<p>The short answer because Jupiter is a gas giant, so it's kind of got a very large atmosphere and atmosphere's aren't very dense.   Also, if you look at your chart, the gravitation inside the Earth increases until you get to the outer core.  This is likely to be much more pronounced on gaseous bodies like gas giants and stars.</p>

<p>Longer answer:</p>

<p>In layman's terms, if we use <a href="http://hyperphysics.phy-astr.gsu.edu/hbase/Mechanics/sphshell2.html" rel="noreferrer">Newton's shell theorem</a>, you can ignore the mass "above" you because the shell of matter with a greater distance from the central point than you has a close to zero gravitational effect on you.   Technically that's a shell of mass around you but I'm going to call it "above" cause it's easier to say.</p>

<p>As a result, looking at the gravitational field, you can just take into account the mass below you and the radius and ignore the shell above you.   </p>

<p>If, for example, you tunnel 10% into a planet, using the law of cubes, 72.9% of the planet's volume is below you, but you're 11.1% closer to the center, using the inverse square law, that's 23.4% greater gravitational pull from the 72.9% of planet that remains below you than you'd get from that 72.9% if you were on the planet's surface.</p>

<p>If the 72.9% of the planet below you weights more than 81% of the planet's mass, then gravity increases.   Lets say it weighs exactly 81%.    81% of the mass x 1.234 greater pull at 10% closer works out exactly the same.   In other words, if the mass above you is light enough, then gravity increases as you tunnel or fall inside a planet.    There's likely a pretty simple logarithmic ration between density ratio and where the gravity stops increasing.   If I can work it out, I'll post it.</p>

<p>In the case of Jupiter and starting near it's equator, with Jupiter's rapid rotation, that should be taken into account as well.  In the case of Earth, Earth's rotation is pretty negligible relative to it's gravity and can be ignored unless you want high accuracy.</p>

<p>With planetary bodies, density can play a bigger role than mass regarding surface gravity.   Mercury, for example, <a href="https://nssdc.gsfc.nasa.gov/planetary/factsheet/planet_table_ratio.html" rel="noreferrer">is about 52% the mass of Mars, but it 38% denser</a>, which enables it to have a slightly higher gravitational pull on it's surface than Mars.</p>

<p>Two people have pointed out that they're not sure the gravity really does increase inside Jupiter.   I'm sure it does because the density deep inside Jupiter likely increases significantly.   We can't get a good look at the inside of Jupiter, so precise numbers are impossible, but it seems very probably to me that the gravitation increases much of the way inside Jupiter, only beginning to decrease when the core reaches considerable density.   </p>

<p>The majority of Jupiter's external layers is hydrogen and hydrogen, even under very high pressure, is not very dense.   At 700 atm, for example, and Earth temperature, not hot temperature deep inside Jupiter, <a href="https://energies.airliquide.com/resources-planet-hydrogen/how-hydrogen-stored" rel="noreferrer">Hydrogen still has a density of less than 1/10th the density of water</a>.    The mass of Jupiter's outer layers are almost certainly too low to have as much gravitational effect as the denser inner parts when you take into account falling into the planet. </p>

<p>Planets like Uranus or Neptune, which much less hydrogen and helium - probably not as much, but for gas giants and most stars, gravity very likely increases considerably for a considerable percentage of their radius, for an object falling inside of them.</p>

CC BY-SA 4.0
2018-12-26T20:37:53.110Z
2018-12-26T20:51:09.283Z
2018-12-26T20:51:09.283Z
userLTK
<p>No idea whether that's true or not for Jupiter. However, it is theoretically possible if the outer layers of the planet are much less dense than the stuff underneath. In that case the shell theorem applied to the outer layers does not remove much gravity, while the gravity from the inner parts increases fast due to the inverse square law.</p>

<p>Even for the image you've posted, gravity increases in the lower half of the lower mantle, close to the outer core - and for the same reason. Stuff in the outer layers is "fluffy", stuff in the core is dense.</p>

CC BY-SA 4.0
2018-12-26T20:10:40.863Z
2018-12-26T20:10:40.863Z
Florin Andrei
<p>You can use <a href="https://en.wikipedia.org/wiki/Gauss%27s_law_for_gravity#Qualitative_statement_of_the_law" rel="noreferrer">Gauss's law for gravitation</a> to work out the gravity as a function of (interior) radius.
<span class="math-container">$$ \oint \vec{g} \cdot d\vec{A} = -4\pi G \int \rho\ dV\ .$$</span>
What this means is that the flux of gravitational field <span class="math-container">$\vec{g}$</span> out of a closed surface is proportional to the mass enclosed within that surface.</p>

<p>Let us assume that the density is a function of (interior) radius, such that <span class="math-container">$\rho = Ar^{\alpha}$</span> and that the planet is spherically sysmmetric, so that the left hand side becomes <span class="math-container">$4 \pi r^2 g(r)$</span> and the volume element <span class="math-container">$dV = 4\pi r^2\ dr$</span>. Then
<span class="math-container">$$ 4 \pi r^2 g(r) = -4\pi G \int_{r=0}^{r} Ar^{\alpha} 4\pi r^2\ dr$$</span>
as long as <span class="math-container">$\alpha \neq -3$</span>.
<span class="math-container">$$ r^2 g(r) = - 4\pi G \frac{Ar^{3+\alpha}}{3+\alpha}$$</span>
<span class="math-container">$$ g(r) = -\frac{4\pi GA }{3+\alpha} r^{1+\alpha}$$</span>
What this shows, is that if <span class="math-container">$\alpha &lt; -1$</span>, then gravity will increase with decreasing radius.</p>

<p>Now we are in a position to answer your question. In the Earth's deep interior <span class="math-container">$\alpha \simeq 0$</span> and the density does not increase greatly as we head towards the centre. This means that <span class="math-container">$g(r) \propto r$</span> as is seen in the plot you reference. In the outer part of the crust <span class="math-container">$\alpha \simeq -1$</span> and <span class="math-container">$g(r)$</span> is roughly constant.</p>

<p>Below is a model density profile for Jupiter (solid line).
<a href="https://i.sstatic.net/M7kRG.jpg" rel="noreferrer"><img src="https://i.sstatic.net/M7kRG.jpg" alt="Jupiter density profile"></a></p>

<p>Just looking at this I would say that <span class="math-container">$\alpha$</span> is just a bit <span class="math-container">$&lt;-1$</span> for most of the interior and so the gravity should be <em>increasing</em> slowly as <span class="math-container">$r$</span> decreases, but there is a sharp uptick in density at the boundary to the core which would see <span class="math-container">$\alpha \ll -1$</span> and the gravity becoming proportional to <span class="math-container">$r$</span> to the power of a negative number (i.e. increasing strongly with decreasing <span class="math-container">$r$</span>), but then <em>must</em> drop to <span class="math-container">$g=0$</span> when <span class="math-container">$r=0$</span> shortly afterwards.</p>

CC BY-SA 4.0
2019-06-22T19:36:14.503Z
Could you tell me the reference for the figure?
Youngsub Yoon
CC BY-SA 4.0
2021-03-04T15:45:27.003Z
@YoungsubYoon It is Fig. 5 of "Giant Planet Formation, Evolution, and Internal Structure" (pre-print https://arxiv.org/abs/1311.1142)  which is a chapter of the book *Protostars and Planets VI* https://muse.jhu.edu/book/35621
DavePhD
CC BY-SA 4.0
2018-12-26T20:18:21.560Z
2018-12-26T20:53:02.047Z
2018-12-26T20:53:02.047Z
ProfRob
<p>The gravitational acceleration, g, inside the Earth generally decreases with decreasing distance to the center:</p>

<p><a href="https://i.sstatic.net/QzqEi.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/QzqEi.jpg" alt="Gravity Acceleration of Earth"></a></p>

<p>However, apparently for Jupiter, the gravitational acceleration only increases with decreasing distance to its center. Why is this?</p>

CC BY-SA 4.0
2018-12-26T19:48:54.447Z
please post the source of your Jovian claim.
Carl Witthoft
CC BY-SA 4.0
2018-12-26T19:49:53.647Z
Similar question here:   https://physics.stackexchange.com/questions/18446/how-does-gravity-work-underground
userLTK
CC BY-SA 4.0
2018-12-26T19:50:38.933Z
Possible duplicate of [Is Jupiter made entirely out of gas?](https://astronomy.stackexchange.com/questions/8251/is-jupiter-made-entirely-out-of-gas)   not an exact duplicate but there is discussion of density vs. radius there
Carl Witthoft
CC BY-SA 4.0
2018-12-26T21:38:33.660Z
@CarlWitthoft I think if it's not an exact dupe it's best to provide the link (as you've done) but not flag for closure - although I'm not saying the VTC was necessarily you. I'm voting to leave the question open.
Chappo Hasn't Forgotten
CC BY-SA 4.0
2018-12-26T21:44:45.987Z
Trivially, the claim that gravitational acceleration only increases as distance to the center of Jupiter decreases is false. The acceleration due to gravity is positive wherever you care to define a "surface" of Jupiter, and 0 at the center, so there must be a net decrease over that distance.
asgallant
CC BY-SA 4.0
2018-12-27T09:38:32.957Z
The answer here and physics SE  clarify the situation. However put it simple that can happens on Earth too considering entering the Earth at atmospheric altitude. Somehow and with exaggeration what happens with gas giants.
Alchimista
CC BY-SA 4.0
2018-12-27T12:54:08.807Z
@immibis All datasets go from 9.8 at the surface to 0 at the center.
Jason C
<gravity><earth><jupiter><newtonian-gravity>
Gravitational acceleration inside a planet
user4437416
# Gravitational acceleration inside a planet

(Asked by: username_0 on 2018-12-26)

The gravitational acceleration, g, inside the Earth generally decreases with decreasing distance to the center:

[![Gravity Acceleration of Earth](https://i.sstatic.net/QzqEi.jpg)](https://i.sstatic.net/QzqEi.jpg)

However, apparently for Jupiter, the gravitational acceleration only increases with decreasing distance to its center. Why is this?

**Tags:** gravity, earth, jupiter, newtonian-gravity
> **Comments:**
> **username_1**: please post the source of your Jovian claim.  
> **username_2**: Similar question here:   https://physics.stackexchange.com/questions/18446/how-does-gravity-work-underground  
> **username_1**: Possible duplicate of [Is Jupiter made entirely out of gas?](https://astronomy.stackexchange.com/questions/8251/is-jupiter-made-entirely-out-of-gas)   not an exact duplicate but there is discussion of density vs. radius there  
> **username_3**: @CarlWitthoft I think if it's not an exact dupe it's best to provide the link (as you've done) but not flag for closure - although I'm not saying the VTC was necessarily you. I'm voting to leave the question open.  
> **username_4**: Trivially, the claim that gravitational acceleration only increases as distance to the center of Jupiter decreases is false. The acceleration due to gravity is positive wherever you care to define a "surface" of Jupiter, and 0 at the center, so there must be a net decrease over that distance.  
> **username_5**: The answer here and physics SE  clarify the situation. However put it simple that can happens on Earth too considering entering the Earth at atmospheric altitude. Somehow and with exaggeration what happens with gas giants.  
> **username_6**: @immibis All datasets go from 9.8 at the surface to 0 at the center.  

---

## 3 answers

### Answer by username_8 on 2018-12-26. Score: 10 | Accepted answer

You can use [Gauss's law for gravitation](https://en.wikipedia.org/wiki/Gauss%27s_law_for_gravity#Qualitative_statement_of_the_law) to work out the gravity as a function of (interior) radius.
$$ \oint \vec{g} \cdot d\vec{A} = -4\pi G \int \rho\ dV\ .$$
What this means is that the flux of gravitational field $\vec{g}$ out of a closed surface is proportional to the mass enclosed within that surface.

Let us assume that the density is a function of (interior) radius, such that $\rho = Ar^{\alpha}$ and that the planet is spherically sysmmetric, so that the left hand side becomes $4 \pi r^2 g(r)$ and the volume element $dV = 4\pi r^2\ dr$. Then
$$ 4 \pi r^2 g(r) = -4\pi G \int\_{r=0}^{r} Ar^{\alpha} 4\pi r^2\ dr$$
as long as $\alpha \neq -3$.
$$ r^2 g(r) = - 4\pi G \frac{Ar^{3+\alpha}}{3+\alpha}$$
$$ g(r) = -\frac{4\pi GA }{3+\alpha} r^{1+\alpha}$$
What this shows, is that if $\alpha < -1$, then gravity will increase with decreasing radius.

Now we are in a position to answer your question. In the Earth's deep interior $\alpha \simeq 0$ and the density does not increase greatly as we head towards the centre. This means that $g(r) \propto r$ as is seen in the plot you reference. In the outer part of the crust $\alpha \simeq -1$ and $g(r)$ is roughly constant.

Below is a model density profile for Jupiter (solid line).
[![Jupiter density profile](https://i.sstatic.net/M7kRG.jpg)](https://i.sstatic.net/M7kRG.jpg)

Just looking at this I would say that $\alpha$ is just a bit $<-1$ for most of the interior and so the gravity should be *increasing* slowly as $r$ decreases, but there is a sharp uptick in density at the boundary to the core which would see $\alpha \ll -1$ and the gravity becoming proportional to $r$ to the power of a negative number (i.e. increasing strongly with decreasing $r$), but then *must* drop to $g=0$ when $r=0$ shortly afterwards.

> **Comments:**
> **username_9**: Could you tell me the reference for the figure?  
> **username_10**: @YoungsubYoon It is Fig. 5 of "Giant Planet Formation, Evolution, and Internal Structure" (pre-print https://arxiv.org/abs/1311.1142)  which is a chapter of the book *Protostars and Planets VI* https://muse.jhu.edu/book/35621  

---

### Answer by username_2 on 2018-12-26. Score: 5

The short answer because Jupiter is a gas giant, so it's kind of got a very large atmosphere and atmosphere's aren't very dense. Also, if you look at your chart, the gravitation inside the Earth increases until you get to the outer core. This is likely to be much more pronounced on gaseous bodies like gas giants and stars.

Longer answer:

In layman's terms, if we use [Newton's shell theorem](http://hyperphysics.phy-astr.gsu.edu/hbase/Mechanics/sphshell2.html), you can ignore the mass "above" you because the shell of matter with a greater distance from the central point than you has a close to zero gravitational effect on you. Technically that's a shell of mass around you but I'm going to call it "above" cause it's easier to say.

As a result, looking at the gravitational field, you can just take into account the mass below you and the radius and ignore the shell above you.

If, for example, you tunnel 10% into a planet, using the law of cubes, 72.9% of the planet's volume is below you, but you're 11.1% closer to the center, using the inverse square law, that's 23.4% greater gravitational pull from the 72.9% of planet that remains below you than you'd get from that 72.9% if you were on the planet's surface.

If the 72.9% of the planet below you weights more than 81% of the planet's mass, then gravity increases. Lets say it weighs exactly 81%. 81% of the mass x 1.234 greater pull at 10% closer works out exactly the same. In other words, if the mass above you is light enough, then gravity increases as you tunnel or fall inside a planet. There's likely a pretty simple logarithmic ration between density ratio and where the gravity stops increasing. If I can work it out, I'll post it.

In the case of Jupiter and starting near it's equator, with Jupiter's rapid rotation, that should be taken into account as well. In the case of Earth, Earth's rotation is pretty negligible relative to it's gravity and can be ignored unless you want high accuracy.

With planetary bodies, density can play a bigger role than mass regarding surface gravity. Mercury, for example, [is about 52% the mass of Mars, but it 38% denser](https://nssdc.gsfc.nasa.gov/planetary/factsheet/planet_table_ratio.html), which enables it to have a slightly higher gravitational pull on it's surface than Mars.

Two people have pointed out that they're not sure the gravity really does increase inside Jupiter. I'm sure it does because the density deep inside Jupiter likely increases significantly. We can't get a good look at the inside of Jupiter, so precise numbers are impossible, but it seems very probably to me that the gravitation increases much of the way inside Jupiter, only beginning to decrease when the core reaches considerable density.

The majority of Jupiter's external layers is hydrogen and hydrogen, even under very high pressure, is not very dense. At 700 atm, for example, and Earth temperature, not hot temperature deep inside Jupiter, [Hydrogen still has a density of less than 1/10th the density of water](https://energies.airliquide.com/resources-planet-hydrogen/how-hydrogen-stored). The mass of Jupiter's outer layers are almost certainly too low to have as much gravitational effect as the denser inner parts when you take into account falling into the planet.

Planets like Uranus or Neptune, which much less hydrogen and helium - probably not as much, but for gas giants and most stars, gravity very likely increases considerably for a considerable percentage of their radius, for an object falling inside of them.


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### Answer by username_7 on 2018-12-26. Score: 2

No idea whether that's true or not for Jupiter. However, it is theoretically possible if the outer layers of the planet are much less dense than the stuff underneath. In that case the shell theorem applied to the outer layers does not remove much gravity, while the gravity from the inner parts increases fast due to the inverse square law.

Even for the image you've posted, gravity increases in the lower half of the lower mantle, close to the outer core - and for the same reason. Stuff in the outer layers is "fluffy", stuff in the core is dense.


---

