<p>This is a companion answer to both Erik's and tfb's answers.</p>

<p>tfb's answer assumes a uniform density throughout the Earth. This is highly inaccurate. Not to disparage tfb's answer; I've run across physics PhDs who thought that this was the correct model of gravitation inside the Earth.</p>

<p>The problem with this model is that the uniform density model is invalid for any object large enough to pull itself into a more or less spherical shape. In the case of our Earth, the mantle is significantly denser than is the crust, and the mantle's density increases with increasing depth due to increasing pressure. The core is much denser than is the mantle, by over a factor of two. The core's density increases with depth, and then jumps even a bit higher at the outer core / inner core boundary.</p>

<p>Erik's answer (which is correct and should be the accepted answer) shows the flaws in this assumption. The uniform density assumption would suggest that gravitational acceleration at the core/mantle boundary should be a bit more than half of the gravitational acceleration at the surface. Erik's answer shows that it is in fact a bit over the gravitational acceleration at the surface. (No halfsies!) Rather than being roughly half the surface acceleration, gravitational acceleration reaches its maximum at the core/mantle boundary.</p>

<p>The graph in Erik's answer shows that in some places (e.g., the Earth's core) gravitational acceleration increases with increasing radial distance from the center of the Earth, but it decreases with increasing radial distance in other places. This answer addresses the conditions that make gravitational acceleration increase or decrease with increasing radial distance from the center of the Earth.</p>

<p>I'll make the same assumption made in the Preliminary Reference Earth Model, which is that density inside the Earth depends only on radial distance from the center of the Earth. This means that Newton's shell theorem applies: The gravitational acceleration at some point inside the Earth depends only on the material closer to the center of the Earth than the point in question. Denoting <span class="math-container">$M(r)$</span> as the mass of all of the material at a radial distance less than <span class="math-container">$r$</span>, then the gravitational acceleration at some point inside the Earth is <span class="math-container">$$g(r) = \frac{G M(r)}{r^2}\tag{1}$$</span>
Differentiating with respect to <span class="math-container">$r$</span> results in <span class="math-container">$$\frac{dg(r)}{dr} = \frac G{r^3}\left(r\frac{dM(r)}{dr} - 2M(r)\right)\tag{2}$$</span>
One way to calculate <span class="math-container">$M(r)$</span> is as a volume integral:<span class="math-container">$$M(r) = \int_0^r 4 \pi x^2 \rho(x) dx$$</span> where <span class="math-container">$\rho(r)$</span> is the local density at a radial distance <span class="math-container">$r$</span> from the center of the Earth. Differentiating with respect to <span class="math-container">$r$</span> results in <span class="math-container">$$\frac{dM(r)}{dr} = 4 \pi r^2 \rho(r)\tag{3}$$</span>
Another way to express <span class="math-container">$M(r)$</span> is via the average density <span class="math-container">$\bar\rho(r)$</span> of all of the material at a radial distance less than <span class="math-container">$r$</span>:<span class="math-container">$$\bar \rho(r) \equiv \frac{M(r)}{V(r)} = \frac {M(r)}{\frac43 \pi r^3}$$</span>
or <span class="math-container">$$M(r) = \frac43 \pi r^3 \bar \rho(r)\tag{4}$$</span>
Applying equations (3) and (4) to (2) yields <span class="math-container">$$\frac{dg(r)}{dr} = 4 \pi G\left(\rho(r) - \frac23\bar\rho(r)\right)\tag{5}$$</span>
In other words, gravitational acceleration increases with increasing distance from the center of a planet in regions where the local density is more than two thirds of the average density of all of the stuff at a lesser distance, but decreases with increasing distance in regions where the local density is less than this limit. In the case of a hypothetical uniform density planet, the local density is equal to the average density and thus gravitational acceleration increases from the center up to the surface.</p>

<p>But in the case of the Earth, there's a very large drop in density at the core-mantle boundary, enough of a drop to make the local density less than the critical two thirds of the average density of all the stuff below. The core mantle boundary is where gravitational acceleration reaches a global maximum. Another marked drop in density occurs at the upper mantle-lower mantle boundary, making gravitational acceleration decrease from there to the surface. The upper mantle-lower mantle boundary is where gravitational acceleration achieves a local (but not global) maximum.</p>

CC BY-SA 4.0
2020-02-09T23:44:40.850Z
2020-02-09T23:44:40.850Z
David Hammen
<p>The below figure, taken from <a href="https://en.wikipedia.org/wiki/Gravity_of_Earth" rel="noreferrer">Wikipedia</a> shows a model of the free fall acceleration, i.e., 'gravity'. The left-most point corresponds to the center of the Earth; then further right at <span class="math-container">$6.3\cdot1000$</span> km you are at the Earth's surface; and then further out you move into space. You can follow the blue line for <a href="http://en.wikipedia.org/wiki/Preliminary_reference_Earth_model" rel="noreferrer">PREM</a> to get an idea of the average (expected) gravity. As you see, the gravity actually increases slightly within the Earth (reaching a maximum at the core-mantle boundary), but tapers down within the core.</p>

<p>To make this kind of calculations, you must think of the Earth like an onion: made up of many concentric spheres. Whenever you move a bit deeper into the Earth, you strip off all the layers you've crossed. As you get closer to the center of the Earth, there are fewer and fewer layers, and eventually, there's nothing left at the center! The reason why gravity goes up ever so slightly within the Earth is that you get close to the much denser core material. If the density of the Earth were constant (per the green 'constant density' line), the gravity would just decrease linearly.</p>

<p><a href="https://i.sstatic.net/Zz4VN.png" rel="noreferrer"><img src="https://i.sstatic.net/Zz4VN.png" alt="Gravity acceleration of Earth"></a></p>

<p>See <a href="https://earthscience.stackexchange.com/a/19140/12155">the other answer</a> and the discussion below for some more details on the math and procedures required to make these calculations.</p>

CC BY-SA 4.0
2020-02-06T09:14:05.130Z
How well-validated is this PREM model?  In other words, how certain are we about the accuracy of the dark blue line?
gerrit
CC BY-SA 4.0
2020-02-06T10:39:05.783Z
@gerrit PREM stands for https://en.wikipedia.org/wiki/Preliminary_reference_Earth_model , it is the most commonly used geophysical model of the Earth. Of course, it's simplified (it's 1-D spherically symmetric), but it matches all known Earth properties very well. See the introduction (or the whole document) here: http://www.geo.uu.nl/~berg/geodynamics/lecturenotes.pdf for more details.
Erik
CC BY-SA 4.0
2020-02-06T17:09:06.110Z
"Whenever you move a bit deeper into the Earth, you strip off all the layers you've crossed." This implies that if you are anywhere inside of a hollow ball, the effect of gravity on you from the ball is always zero, irregardless of the thickness of the material that the ball is made of. Is this true? Would there be zero gravity from the ball, even if you were touching the inside of it?
John Pankowicz
CC BY-SA 4.0
2020-02-06T17:24:48.470Z
@JohnPankowicz Yep! I think it's a pretty cool result! See: [Shell Theorem](https://en.wikipedia.org/wiki/Shell_theorem#Inside_a_shell)
Mark Omo
CC BY-SA 4.0
2020-02-06T19:00:57.933Z
I always thought it was interesting that you can enjoy a zero-G environment either by going up (into orbit), or by going down (into the center of the earth).
Nuclear Hoagie
CC BY-SA 4.0
2020-02-06T19:49:14.933Z
@NuclearWang I'm not sure "enjoy" would be the right word here - unless you take pleasure from being crushed inside a molten ball of nickel-iron alloy...
Darrel Hoffman
CC BY-SA 4.0
2020-02-06T22:24:22.740Z
@DarrelHoffman zero-g and under immense pressure? Sounds like a great way to relieve my back pain. Time to fire my physical therapist and take a trip to the center of the earth...
asgallant
CC BY-SA 4.0
2020-02-07T00:25:30.777Z
"Whenever you move a bit deeper into the Earth, you strip off all the layers you've crossed" -- I think the question was made with the assumption that they wouldn't be stripped off, as if the person falling could pass through the mass but still be affected by its gravity, or is going through a tunnel that slowly spirals into the center of the earth. The question body says "Or does the mantle and crust above you counteract". I imagine taking that into consideration would change the shape of the blue line quite drastically.
JoL
CC BY-SA 4.0
2020-02-07T02:33:06.400Z
@JoL Erik's statement isn't implying that the model is ignoring the gravity of the spherical layers above you. Rather, gravity (and electromagnetism) has the property that the pull in all directions from the spherical layers above you exactly cancels out, so these layers contribute nothing to the force you feel! See tfb's answer or Mark Omo's link to the Shell Theorem above. The blue line incorporates the pull from all of the Earth's mass; it's just that only the mass below you has a nonzero effect.
jawheele
CC BY-SA 4.0
2020-02-07T04:11:22.257Z
@JoL indeed I meant it the way that jawheele clarifies it. It's a mathematical coincidence that it makes no difference whether we take the tunnel that you describe, or literally strip the Earth's layers! Say we take the tunnel approach, and are at the core-mantle boundary. Then the bit of mantle above you 'pulls you up' exactly as much as all the remaining mantle (on the other side of the Earth) 'pulls you down': https://qph.fs.quoracdn.net/main-qimg-d7001dd9dfefa24681878ac25011bcbb The arrows sum to a net zero effect! So you can ignore all the (spherical) layers that you have crossed!
Erik
CC BY-SA 4.0
2020-02-07T12:53:38.173Z
@asgallant Indeed I can assure you that if you went to the core, you'd feel no back pain anymore.
JiK
CC BY-SA 4.0
2020-02-08T16:59:39.580Z
I am not sure I interpret the image correctly; but if yes, then the image is false. The gravitational acceleration start with 9.81m/s^2 on the surface, and not from 0. Then it goes up until 10.3m/s^2, and then decreases to zero in the center, that it already okay.
peterh
CC BY-SA 4.0
2020-02-09T00:09:46.460Z
@peterh-ReinstateMonica: I think you must be misinterpreting the image, because it shows the same thing that you say is the case. (The vertical gray separator at about 6.4 × 1000 km represents the surface of the Earth; everything to the left is below the surface, everything to the right is above the surface.)
ruakh
CC BY-SA 4.0
2020-02-09T00:39:32.633Z
@ruakh Right! I misinterpreted. I tought the horizontal axe shows the distance from the surface, and not from the core.
peterh
CC BY-SA 4.0
2020-02-06T08:23:04.990Z
2020-02-08T20:24:57.110Z
2020-02-08T20:24:57.110Z
Erik
<p>Or does the mantle and crust above you counteract the increase at one point and it actually decreases?</p>

CC BY-SA 4.0
2020-02-07T17:43:03.577Z
[What matters is the mass beneath you](https://en.wikipedia.org/wiki/Gauss%27s_law_for_gravity); the mass above neither increases nor reduces the local gravity.
J.G.
CC BY-SA 4.0
2021-07-27T08:09:24.993Z
That doesn't make any sense. Any nearby mass has an effect on local gravity. That's why the moon, when it is above you has an effect, too.
stippi
<geology><geophysics><earth-observation><gravity>
Does gravity increase the closer to the core you get?
HingedHD
# Does gravity increase the closer to the core you get?

(Asked by: username_0 on 2020-02-06)

Or does the mantle and crust above you counteract the increase at one point and it actually decreases?

**Tags:** geology, geophysics, earth-observation, gravity

> **Comments:**
> **username_1**: [What matters is the mass beneath you](https://en.wikipedia.org/wiki/Gauss%27s_law_for_gravity); the mass above neither increases nor reduces the local gravity.  
> **username_2**: That doesn't make any sense. Any nearby mass has an effect on local gravity. That's why the moon, when it is above you has an effect, too.  

---

## 2 answers

### Answer by username_4 on 2020-02-06. Score: 54 | Accepted answer

The below figure, taken from [Wikipedia](https://en.wikipedia.org/wiki/Gravity_of_Earth) shows a model of the free fall acceleration, i.e., 'gravity'. The left-most point corresponds to the center of the Earth; then further right at $6.3\cdot1000$ km you are at the Earth's surface; and then further out you move into space. You can follow the blue line for [PREM](http://en.wikipedia.org/wiki/Preliminary_reference_Earth_model) to get an idea of the average (expected) gravity. As you see, the gravity actually increases slightly within the Earth (reaching a maximum at the core-mantle boundary), but tapers down within the core.

To make this kind of calculations, you must think of the Earth like an onion: made up of many concentric spheres. Whenever you move a bit deeper into the Earth, you strip off all the layers you've crossed. As you get closer to the center of the Earth, there are fewer and fewer layers, and eventually, there's nothing left at the center! The reason why gravity goes up ever so slightly within the Earth is that you get close to the much denser core material. If the density of the Earth were constant (per the green 'constant density' line), the gravity would just decrease linearly.

[![Gravity acceleration of Earth](https://i.sstatic.net/Zz4VN.png)](https://i.sstatic.net/Zz4VN.png)

See [the other answer](https://earthscience.stackexchange.com/a/19140/12155) and the discussion below for some more details on the math and procedures required to make these calculations.

> **Comments:**
> **username_5**: How well-validated is this PREM model?  In other words, how certain are we about the accuracy of the dark blue line?  
> **username_4**: @username_5 PREM stands for https://en.wikipedia.org/wiki/Preliminary_reference_Earth_model , it is the most commonly used geophysical model of the Earth. Of course, it's simplified (it's 1-D spherically symmetric), but it matches all known Earth properties very well. See the introduction (or the whole document) here: http://www.geo.uu.nl/~berg/geodynamics/lecturenotes.pdf for more details.  
> **username_6**: "Whenever you move a bit deeper into the Earth, you strip off all the layers you've crossed." This implies that if you are anywhere inside of a hollow ball, the effect of gravity on you from the ball is always zero, irregardless of the thickness of the material that the ball is made of. Is this true? Would there be zero gravity from the ball, even if you were touching the inside of it?  
> **username_7**: @username_6 Yep! I think it's a pretty cool result! See: [Shell Theorem](https://en.wikipedia.org/wiki/Shell_theorem#Inside_a_shell)  
> **username_8**: I always thought it was interesting that you can enjoy a zero-G environment either by going up (into orbit), or by going down (into the center of the earth).  
> **username_9**: @NuclearWang I'm not sure "enjoy" would be the right word here - unless you take pleasure from being crushed inside a molten ball of nickel-iron alloy...  
> **username_10**: @username_9 zero-g and under immense pressure? Sounds like a great way to relieve my back pain. Time to fire my physical therapist and take a trip to the center of the earth...  
> **username_11**: "Whenever you move a bit deeper into the Earth, you strip off all the layers you've crossed" -- I think the question was made with the assumption that they wouldn't be stripped off, as if the person falling could pass through the mass but still be affected by its gravity, or is going through a tunnel that slowly spirals into the center of the earth. The question body says "Or does the mantle and crust above you counteract". I imagine taking that into consideration would change the shape of the blue line quite drastically.  
> **username_12**: @username_11 username_4's statement isn't implying that the model is ignoring the gravity of the spherical layers above you. Rather, gravity (and electromagnetism) has the property that the pull in all directions from the spherical layers above you exactly cancels out, so these layers contribute nothing to the force you feel! See tfb's answer or username_7's link to the Shell Theorem above. The blue line incorporates the pull from all of the Earth's mass; it's just that only the mass below you has a nonzero effect.  
> **username_4**: @username_11 indeed I meant it the way that username_12 clarifies it. It's a mathematical coincidence that it makes no difference whether we take the tunnel that you describe, or literally strip the Earth's layers! Say we take the tunnel approach, and are at the core-mantle boundary. Then the bit of mantle above you 'pulls you up' exactly as much as all the remaining mantle (on the other side of the Earth) 'pulls you down': https://qph.fs.quoracdn.net/main-qimg-d7001dd9dfefa24681878ac25011bcbb The arrows sum to a net zero effect! So you can ignore all the (spherical) layers that you have crossed!  
> **username_13**: @username_10 Indeed I can assure you that if you went to the core, you'd feel no back pain anymore.  
> **username_14**: I am not sure I interpret the image correctly; but if yes, then the image is false. The gravitational acceleration start with 9.81m/s^2 on the surface, and not from 0. Then it goes up until 10.3m/s^2, and then decreases to zero in the center, that it already okay.  
> **username_15**: @username_14-ReinstateMonica: I think you must be misinterpreting the image, because it shows the same thing that you say is the case. (The vertical gray separator at about 6.4 × 1000 km represents the surface of the Earth; everything to the left is below the surface, everything to the right is above the surface.)  
> **username_14**: @username_15 Right! I misinterpreted. I tought the horizontal axe shows the distance from the surface, and not from the core.  

---

### Answer by username_3 on 2020-02-09. Score: 6

This is a companion answer to both username_4's and tfb's answers.

tfb's answer assumes a uniform density throughout the Earth. This is highly inaccurate. Not to disparage tfb's answer; I've run across physics PhDs who thought that this was the correct model of gravitation inside the Earth.

The problem with this model is that the uniform density model is invalid for any object large enough to pull itself into a more or less spherical shape. In the case of our Earth, the mantle is significantly denser than is the crust, and the mantle's density increases with increasing depth due to increasing pressure. The core is much denser than is the mantle, by over a factor of two. The core's density increases with depth, and then jumps even a bit higher at the outer core / inner core boundary.

username_4's answer (which is correct and should be the accepted answer) shows the flaws in this assumption. The uniform density assumption would suggest that gravitational acceleration at the core/mantle boundary should be a bit more than half of the gravitational acceleration at the surface. username_4's answer shows that it is in fact a bit over the gravitational acceleration at the surface. (No halfsies!) Rather than being roughly half the surface acceleration, gravitational acceleration reaches its maximum at the core/mantle boundary.

The graph in username_4's answer shows that in some places (e.g., the Earth's core) gravitational acceleration increases with increasing radial distance from the center of the Earth, but it decreases with increasing radial distance in other places. This answer addresses the conditions that make gravitational acceleration increase or decrease with increasing radial distance from the center of the Earth.

I'll make the same assumption made in the Preliminary Reference Earth Model, which is that density inside the Earth depends only on radial distance from the center of the Earth. This means that Newton's shell theorem applies: The gravitational acceleration at some point inside the Earth depends only on the material closer to the center of the Earth than the point in question. Denoting $M(r)$ as the mass of all of the material at a radial distance less than $r$, then the gravitational acceleration at some point inside the Earth is $$g(r) = \frac{G M(r)}{r^2}\tag{1}$$
Differentiating with respect to $r$ results in $$\frac{dg(r)}{dr} = \frac G{r^3}\left(r\frac{dM(r)}{dr} - 2M(r)\right)\tag{2}$$
One way to calculate $M(r)$ is as a volume integral:$$M(r) = \int\_0^r 4 \pi x^2 \rho(x) dx$$ where $\rho(r)$ is the local density at a radial distance $r$ from the center of the Earth. Differentiating with respect to $r$ results in $$\frac{dM(r)}{dr} = 4 \pi r^2 \rho(r)\tag{3}$$
Another way to express $M(r)$ is via the average density $\bar\rho(r)$ of all of the material at a radial distance less than $r$:$$\bar \rho(r) \equiv \frac{M(r)}{V(r)} = \frac {M(r)}{\frac43 \pi r^3}$$
or $$M(r) = \frac43 \pi r^3 \bar \rho(r)\tag{4}$$
Applying equations (3) and (4) to (2) yields $$\frac{dg(r)}{dr} = 4 \pi G\left(\rho(r) - \frac23\bar\rho(r)\right)\tag{5}$$
In other words, gravitational acceleration increases with increasing distance from the center of a planet in regions where the local density is more than two thirds of the average density of all of the stuff at a lesser distance, but decreases with increasing distance in regions where the local density is less than this limit. In the case of a hypothetical uniform density planet, the local density is equal to the average density and thus gravitational acceleration increases from the center up to the surface.

But in the case of the Earth, there's a very large drop in density at the core-mantle boundary, enough of a drop to make the local density less than the critical two thirds of the average density of all the stuff below. The core mantle boundary is where gravitational acceleration reaches a global maximum. Another marked drop in density occurs at the upper mantle-lower mantle boundary, making gravitational acceleration decrease from there to the surface. The upper mantle-lower mantle boundary is where gravitational acceleration achieves a local (but not global) maximum.


---

