If I was involved in "journey to the centre of the earth", and I reached a cavern where there was as much of the earth's mass above me as below me, and for that matter, all around me, how would my body and equipment be affected by gravity??
Newbie
If I was involved in "journey to the centre of the earth", and I reached a cavern where there was as much of the earth's mass above me as below me, and for that matter, all around me, how would my body and equipment be affected by gravity??
Order of Kilopi
At the exact center? you'd be effectively weightless-- equal pull from all sides.
(You'd probably also be crushed to a pinpoint by the pressure. But I'm assuming something-- let's call it "hypotheticum"-- holds this cavern open against the literal weight of the world.)
STARGAZING: All I see are the lights of a billion places I'll never go. --Howard Tayler, Schlock Mercenary
Established Member
Assuming the Earth is of uniform density throughout (which it ain't) gravity decreases linearly from 1g at the surface to 0 at the centre. At caving, drilling, mining depths, the decrease is negligible, eg, at 70km depth g= 6300/6370 = 98.8%. However, atmospheric pressure will increase exponentially. From memory, I think it doubles every 16km.
Established Member
You would be pretty much weightless since there would be (pretty much) equal amounts of mass pulling in all directions for a net force of zero.
Established Member
Wouldn't that result in a naturally hollow center?
Think about it, there is gravity a short distance away in any direction. Wouldn't that be an incentive for matter to leave the center and compact towards the Earth's upper surface?
Wouldn't gravity be at maximum half way to the center of the Earth (assuming a hollow center)?
Order of Kilopi
Gravity all pulls inward, toward the center of mass. That's where the pressure is highest, too.
STARGAZING: All I see are the lights of a billion places I'll never go. --Howard Tayler, Schlock Mercenary
Banned
You'd be weightless. If, however, you fell down this hole, your Speed would increase at a certain acceleration.Originally Posted by billrot
If I was involved in "journey to the centre of the earth", and I reached a cavern where there was as much of the earth's mass above me as below me, and for that matter, all around me, how would my body and equipment be affected by gravity??
As you fell your Speed would increase, but your rate of Accelleration would begin to fall, so that by the time you arrived at the center, your speed wopuld be at a maximum, but you would no longer have any acceleration, until , of course, you passed thru' the center and began to 'rise' up the remainder of the tunnel.
Slowing.
Finally, with no more speed and no more acceleration, you would pop out at the surface, much to the amazement of your friends and step into their world, easily and narturally.
Order of Kilopi
Minus speed lost to friction, of course. You'd need to be in a vacuum to travel that far without losing energy.
STARGAZING: All I see are the lights of a billion places I'll never go. --Howard Tayler, Schlock Mercenary
Banned
As you fell your Speed would increase, but your rate of Accelleration would begin to fall, so that by the time you arrived at the center, your speed wopuld be at a maximum, but you would no longer have any acceleration, until , of course, you passed thru' the center and began to 'rise' up the remainder of the tunnel.
Slowing.
Finally, with no more speed and no more acceleration, you would pop out at the surface, much to the amazement of your friends and step into their world, easily and naturally.
Established Member
Ok, So I made a wrong assumption there would not be gravity a short distance away. A hollow Earth would be at 0 gravity on the inner surface except for slight gravity due to Earth not being perfectly spherical. So there would be no incentive for matter to compact towards the upper crust:
Gravity and a Hollow Earth
But then again wouldn't the slight outward gravity result in a small hollow cavity?
Order of Kilopi
What's "outward" gravity??
Also, pressure at that depth would squish any cavity out of existence.
STARGAZING: All I see are the lights of a billion places I'll never go. --Howard Tayler, Schlock Mercenary
Order of Kilopi
If you mean gravity from the rest of the Earth's mass, that equalizes at the center point; no net pull.
STARGAZING: All I see are the lights of a billion places I'll never go. --Howard Tayler, Schlock Mercenary
Established Member
Outward gravity would be a pull towards the surface of Earth mostly due to centrifugal forces at equator which according to the link in my last post is at 1/300 of ordinary Earth (upper surface) gravity.
Order of Kilopi
At roughly 3 million atmospheres of pressure pushing in, that small amount of centrifugal force pushing out is not going to make a difference.
STARGAZING: All I see are the lights of a billion places I'll never go. --Howard Tayler, Schlock Mercenary
Established Member
Why would there be any pressure in a weightless center? It seems to me pressure is a product of gravitational forces.
Weightless => no gravity => no pressure
Order of Kilopi
Don't forget, all the matter around that weightless point is under gravity, all being pulled in ("down") with tremendous weight. The whole planet isn't weightless, just one tiny spot in the center.
ADDED: It's not actually a "zero" gravity point, it's gravity pulling every-which-way at once, leaving you suspended in the middle.
STARGAZING: All I see are the lights of a billion places I'll never go. --Howard Tayler, Schlock Mercenary
Order of Kilopi
Think of this analogy: tie two people to a tree tightly, with the first person between the tree and the second person. No ropes are pulling or pressing on the first person, but they feel the pressure of the second person. Does that make any sense?
Some would say that that is actually zero gravity
Established Member
Here is how I figure:
* On Earth's surface gravity is at 1g.
* Gravity is reduced as we move above the surface.
* Gravity would increase to a value higher than 1g if we moved towards the center of Earth. It would increase to a maximum and then start reducing as we move further towards the center where it reaches 0g.
* As such there is an inner spherical zone where gravity is maximal. This sphere has a radius less than that of Earth but bigger than 0. This imaginary spherical zone is where all Earth matter should pull towards and compress about.
* If I am wrong then that's normal
Order of Kilopi
That's pretty much what happens. The maximum is about half way down.
No, gravity on both sides of the zone still pulls towards the center of the earth.This imaginary spherical zone is where all Earth matter should pull towards and compress about.Good to be normal* If I am wrong then that's normal
Established Member
You are right. The direction is always towards the center.
I can't visualize what would happen to density of the earth below the maximal zone. I assume it should reduce as the force of gravity is reduced. In other words Earth is at its densest at the "maximal sphere". But that's not like the cut section depictions I have seen of Earth.
Established Member
The deepest mines are about 8 miles = 12 kilometers. Most have slightly less gravity but some have slightly more gravity due to the equitoral bulge, the surface being wrinkled with mountains and the crust varying wildly in density. Mathematically; we are quite sure gravity decreases with depth, a bit faster than linear at depths of about 70 kilometers and deeper. Neil
Order of Kilopi
For good reason
Why would it be densest there? There is less stuff on top of it, than on top of the material below it--all the stuff that is pushing down on the maximal zone is also pushing down on the lower stuff.
Actually, what happens at the maximal zone, is the composition of the earth changes from lower density solid to high density liquid, at the core-mantle boundary.
Established Member
I'm glad I participated. Thanks for teaching an old dog a new trick.
Established Member
Another way of looking at it: (ignoring the fact that we're dealing with a 3D object) density increases as you go to the centre from the maximum gravity point, but not as quickly as it was increasing as you went from the surface to the max gravity point.
It's like driving a car. You start accellerating, and you move some distance north. The increase in speed is gravity increasing as you fall, and the motion is density increasing. But then you hit the brakes and slow down. You still move north (density still increasing), but you're not driving as quickly (the magnitude of gravity is less).
Established Member
Hi Gents,
I really am not trying to be a pain, but something doesn't add up for me.
0 is a funny number and makes any function that limits to it a funny function.
I will try to explain with my very limited expressive abilities:
* f(x)= function of gravity versus distance from the center
* Lower limit of x=0, upper limit of x=distance from center to core-mantle boundary
* f(0) = 0g (Any mass will weigh 0 kg)
* f(x) has a solution such that any given mass m will weigh any desired value W where W < m
* There is an equilibrium distance x<>0 for any given centrifugal force where this force is equal to the weight of the mass of the earth above it.
I think it does make sense.
Last edited by a1call; 2007-May-28 at 03:21 PM.
Order of Kilopi
Since you are talking about weight, f must be a force, so that should be f(0) = 0N (N is Newtons)Gravity is a very weak force, but the centrifugal force is small too--the earth only rotates once per day. So, no, that's not going to happen. Just imagine how hard it would be to shake Brad Pitt off if you were turning around once per day.* There is an equilibrium distance x<>0 for any given centrifugal force where this force is equal to the weight of the mass of the earth above it.
Established Member
I got a strange figure
The location of the balance pointHi Gents,
I really am not trying to be a pain, but something doesn't add up for me.
0 is a funny number and makes any function that limits to it a funny function.
I will try to explain with my very limited expressive abilities:
* f(x)= function of gravity versus distance from the center
* Lower limit of x=0, upper limit of x=distance from center to core-mantle boundary
* f(0) = 0g (Any mass will weigh 0 kg)
* f(x) has a solution such that any given mass m will weigh any desired value W where W < m
* There is an equilibrium distance x<>0 for any given centrifugal force where this force is equal to the weight of the mass of the earth above it.
I think it does make sense.
With gravity for balance would there need to be an equal volume of the world from the centre at which all points pull apart to a neutral section to balance the surface gravity down.
Using the formula of a cone 1/3 * pi * r^2 * h
where r^2 is the radius of the cone
and h is the height of the cone in this case the radius of the earth
We get large ratio : small ratio
=> (1/3 * pi * r^2 * h) - (1/3 * pi * (r/2)^2 * h) : 1/3 * pi * (r/2)^2 * h
if we divide by 1/3 * pi * h we get r^2 - (r/2)^2 : (r/2)^2
=> 4r^2 - r^2 : r^2
=> 3 : 1
Wouldn't this indicate the balance point is a fair bit closer to the surface of the planet than at a point halfway down.
I really need help here as my math ability is to say the least highly questionable.
Cheers Mike
Order of Kilopi
Yes, and there should be less compaction towards the center due to gravity; Gravity is actually strongest very near the surface. However, as others have pointed out, the atmospheric pressure gradient maintains a great deal of pressure towards the center.
The total effect is rather odd: if you assumed the center ~200 meters of the earth is hollow and filled with nitrogen & oxygen, (unlikely, but not totally unreasonable), the gravity would be virtually none existant, but the dense air very viscous - you could easily fly by swimming through the air! The rotational effects near the center would be minimal, but there would be a very slight pull towards tidal foci (moon and sun), and towards the equatorial bulge.
Order of Kilopi
Balance point of what?
Gravity "balances" at the center of the earth. At the surface, it does not. Gravity balances with centrifugal force at the geostationary orbit.
Order of Kilopi
No, it's strongest near the core-mantle boundary, about half way downWhat's a tidal focus? Is the moon itself a tidal focus, is that what is meant?The rotational effects near the center would be minimal, but there would be a very slight pull towards tidal foci (moon and sun), and towards the equatorial bulge.
Posting Permissions
Reply With Quote