1. Newbie
Join Date
May 2007
Posts
2

local gravity

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??

2. 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.)

3. Established Member
Join Date
Jul 2002
Posts
330
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.

4. Established Member
Join Date
Oct 2006
Posts
113
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.

5. 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)?

6. Gravity all pulls inward, toward the center of mass. That's where the pressure is highest, too.

7. 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??
You'd be weightless. If, however, you fell down this hole, your Speed would increase at a certain acceleration.

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.

8. Minus speed lost to friction, of course. You'd need to be in a vacuum to travel that far without losing energy.

9. Originally Posted by astrocat
You'd be weightless. If, however, you fell down this hole, your Speed would increase at a certain acceleration.
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.

10. 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?

11. What's "outward" gravity??

Also, pressure at that depth would squish any cavity out of existence.

12. If you mean gravity from the rest of the Earth's mass, that equalizes at the center point; no net pull.

13. 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.

14. At roughly 3 million atmospheres of pressure pushing in, that small amount of centrifugal force pushing out is not going to make a difference.

15. 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

16. 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.

17. Originally Posted by a1call
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
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?

Originally Posted by Noclevername
ADDED: It's not actually a "zero" gravity point, it's gravity pulling every-which-way at once, leaving you suspended in the middle.
Some would say that that is actually zero gravity

18. 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

19. Originally Posted by a1call
It would increase to a maximum and then start reducing as we move further towards the center where it reaches 0g.
That's pretty much what happens. The maximum is about half way down.
This imaginary spherical zone is where all Earth matter should pull towards and compress about.
No, gravity on both sides of the zone still pulls towards the center of the earth.
* If I am wrong then that's normal
Good to be normal

20. Originally Posted by hhEb09'1
No, gravity on both sides of the zone still pulls towards the center of the earth.
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.

21. Established Member
Join Date
Nov 2005
Posts
2,967
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

22. Originally Posted by a1call
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.
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.

23. I'm glad I participated. Thanks for teaching an old dog a new trick.

24. Originally Posted by a1call
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.
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).

25. 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.

26. Originally Posted by a1call
* f(0) = 0g (Any mass will weigh 0 kg)
Since you are talking about weight, f must be a force, so that should be f(0) = 0N (N is Newtons)
* 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.
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.

27. I got a strange figure

Originally Posted by a1call
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.
The location of the balance point

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

28. Originally Posted by Michael Noonan
Wouldn't this indicate the balance point is a fair bit closer to the surface of the planet than at a point halfway down.
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.

29. Originally Posted by Michael Noonan
Wouldn't this indicate the balance point is a fair bit closer to the surface of the planet than at a point halfway down.
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.

30. Originally Posted by Jerry
Yes, and there should be less compaction towards the center due to gravity; Gravity is actually strongest very near the surface.
No, it's strongest near the core-mantle boundary, about half way down
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.
What's a tidal focus? Is the moon itself a tidal focus, is that what is meant?

Posting Permissions

• You may not post new threads
• You may not post replies
• You may not post attachments
• You may not edit your posts
•