you get there or not ... unless the pressure gets so high that the density of the air becomes greater then the density of whatever is falling such that boyancy becomes an issue, I don't see how you wouldn't. Drag alone wouldn't prevent it because drag is proportional to velocity, and we're talking about aerodynamic drag, not static friction ... as long as there is a force on the body, there will be motion, and as long as you aren't exactly at the center of the earth there will be a force due to gravity.

It's not just a foce, but a net force on a body which will produce acceleration and motion. At super slow speeds, is the force of friction of the air equal to the gravitational force of attraction? If so, it may stop before it gets to the center.

Fluid drag is proportional to velocity, so as velocity approaches 0, so will the force due to the drag. So, if there is a gravitational force, no matter how small, then the velocity has to be non-zero, right? It will be at critical velocity ... Fdrag = Fgravity, and for there to be a drag force, there must be a velocity. No velocity = no drag = unbalanced force

Here's a program to describe this motion better... is the air resistance enough to cause critical damping?
here
(think of the 0 on the y-axis as being the center of the earth)

Meh ... need some plugin that I don't have installed. Here's a static graph of what I think it likely shows: http://hyperphysics.phy-astr.gsu.edu/hbase/oscda2.html

Maybe I see what you're getting at ... if the system is overdamped, it will approach, but never quite reach 0. In most systems where there is a component of mechanical friction involved as well, the velocity will go to 0 and you won't get any closer, but in this case, since we only have fluid friction, I think that your velocity will never go precisely to 0 ... it will continue to approach the 0 position indefinitely.

Note that an overdamped oscilator with a high enough initial velocity can overshoot 0 once, and then will approach from the other side.

Bingo! Now you see what I'm talking about.
I've seen the trivia problem multiple times, the answer of which is "you'd come out at the other side of the earth, then fall back through again back and forth indefinitely" But, that's only if you could ignore air resistance. I've just always wanted to have an accurate answer for that question

It's been a while since I did any actual calculations, but I remember thinking "wow, it'd take a long time before you were even close to the center" If anyone wants to have fun, just try deriving an equation for the acceleration of gravity (or the force) as a function of distance from the center of the earth. I'm sure there's got to be an easier way than the way I started. I started with a circle, and as a point on the surface moves inward, I subtracted the area "behind it" from the area "in front of it", creating a crescent. Then, center of mass of the crescent, distance to the center.... it got messy quickly enough that I quit, waiting for a more elegant and simple solution.

Don't you simply neglect the mass of the spherical shell above you, since the force due to gravity inside a spherical shell is 0?
So in the gravitational equation F = GMm/r^2, the primary mass M is now a function of r, specifically M = 4/3 pi r ^3 d where d is the density.

So now we have F = G 4/3 pi r d m

Ah ha! Now I remember... I knew there was something simpler the last time I did that. But, I remember that the simplification that I used - using circles - didn't work.

whoa.. This was unexpected...
Assuming a uniform density (which we know isn't true) and realizing that we can neglect the hollow sphere above us, use

F = Gm1m2/r^2
m1 = mass of you
m2 equals mass of the sphere of the earth below you after neglecting the outer sphere above you
r=distance from center of the earth
R=radius of earth (assume spherical)
me = mass of the earth

m2 is the volumetric proportion of the earth remaining times the mass of the earth
or (4/3 Pi r^3)/(4/3 Pi R^3) * me
a little cancelling
m2 = r^3 * me / R^3

substitute:
F = G m1 r^3 *me / (r^2 R^3)

rearrange and separate out G m1 me / R*2 (which is equal to the force at the surface of the earth)

and you're left with r/R

So, the force as a function of distance from the center of the earth
F(r) = Force at the surface * r/R, where R = radius of the earth.

So, at a radius of 1/10 the radius of the earth, the force would equal 1/10 the force on the surface.

progress so far... unless I've made a mistake?

 

[Armitage]

Originally posted by: DrPizza

Originally posted by: Armitage

Originally posted by: DrPizza

Originally posted by: Armitage

Originally posted by: DrPizza

Originally posted by: Armitage
The air pressure would certainly be high ... the weight of a column of air about 6,600 Km high! But the weight of a given volume of air would change through the distance ... increase due to pressure as you go down, but decrease due to the gravity gradient inside the earth. Sounds like a fun problem in integral calculus! Matbe I'll fire up Mathematica over lunch and have a wack at it.

As to whether you get there or not ... unless the pressure gets so high that the density of the air becomes greater then the density of whatever is falling such that boyancy becomes an issue, I don't see how you wouldn't. Drag alone wouldn't prevent it because drag is proportional to velocity, and we're talking about aerodynamic drag, not static friction ... as long as there is a force on the body, there will be motion, and as long as you aren't exactly at the center of the earth there will be a force due to gravity.

It's not just a foce, but a net force on a body which will produce acceleration and motion. At super slow speeds, is the force of friction of the air equal to the gravitational force of attraction? If so, it may stop before it gets to the center.

Fluid drag is proportional to velocity, so as velocity approaches 0, so will the force due to the drag. So, if there is a gravitational force, no matter how small, then the velocity has to be non-zero, right? It will be at critical velocity ... Fdrag = Fgravity, and for there to be a drag force, there must be a velocity. No velocity = no drag = unbalanced force

Here's a program to describe this motion better... is the air resistance enough to cause critical damping?
here
(think of the 0 on the y-axis as being the center of the earth)

Meh ... need some plugin that I don't have installed. Here's a static graph of what I think it likely shows: http://hyperphysics.phy-astr.gsu.edu/hbase/oscda2.html

Maybe I see what you're getting at ... if the system is overdamped, it will approach, but never quite reach 0. In most systems where there is a component of mechanical friction involved as well, the velocity will go to 0 and you won't get any closer, but in this case, since we only have fluid friction, I think that your velocity will never go precisely to 0 ... it will continue to approach the 0 position indefinitely.

Note that an overdamped oscilator with a high enough initial velocity can overshoot 0 once, and then will approach from the other side.

Bingo! Now you see what I'm talking about.
I've seen the trivia problem multiple times, the answer of which is "you'd come out at the other side of the earth, then fall back through again back and forth indefinitely" But, that's only if you could ignore air resistance. I've just always wanted to have an accurate answer for that question

It's been a while since I did any actual calculations, but I remember thinking "wow, it'd take a long time before you were even close to the center" If anyone wants to have fun, just try deriving an equation for the acceleration of gravity (or the force) as a function of distance from the center of the earth. I'm sure there's got to be an easier way than the way I started. I started with a circle, and as a point on the surface moves inward, I subtracted the area "behind it" from the area "in front of it", creating a crescent. Then, center of mass of the crescent, distance to the center.... it got messy quickly enough that I quit, waiting for a more elegant and simple solution.

Don't you simply neglect the mass of the spherical shell above you, since the force due to gravity inside a spherical shell is 0?
So in the gravitational equation F = GMm/r^2, the primary mass M is now a function of r, specifically M = 4/3 pi r ^3 d where d is the density.

So now we have F = G 4/3 pi r d m

Ah ha! Now I remember... I knew there was something simpler the last time I did that. But, I remember that the simplification that I used - using circles - didn't work.

whoa.. This was unexpected...
Assuming a uniform density (which we know isn't true) and realizing that we can neglect the hollow sphere above us, use

F = Gm1m2/r^2
m1 = mass of you
m2 equals mass of the sphere of the earth below you after neglecting the outer sphere above you
r=distance from center of the earth
R=radius of earth (assume spherical)
me = mass of the earth

m2 is the volumetric proportion of the earth remaining times the mass of the earth
or (4/3 Pi r^3)/(4/3 Pi R^3) * me
a little cancelling
m2 = r^3 * me / R^3

substitute:
F = G m1 r^3 *me / (r^2 R^3)

rearrange and separate out G m1 me / R*2 (which is equal to the force at the surface of the earth)

and you're left with r/R

So, the force as a function of distance from the center of the earth
F(r) = Force at the surface * r/R, where R = radius of the earth.

So, at a radius of 1/10 the radius of the earth, the force would equal 1/10 the force on the surface.

progress so far... unless I've made a mistake?

Looks ok ... I was messing around with a derivation of pressure vs. r over lunch. Not ready for primetime just yet.

 

[CycloWizard]

Originally posted by: Armitage
Not following you there? What do you mean by "highest gravity field"? That is at the surface of the earth of course, but that isn't the bottom of the air column in any way that I can see.

The field would be proportional to the mass of earth in a given direction. I've been solving similar equations all day and I think my head might explode if I do any more math. Maybe tomorrow.

 

[Mr Nasty]

If you use :F = GMeMm/r^2
where r is the distance from centre of a mass
G is the universal gravitational constant
Me is the mas of the Earth
Mm is the mass of you

if r = 0 (because you are at the centre of the Earth)
you would get F = GMeMm/0^2
anything over 0 is infinite thusly without subbing in any values other than 'r', at the centre of the Earth, the force exerted on Mm by Me infinite (although this is unlikely because gravity that large would suggest a black hole), anyway it shows us that gravity is at its maximum at the centre of an object.

Now g (gravity) is a uniform constant acceleration (9.81ms^-2) so you would fall towards the centre of the Earth with an increasing velocity (acceleration is velocity that changes its unit vector) with your maximum velocity being reached at the centre regardless of air resistance. A man that falls from a 40 storey building with impact with the ground at a higher velocity than someone who fell from a 4 storey building. What the extra air resistance will do is rob this system of energy and you will probably not have enough momentum to reach the other end of the tube.

 

[Armitage]

Originally posted by: Mr Nasty
If you use :F = GMeMm/r^2
where r is the distance from centre of a mass
G is the universal gravitational constant
Me is the mas of the Earth
Mm is the mass of you

if r = 0 (because you are at the centre of the Earth)
you would get F = GMeMm/0^2
anything over 0 is infinite thusly without subbing in any values other than 'r', at the centre of the Earth, the force exerted on Mm by Me infinite (although this is unlikely because gravity that large would suggest a black hole), anyway it shows us that gravity is at its maximum at the centre of an object.

This equation refers to the gravitational attraction between two point masses, but if you work the calculus, it works for a sphere as well, so long as you are outside the sphere. As discussed earlier, the gravitational attraction of a spherical shell on a point inside the shell cancels out. So we only consider the mass of the sphere of radius r. So Me = 4/3 rho pi r^3 where rho is the density of the earth (assumed uniform for this exercise).

Then
F = 4/3 G rho pi r Mm
At r=0 you have F = 0

Now g (gravity) is a uniform constant acceleration (9.81ms^-2)

This is only true at the radius of the earth ... again, as discussed before g = 9.81 m/s^2 = G Me/Re^2 where Re is the radius of the earth.

so you would fall towards the centre of the Earth with an increasing velocity (acceleration is velocity that changes its unit vector) with your maximum velocity being reached at the centre regardless of air resistance.

No, not regardless of air resistance ... that's the point of this thread. Actually, the maximum velocity with air resistance will almost certainly not occur at the center of the earth because, the closer you move to the center, the higher the air density, and the lower the force due to gravity. The critical velocity is the velocity at which the force of gravity (Fg) equal the force due to drag (Fd). Fg will decrease toward the center as shown above, while Fd will increase toward the center (increased air density), so as you head toward the center your critical velocity will be dropping.

A man that falls from a 40 storey building with impact with the ground at a higher velocity than someone who fell from a 4 storey building. What the extra air resistance will do is rob this system of energy and you will probably not have enough momentum to reach the other end of the tube.

If the air resistance is high enough such that the system is over-damped, you might not even reach the center.

 

[jagec]

Originally posted by: Armitage

This is only true at the radius of the earth ... again, as discussed before g = 9.81 m/s^2 = G Me/Re^2 where Re is the radius of the earth.

Re is Reynolds number dang it, change your nomenclature:evil:

Yeah, off the top of my head I think you'd reach maximum velocity about 10 seconds after the initial drop. From there on, the force of gravity would steadily weaken and the density of air would increase, slowing you down.

 

[sao123]

Did anyone see that dodgeball movie? I love when he throws that wrench at the guy.
Heres one that I'll throw at you now...


With the gravitational profile from the north pole to the south pole starting at g = 9.8 m/s^2 then decreasing as g = G * Me / Re^2 to 0 through the center of the earth, then increasing again back to 9.8 m/s^2 at the other side...

Ok now im outta breath, that was long...

Can we be sure that the atmospheric pressure profile will be as were predicting...IE will the max air resistance be at the center of the earth? If gravity is so weak down there...can we be sure that the air will be pulled or even forced down that far? Or will the maximum air density be at some R > 0 ?

 

[DrPizza]

So... the new question becomes:

find a function for the air pressure at a distance r from the center of the earth inside a hole drilled through the earth.

(Oh no! if you puncture the earth, all the gravity will leak out!)

I'm fairly certain the highest pressure would be at the center.