What's gravity like? at center of the earth

  • Thread starter Chitose
  • Start date
  • #1
71
0

Main Question or Discussion Point

Hello, Chitose wonder chick here.

I don't know which where to put this question... earth or physic, so I place it here.

....................................................

Since earth is big, most common people know that gravity is pull us stick on earth, or rather Space itself push us to stick to earth. either way.

but I'm curious, Assume that center of earth is giant hole, so what gravity at center of earth gonna looks like?

(Woh there, cut of too Heat and too much pressure out :))

.......................................................

English is not my native language, forgive me If I'm wrong in spelling and gamma.
 

Answers and Replies

  • #2
Evo
Mentor
23,113
2,472
but I'm curious, Assume that center of earth is giant hole, so what gravity at center of earth gonna looks like?
Here is the answer.

But the value of g is only zero at the center of the earth. As you move away from the center, the value of g increases until it reaches its maximum value at the earth's surface.

Each layer of earth must support the layers above it. Only the very center is 'weightless'.
 
  • #3
4,464
65
Here is the answer.

..
But the value of g is only zero at the center of the earth. As you move away from the center, the value of g increases until it reaches its maximum value at the earth's surface.
which is likely not entirely correct. This assumes homogenous density of the earth. However the density of the cores is much higher than the mantle. When you do the calc's using the standard values, you'll see that gravity is highest at the mantle core boundary.
 
Last edited:
  • #4
115
1
I remember asking my primary school teacher this, and watching her get very flustered. But yeah, you would be "weightless" in the centre of the earth.
 
Last edited:
  • #5
4,464
65
As I said, gravity is highest at the mantle core boundary, check this post
 
  • #6
Redbelly98
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
12,098
129
But the value of g is only zero at the center of the earth. As you move away from the center, the value of g increases until it reaches its maximum value at the earth's surface.
which is likely not entirely correct. This assumes homogenous density of the earth. However the density of the cores is much higher than the mantle. When you do the calc's using the standard values, you'll see that gravity is highest at the mantle core boundary.
Interesting. But a homogeneous density would imply that g is proportional to the distance from the center, and Doc Al never said that. He only said that g increases with distance from the center up to the surface.

Doc Al's statement would place a restriction on how the density can vary, but Andre's link suggests that restriction is not met. We can work out what this restriction on the Earth's density is ... skip down to "Final Result" if you want to skip all the math.

When increasing the distance from the center, the change in g is:

Δg = Gm(1/(r+Δr)2 - 1/r2) + G Δm/(r+Δr)2

where m is the mass contained within r, and Δm is the mass in the shell between r and r+Δr. This equation says that there are two terms involved in evaluating Δg:

1. The increased distance from the mass m, which tends to lower g, and
2. The added mass located between r and r+Δr, which tends to increase g.

Continuing, approximating to first order in Δr and writing Δm in terms of the density:

Δg ≈ -2Gm Δr/r3 + G 4π r2 Δr ρ / r2

= (-2m/r3 + 4πρ) GΔr​

So we can say that

dg/dr = (4πρ - 2m/r3) G​

We can write m in terms of [ρ], the average density contained within r:

dg/dr = (4πρ - (2/r3)(4π/3)r3 [ρ]) G

= (4πρ - (8π/3) [ρ]) G
So g will decrease with increasing r when

4πρ - (8π/3) [ρ] < 0​

or

Final result: ρ < (2/3) [ρ]​

In other words, if the density at some distance from the Earth's center is less than 2/3 of the average density within that distance, then g will decrease with increasing distance. This is obviously true at the Earth's surface, but could be true somewhere within the Earth as well -- and is true at the mantle/core boundary, according to http://pubs.usgs.gov/gip/interior/" [Broken]

p.s. I hope there are no math errors here -- any corrections would be welcome.
 
Last edited by a moderator:
  • #7
4,464
65
He only said that g increases with distance from the center up to the surface.
I was refering to this:

...the value of g increases until it reaches its maximum value at the earth's surface.
The maximum is likely to be at the core mantle boundary, even the value at the crust bantle boundary is higher. See my spreadsheet.

Obviously you dont need the earth to be homogeous, that was too quick.
 
  • #8
Redbelly98
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
12,098
129
Yes, agreed. Even though the Earth does not have a uniform density, it is commonly assumed that g still increases with distance. Very interesting to find that isn't so.

p.s. BTW, I've been editing my Post #6 as you were posting your response.
 
Last edited:
  • #9
Interesting mental image. If there were an empty sphere 10 feet in diameter at the center of the earth and a person was inside it, what would happen? Would that person be pulled in all directions towards the mass of the surrounding earth? Would they be pulled in the direction of the sun or moon, scrapping along that part of the surface of the inside of the sphere as the earth rotated on its axis? Or should they just wait in wait-lessness for a physicist to tell them what to do?
 
  • #10
Redbelly98
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
12,098
129
One would be weightless inside an empty sphere -- neglecting the effects of the sun, moon, and other planets, and neglecting the fact that the Earth is not a perfect sphere. And I doubt one could notice these other effects.
 
  • #11
765
15
Interesting. But a homogeneous density would imply that g is proportional to the distance from the center, and Doc Al never said that. He only said that g increases with distance from the center up to the surface.

Doc Al's statement would place a restriction on how the density can vary, but Andre's link suggests that restriction is not met. We can work out what this restriction on the Earth's density is ... skip down to "Final Result" if you want to skip all the math.

When increasing the distance from the center, the change in g is:

Δg = Gm(1/(r+Δr)2 - 1/r2) + G Δm/(r+Δr)2

where m is the mass contained within r, and Δm is the mass in the shell between r and r+Δr. This equation says that there are two terms involved in evaluating Δg:

1. The increased distance from the mass m, which tends to lower g, and
2. The added mass located between r and r+Δr, which tends to increase g.

Continuing, approximating to first order in Δr and writing Δm in terms of the density:

Δg ≈ -2Gm Δr/r3 + G 4π r2 Δr ρ / r2

= (-2m/r3 + 4πρ) GΔr​

So we can say that

dg/dr = (4πρ - 2m/r3) G​

We can write m in terms of [ρ], the average density contained within r:

dg/dr = (4πρ - (2/r3)(4π/3)r3 [ρ]) G

= (4πρ - (8π/3) [ρ]) G
So g will decrease with increasing r when

4πρ - (8π/3) [ρ] < 0​

or

Final result: ρ < (2/3) [ρ]​

In other words, if the density at some distance from the Earth's center is less than 2/3 of the average density within that distance, then g will decrease with increasing distance. This is obviously true at the Earth's surface, but could be true somewhere within the Earth as well -- and is true at the mantle/core boundary, according to http://pubs.usgs.gov/gip/interior/" [Broken]

p.s. I hope there are no math errors here -- any corrections would be welcome.
Great post! Very elegant final result.

Unfortunately I got a bit lost in the maths. I agree with you with this result.

When increasing the distance from the center, the change in g is:

Δg = Gm(1/(r+Δr)2 - 1/r2) + G Δm/(r+Δr)2
However I don't see this:

Continuing, approximating to first order in Δr and writing Δm in terms of the density:

Δg ≈ -2Gm Δr/r3 + G 4π r2 Δr ρ / r2
Rather, my maths is giving me:

Δg ≈ -2Gm Δr/r2(r2 + 2Δr) + G 4π r2 Δr ρ / (r2 + 2Δr)​

If I'm right I think that spoils the elegance of your result. I hope I'm wrong!
 
Last edited by a moderator:
  • #12
Redbelly98
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
12,098
129
... my maths is giving me:

Δg ≈ -2Gm Δr/r2(r2 + 2Δr) + G 4π r2 Δr ρ / (r2 + 2Δr)​

If I'm right I think that spoils the elegance of your result. I hope I'm wrong!
One problem with your calculation is that you can't add r2 and 2Δr together, as is being done in these expressions. The units would be m2 and m, which are inconsistent for the purposes of adding or subtracting terms.
 
  • #13
765
15
One problem with your calculation is that you can't add r2 and 2Δr together, as is being done in these expressions. The units would be m2 and m, which are inconsistent for the purposes of adding or subtracting terms.
Agreed. I checked my equations. My equation should read:

Δg ≈ -2Gm Δr/r(r2 + 2rΔr) + G 4π r2 Δr ρ / (r2 + 2rΔr)​

Which is in contrast to this:

Δg ≈ -2Gm Δr/r3 + G 4π r2 Δr ρ / r2
How did you get the Δr out of the denominators?
 
  • #14
D H
Staff Emeritus
Science Advisor
Insights Author
15,393
682
In other words, if the density at some distance from the Earth's center is less than 2/3 of the average density within that distance, then g will decrease with increasing distance.
Correct.

Another way to look at this is that g increases (decreases) with increasing depth if the local density is less than (exceeds) 2/3 of the average density of the material deeper than the depth in question.

Starting from the surface, gravity inside the Earth increases initially, reaching a local max at the top of the lower mantle. It drops slightly in the lower mantle before starting to rise again, reaching a global max at the core/mantle boundary. From that point inward g drops monotonically to zero. The local min in the lower mantle is just a point at which the local density is 2/3 of the average density of the deeper material. There isn't anything physical going on at that point. Those maxima at the top and bottom of the lower mantle on the other hand represent very real discontinuities in density.

Here's another link: http://geophysics.ou.edu/solid_earth/prem.html [Broken], based on the 1981 "Preliminary Reference Earth Model". (A new and improved reference Earth model is apparently in the works.)
 
Last edited by a moderator:
  • #15
My head is spinning. What a rush. Even better, let's go on down, and check it out! (jp of course ;)

The equations you all have posted here are just so artful.
 
  • #16
Redbelly98
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
12,098
129
Agreed. I checked my equations. My equation should read:

Δg ≈ -2Gm Δr/r(r2 + 2rΔr) + G 4π r2 Δr ρ / (r2 + 2rΔr)​
.
.
.
How did you get the Δr out of the denominators?
Here are some intermediate steps I left out of my earlier calculation:

We had:
Δg = Gm(1/(r+Δr)2 - 1/r2) + G Δm/(r+Δr)2
Let's get an approximation for the 1/(r+Δr)2 term, which appears twice in the above expression.
1/(r+Δr)2 = 1/(r2 + 2rΔr +Δr2)
= 1 / r2(1 + 2 Δr/r + Δr2/r2)
≈ 1 / (r2(1 + 2 Δr/r))
= (1/r2) (1 + 2 Δr/r)-1
≈ (1/r2) (1 - 2 Δr/r)
= (1/r2) - (2Δr/r3)
Substituting this into the expression for Δg gives
Δg ≈ - 2GmΔr/r3 + GΔm/r2 - 2GΔmΔr/r3
Since Δm=4πr2Δrρ is proportional to Δr, this last term is of order Δr2 and may be neglected, leaving us with
Δg ≈ - 2GmΔr/r3 + GΔm/r2
Substitute Δm=4πr2 Δr ρ, and you'll get the result I had earlier.

p.s. It may have been easier to follow had I originally written g as a function of r and m, and then differentiated. Oh well.
 
  • #17
765
15
Let's get an approximation for the 1/(r+Δr)2 term, which appears twice in the above expression.

(1/r2) (1 + 2 Δr/r)-1 ≈ (1/r2) (1 - 2 Δr/r)
Thanks, I get it now. :smile:

I was missing, the approximation step, but I see how you can get that with the binomial series expansion.
 
  • #18
103
1
One would be weightless inside an empty sphere -- neglecting the effects of the sun, moon, and other planets, and neglecting the fact that the Earth is not a perfect sphere. And I doubt one could notice these other effects.
Do we really need to neglect the rest of the solar system/galaxy/universe? Earth is in free-fall with respect to their combined effect, after all.

I suppose that if the final approximation (treating Earth as a perfect sphere) needs to be made, this implies that the hypothetical empty sphere is at the geometric centre, not the centre of mass. So the observer wouldn't be quite in free-fall with respect to Earth, and thus not quite in free-fall with respect to everything else, either. But even then, it's only a correction to a correction, with a magnitude likely to be closer to that of tidal effects across one's body than to the first-order approximation.
 
  • #19
Last edited by a moderator:
  • #20
Redbelly98
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
12,098
129
Just FYI, we like to limit displayed image widths to about 650 pixels or less. Otherwise it messes up the display in the browser window. Either resizing the image, or simply posting the URL, are acceptable.

Thanks and regards,

RB
 
  • #21
Sorry about that, it bothers me when others do that. Alright here is the same image, much reduced:
300px-EarthGravityPREM.jpg


And the accompanying image of density as you go down to the center:
300px-RadialDensityPREM.jpg


I have done the calculation on gravity at the outer core mantle boundary several times myself. I was happy when I found these graphs in a Wiki article about the Structure of the Earth.

ETA: Unfortunately they use different horizontal scales. The gravity graph has the effect as you leave the surface of the Earth and go out into space. The density graph is from the surface down.
 
  • #22
71
0
Thanks,
sorry to say that I'm just a common people so... I can't understand any math formula in this post at all (Wahhhh)

put is simple, gravity increase when we go deep into earth and till reach certain dept It start to decrease and 0 at core ... right?

.............................................


I was just thinking ... dose that mean center of black hole is 0 gravity too?
 
  • #23
115
1
I was just thinking ... dose that mean center of black hole is 0 gravity too?
No it doesn't. Remember, the reason we get zero gravity at the centre of the Earth is because we have an equal attraction from the bodies of mass on all sides, in a black hole however, the mass in concentrated in a singularity with 0 volume and 0 radius, and so you cannot have a point inside it where the grvaitational attraction from all sides balances out.
 

Related Threads for: What's gravity like? at center of the earth

  • Last Post
Replies
1
Views
64K
  • Last Post
Replies
14
Views
5K
Replies
44
Views
44K
Replies
6
Views
7K
Replies
5
Views
6K
Replies
8
Views
13K
Replies
4
Views
2K
Replies
17
Views
8K
Top