# Thread: Gravity at center of the earth

1. Newbie
Join Date
Nov 2004
Posts
4

## Gravity at center of the earth

As I move away from the earth, the effect of gravity is reduced. At the earth's surface, gravity is greatest on me. Will gravity decrease if I go below ground and closer to the center of the Earth?
If I could stand at the center of the earth, wouldn't I be weightless because there is no mass at the center but the earth's mass would be all around me? Would I be pulled in all directions and blow up like a balloon?

Anybody seen any web sites that discuss this topic?

2. Established Member
Join Date
Sep 2004
Posts
1,550
At the center of the Earth you would be weightless for the reason you surmise. In fact, if the Earth were made of a thin shell of neutronium and were otherwise completely hollow you would be weightless at all points within. You wouldn't blow up (supposing normal atmosphere).

3. For this discussion, let's assume that earth is perfectly spherical and has its mass perfectly distributed.

With these assumptions, if you could descend a magic elevator to the center of the earth, you'd weigh less and less as you descended. That's because it is only the amount of earth below you that has a net gravity force. When you got to the exact center, you would be "pulled in all directions equally", but this would result in weightlessness.

Edited to add: Drat you, Evan --- I need to type faster

4. Member
Join Date
Nov 2004
Posts
50

## Other forces?

Physics isn't my strong point so forgive a naive question. Wouldn't gravity at the center of the Earth be equal in all directions only if the Earth itself was all of existence? What about the pulling of the Moon, Sun, and the rest of the Universe? The gravity acting on the Earth is not equal from all directions.

5. ## Re: Other forces?

Originally Posted by fossilnut
Physics isn't my strong point so forgive a naive question. Wouldn't gravity at the center of the Earth be equal in all directions only if the Earth itself was all of existence? What about the pulling of the Moon, Sun, and the rest of the Universe? The gravity acting on the Earth is not equal from all directions.
Your thinking is correct. It is the sum of all the forces which produces the final result.

However, the force you feel from distance objects are quite weak. The gravitational force gets much weaker for every increase in distance. The inverse square law applies. If you double the distance of an object from you, you will feel (1/2)^2 (or 1/4) the force. Therefore, the distance of celestial bodies are not felt hardly at all.

6. Newbie
Join Date
Nov 2004
Posts
4
I think that if you were in the middle of a hollow planet which consisted of material with black hole strength gravity, your body blow up like a balloon or would be pulled apart if the forces were close enough.

This is because (I think) that the gravity acting on one side of your body would be stronger than the far side (tidal effect). This would have minimul effect in the middle of the earth, but in my hypothetical black-hole-strength case, I believe your body would experience some expansion. Even in the middle of the earth, there would probably be some calculable expansion - theoretically anyway.

7. Originally Posted by TBoz
I think that if you were in the middle of a hollow planet which consisted of material with black hole strength gravity, your body blow up like a balloon or would be pulled apart if the forces were close enough.
Wow. You just threw the cow into the "glass of milk" discussion.

This is because (I think) that the gravity acting on one side of your body would stronger than the far side (tidal effect). This would have minimul effect in the middle of the earth, but in my hypothetical black-hole-strenght case, I believe your body would experience some expansion. Even in the middle of the earth, there would probably be some calculable expansion - theoretically anyway.
You are correct in a big way. Getting close to a black hole has huge tidal consequences. Once you are in the center of the black hole, however, well...you won't really be able to say much at that "point" (pun intended - it's Friday) :wink:

BTW, welcome to the board. =D>

8. AT
Member
Join Date
Nov 2004
Posts
46

## woo!

Yay! A physics questions I can answer! .... Wait.... Somone already did.

However, the last question:
I think that if you were in the middle of a hollow planet which consisted of material with black hole strength gravity, your body blow up like a balloon or would be pulled apart if the forces were close enough.
Well, yes or no. The part of me that has taken physics says "If it is a perfect sphere, and you are a point mass, then no." However, I don't know many people that are really point-masses.

9. Established Member
Join Date
Sep 2004
Posts
1,550
The gravity acting on the Earth is not equal from all directions.
Actually it is. The Earth is in free fall therefore the forces must be balanced.

10. Established Member
Join Date
Oct 2001
Posts
403
Originally Posted by TBoz
I think that if you were in the middle of a hollow planet which consisted of material with black hole strength gravity, your body blow up like a balloon or would be pulled apart if the forces were close enough.
You would be wrong. It doesn't matter how dense the material making up the shell is. The net gravitational effect sums to zero for every point inside the shell. You won't be pulled apart or blow up like a baloon, you'll be in free fall, feeling no gravitational effects at all.

11. Established Member
Join Date
Sep 2004
Posts
1,550
Think in terms of the rubber sheet analogy with a weight depressing it to make a funnel shape. For a hollow planet you could replace that weight with a ring. The field lines would slope down to the ring edge but inside the ring it would be perfectly flat.

12. Order of Kilopi
Join Date
May 2004
Posts
6,971
Welcome to BABB TBoz,

I heard if you could dig a whole straight though the center on the earth and jumped in once you were at wou would have accerlerated to 11 km sec so therefore you would keep going and actully reach the other side only to fall back down again sounds like fun though.

13. Established Member
Join Date
Nov 2004
Posts
1,534
From what I recall, gravity is greatest at the surface of Earth, dropping off according to the inverse square of the distance as you leave the surface. However, as you descend towards the centre of the Earth, gravity will drop off in a linear way instead.

This is from memory btw, I'd try and prove it but I've been up all night doing maths and I'm pretty much mathed out :-?

14. Established Member
Join Date
Feb 2004
Posts
1,992
Actually it is. The Earth is in free fall therefore the forces must be balanced.
I thought objects were stationary when forces were balanced.

15. Established Member
Join Date
Sep 2004
Posts
1,550
???

"An object at rest tends to stay at rest and an object in motion tends to stay in motion with the same speed and in the same direction unless acted upon by an unbalanced force."

Newton's first law.

See here

The Earth is falling around the sun. It is following a straight line in the gravitational potential field of the space around the sun (including the myriad of other minor gravitational forces from planets etc.). I did neglect to mention that within a hollow sphere there would still be tidal forces since matter of any kind, even hypothetical neutronium, is not a "gravity shield". If you were drifting within a hollow Earth orbiting the sun the tidal forces (clarification: from external objects) would cause you to drift towards the center of mass, the center of the hollow Earth.

16. AT
Member
Join Date
Nov 2004
Posts
46
Originally Posted by TinFoilHat
You would be wrong. It doesn't matter how dense the material making up the shell is. The net gravitational effect sums to zero for every point inside the shell. You won't be pulled apart or blow up like a baloon, you'll be in free fall, feeling no gravitational effects at all.
True, if you are a point mass; the net gravitational force on your body is zero; however, if you are on a scale where arn't a point mass, you'd have problems, as the net force on your left hand would be opposite than on your right. I'm told by Bio majors that connective tissue isn't infinitly strong, so.....

17. Inside a spherical shell, the net gravitational force from the shell's mass is zero at all points on your body so tidal forces are not an issue. Tidal forces are the result of gravitational forces being different and different points on a body because the body takes up space and so parts are significantly further away than others, hence differential weight. However, in the spherical shell, it's all zero so no tidal forces.

18. Established Member
Join Date
Sep 2004
Posts
1,550
Glom is correct, the only tidal forces (as I mentioned above) are from other gravitational sources external to the shell.

19. Newbie
Join Date
Nov 2004
Posts
4
Originally Posted by Glom
Inside a spherical shell, the net gravitational force from the shell's mass is zero at all points on your body so tidal forces are not an issue.
So as I drifted around inside this hypothetical hollow sphere and moved real close to the "side," I wouldn't experience any more gravitational pull on the side of my body closest to that edge? It would seem that as I drifted close to a side, that now closer side would have an increased gravitational pull and the sides further away would have less pull, in essence "sticking" me to that side, or at least having SOME tidal effect on my body. This is all assuming no outside forces.

Where am I wrong?

20. Consider a uniform spherical shell. A mass, m, is inside at a distance d from a small segment of the shell, dA, subtending a solid angle, dS, at the mass. dA = dS × d². Let's say that the shell has a density, s. The gravitational attraction on m is Gm.s.dS.d²/d² = Gm.dS. In other words, gravitational attraction from a segment of the shell is independent of the position of the mass. You can think of it as the far side may have a weaker gravitational attraction per unit mass, but there's more mass.

21. Newbie
Join Date
Nov 2004
Posts
4
Originally Posted by Glom
You can think of it as the far side may have a weaker gravitational attraction per unit mass, but there's more mass.
That's the part of your explanation I understood . . . thanks for the clarification.

What a great forum I have stumbled onto.

22. Established Member
Join Date
Sep 2004
Posts
1,550
Here is a graphical illustration of why it is so. The closer you get to the shell the less mass there is "below" the line and the more above it. The line intersects the shell where the shell mass is pulling the object "left" and "right" in both cases.

23. Originally Posted by Evan
???

"An object at rest tends to stay at rest and an object in motion tends to stay in motion with the same speed and in the same direction unless acted upon by an unbalanced force."

Newton's first law.

See here

The Earth is falling around the sun. It is following a straight line in the gravitational potential field of the space around the sun (including the myriad of other minor gravitational forces from planets etc.). I did neglect to mention that within a hollow sphere there would still be tidal forces since matter of any kind, even hypothetical neutronium, is not a "gravity shield". If you were drifting within a hollow Earth orbiting the sun the tidal forces (clarification: from external objects) would cause you to drift towards the center of mass, the center of the hollow Earth.
But still, Evan, your implication that all gravitational forces are balanced is incorrect. The earth's centrifugal force is is not gravitational (at least not in origin, coming from the spin of the molecular cloud), and this is one of the forces balanced by the others.

24. Established Member
Join Date
Sep 2004
Posts
1,550
What? Centrifugal "force" is not a force.

"spin of the molecular cloud"?????

25. I meant to say centripetal force. ops:

In any case, my objection is that not all forces acting on the earth are gravitational. Our motion around the sun is associated with another force, a non-gravitational force. Therefore, the gravity acting on Earth is *not* equal from all directions. This is the point that you made earlier which I am refuting. Think of a 2-body system. Gravity is definitely not the only force involved there. Add bodies and my point still remains valid.

26. Ut
Established Member
Join Date
Feb 2004
Posts
2,494
Originally Posted by Tobin Dax
I meant to say centripetal force. ops:

In any case, my objection is that not all forces acting on the earth are gravitational. Our motion around the sun is associated with another force, a non-gravitational force. Therefore, the gravity acting on Earth is *not* equal from all directions. This is the point that you made earlier which I am refuting. Think of a 2-body system. Gravity is definitely not the only force involved there. Add bodies and my point still remains valid.
Err, uhh...

Assuming both bodies in a 2 body system are electrically neutral, then gravity is the only force at play. Gravity is the centripital force you speak of.

I think the point you're trying to make, and that you're picking nits with, is that there are other bodies contributing to the net force.

27. Established Member
Join Date
Sep 2004
Posts
1,550
Centripetal, centrifugal. They are not forces. There is gravitation. There is the electroweak force. There is the strong force. That's it to our knowledge. The so called centripetal "force" is the consequence of an object being perturbed from the path it would take if not so perturbed. The Earth in orbit around the sun is not so perturbed.

28. Think of a 2-body system. Gravity is definitely not the only force involved there.
Wrong!!! [-X Momentum is not a force!

29. Originally Posted by Ut
Originally Posted by Tobin Dax
I meant to say centripetal force. ops:

In any case, my objection is that not all forces acting on the earth are gravitational. Our motion around the sun is associated with another force, a non-gravitational force. Therefore, the gravity acting on Earth is *not* equal from all directions. This is the point that you made earlier which I am refuting. Think of a 2-body system. Gravity is definitely not the only force involved there. Add bodies and my point still remains valid.
Err, uhh...

Assuming both bodies in a 2 body system are electrically neutral, then gravity is the only force at play. Gravity is the centripital force you speak of.

I think the point you're trying to make, and that you're picking nits with, is that there are other bodies contributing to the net force.
Well that's what caused my confusion, but I'm still wrong. ops: (My brain must already be on Thanksgiving vacation.) A five-second force diagram in my head didn't balance with other planets involved, but I see my mistake now. That's what I get for not thinking it through.

30. Newbie
Join Date
Jun 2004
Posts
5
Here is a plot of the earth's gravity as a function of distance from the earth's center, calculated using the PREM model of the earth's density distribution:

gravity.png

It's notable how constant the gravitational pull is through most of the mantle.

#### Posting Permissions

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