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The Earth's gravitational pull in its interior looks like this:

Gravitational acceleration due to the earth as a function of radial distance from the center of the Earth

The pull remains about the same and increases even a bit till the outer core from where on it starts getting weaker till 0g in the core's very center. My question is what such a model looks like in a gas giant.

The four gaseous planets don't have solid surfaces (except for their unreachable cores perhaps) and therefore may be explored by airborne craft, including crewed airships and someday maybe cloud cities. Such a crewed airship plan is the HAVOC for Venus. The airship would float at about 33 mi (53 km) above the equator where the gravity is 8.73 m/s² or 0.89g compared to the 0.905g on the surface.

Now let's imagine a similar scenario on Uranus. On the level of Earth-like air pressure, Uranus has an equatorial gravity of 8.69 m/s² (0.886 g). If one went deeper, would gravity increase to eventually 1 g? If so, at what depth/distance from the planet's center would that be? The same question might apply to Saturn which has an equatorial gravity of 0.93 g but it might be hard to go around the rings. On Neptune, the opposite might be the case: Neptune's equatorial gravity is 1.122 g but if you went higher, would you reach a level of 1 g still within Neptune's atmosphere?

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  • $\begingroup$ Thank you for the improvement, Mr Hammen, but why wasn't I asked to confirm or reject the edit? $\endgroup$ – LoveForChrist Jun 27 at 17:55
  • $\begingroup$ @ LoveForChrist -- I have enough points that any edits I make are automatically approved. As this is a beta site, only 4000 reputation points are needed to attain the level where edits are automatically approved. To me, this means that I need to be very, very careful when making edits. Even if I didn't have that many points, anyone with 100 reputation points could have participated in approving / rejecting my edit. Author consultation is not required. Stack Exchange is a community effort. The author however can always revert edits. $\endgroup$ – David Hammen Jun 27 at 18:28
  • $\begingroup$ One reason author approval is not required is the problem of drive-by posters (you obviously are not in this class). Sometimes people post interesting but flawed questions, and then never come back. The flaws could never be addressed if author approval was required, As a community resource, the original author of a question or an answer has limited rights. The questions and answers belong to the community as a whole rather than to the original author. I sometimes make mistakes in my answers, and I am almost always grateful when those answers are edited without my approval. $\endgroup$ – David Hammen Jun 27 at 18:40
  • $\begingroup$ @DavidHammen Me too, I just was surprised that I'm discovering my question edit "by accident". Perhaps there should be some notation in the upper right corner saying "an edit was done to your question x". Again, thanks for the improvement of my question body, Mr Hammen. $\endgroup$ – LoveForChrist Jun 28 at 4:56
  • $\begingroup$ I suggest removing the balloons tag, for two reasons. One is that balloons won't work on the giant planets, which is the primary subject of this question. The atmospheres of the giant planets are mostly hydrogen and helium, pretty much ruling out lighter than air balloons filled with hydrogen or helium. A hot air balloon might work, but how to fill a Jovian balloon with hot air is problematic. The other is a stackexchange glitch, which ends up tagging the title of the question as balloons rather than gravity. $\endgroup$ – David Hammen Jul 6 at 6:07
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I'll make a simplifying assumptions that the giant planets are very close to spherical and that the density inside these planets depends only on radial distance from the center of the planet. (This is not quite correct as the giant planets rotate rather quickly, making the planets oblate spheroids rather than spheres. But the differences are small.)

These assumptions mean Newton's shell theorem applies and that the gravitational acceleration inside the planet at some distance $r$ from the center of the planet is $$g(r) = \frac{GM(r)}{r^2}\tag{1}$$ where $g(r)$ is the gravitational acceleration at a distance $r$ from the center of the planet, $G$ is the Newtonian gravitational constant, and $M(r)$ is the mass of all the stuff closer to the center of planet than is the point of interest.

Differentiating equation (1) with respect to radial distance yields $$\frac{dg(r)}{dr} = G\left(\frac1{r^2}\frac{dM(r)}{dr} - \frac2{r^3}M(r)\right)\tag{2}$$

Expressions for $M(r)$ are needed to make sense of equation (2). One such expression uses the average density of all of the material at a radial distance less than or equal to $r$: $$\bar{\rho}(r) \equiv \frac{M(r)}{\frac43\pi r^3} \quad\implies\quad \frac2{r^3} M(r) = \frac83\pi \bar{\rho}(r)\tag{3}$$ Another expression for $M(r)$ results from integrating local density $\rho(r)$ from the center to the radius $r$: $$M(r) = \int_0^r 4\pi x^2 \rho(x)\, dx \quad\implies\quad \frac1{r^2}\frac{dM(r)}{dr} = 4\pi \rho(r)\tag{4}$$

Applying equations (3) and (4) to equation (2) results in $$\frac{dg(r)}{dr} = 4\pi G\left(\rho(r) - \frac23 \bar{\rho}(r)\right)\tag{5}$$

The sign of the term $\rho(r) - \frac23 \bar{\rho}(r)$ dictates whether gravitation inside a planet is locally increasing or decreasing. In rocky planets with a dense metallic core surrounded by a less dense rocky mantle, the marked density transition at the core-mantle boundary makes $dg/dr$ change from positive to negative at that boundary. In the Earth, the core-mantle boundary is the place where gravitation reaches its maximum.

What about the giant planets? At the very center, the local density and the average density are one and the same ($\rho(0) = \bar{\rho}(0)$), so gravitational acceleration initially increases with increasing distance from the center. (This only makes sense; gravitational acceleration is zero at the center; it can only go up.) At the other extreme, where the tenuous upper atmosphere gives way to space, the local density $\rho(r)$ is nearly zero while the average density $\bar{\rho}(r)$ is non-zero -- much more than 1.5 times the local density. Gravitational acceleration drops off at $GM/r^2$ beyond this point.

Somewhere in between there exists a global maximum, and its beyond ludicrous to think that this magically occurs at the one bar level that we have arbitrarily defined as the "surface". It almost certainly occurs well below the one bar level as the density is rather low at that level, much less than the average density of the giant planets (which is on the order of density of liquid water).

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  • $\begingroup$ Thank you. Therefore, you do assume that Uranus has 1g deeper beneath its "surface", right? $\endgroup$ – LoveForChrist Jul 6 at 6:30
  • $\begingroup$ As an example of realistic modeling, see arxiv.org/abs/1707.01997 . The models of Jupiter seem pretty believable, because they are tightly constrained by the Juno spacecraft's measurements of moments of the field. People making models mostly seem to give graphs and tables involving density and pressure, since those are the variables in the equation of state. Anyone who was interested enough could integrate a $\rho(r)$ tabulation to find $g(r)$, as described in your equations. $\endgroup$ – Ben Crowell Jul 6 at 14:37
  • $\begingroup$ The derivation presented here relies on a strange trick: Eqn. (2) should reproduce the Poisson-equation without the second term on the r.h.s. After all, Eqn. (1) results from integrating the Poisson equation radially. This back-and-forth then produces the extra term. This seems inconsistent to me. $\endgroup$ – AtmosphericPrisonEscape Jul 6 at 15:44
  • $\begingroup$ @AtmosphericPrisonEscape - Equation (2) should have been $\frac{dg(r)}{dr}$ rather than $\frac{g(r)}{dr}$. The latter makes no sense; the former is the derivative of the internal gravitational acceleration with respect to $r$. The right hand side of equation (2) has nothing to do with the Poisson equation. It is simply the chain rule applied to equation (1). The Poisson equation does come into play here, it's implicit in equation (1). $\endgroup$ – David Hammen Jul 6 at 16:16
  • $\begingroup$ @LoveForChrist - Almost certainly. There's no way the density inside the giant planets is constant. $\endgroup$ – David Hammen Jul 6 at 17:33
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That's a tricky question, if you want to go into enough detail.

Generally, the gas giants consist mostly of gas, and you can derive the density and thus gravity if you know the equation of state of the gas (that's the tricky bit). A random treatise on the exploration of Saturn's internal structure based on gravity data could be this article as found in ArXiv. You will need to derive the radius-dependent gravitational acceleration from the density-radius relation.

As to your last paragraph... that's totally unrelated in my eyes... however the scale height of Neptunes atmosphere is about 20km. The scale height is the length at which the pressure reduces to about 37% when comparing bottom and top of the column.

However the top of the atmosphere is usually defined at around 1 bar pressure or an opacity of $\tau \approx 1$ ... that's not a place you will want to place anything long-term unless you want to spend insane amounts of energy just to keep your place in orbit.

Edited to add: for humans it's no big deal to live at different gravity... 0.3 - 1.5g likely is the range which one will do fine also long-term (problematic is rather the 1.5g than the 0.3g end of that range; c.f. Mars has a surface gravity of 0.38g, Moon 0.16g).

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  • $\begingroup$ This paper may also be useful for gas giants. $\endgroup$ – tfb Jun 26 at 14:55
  • $\begingroup$ Do you have a source that 0.3-1.5g would be alright long-term? The three gas giants other than Jupiter anyway have gravities similar to Earth, but it would be best if one would reach 1g of course. Saturn actually has exactly 1g at a certain latitude but an airship or cloud city can float above their equator only I think. $\endgroup$ – LoveForChrist Jun 26 at 15:06
  • $\begingroup$ No linked source, but this reasoning: astronauts do fine on the ISS for many months to years.But 0g is not exactly nice and needs special precautions for lots of things (like liquid flows, showers,...), thus some gravity is very helpful. And from parabolic flights I know from own experience that one can somewhat work normally at each of 0.16, 0.38 and 1.5g. But 1.5g...1.8g gets really exhausting, especially walking. But then... our body can adapt and will produce more muscle, if exposed long-term (again: ISS studies on physiology) $\endgroup$ – planetmaker Jun 26 at 15:14
  • $\begingroup$ Well, in the Moon's gravity the astronauts often fell because of losing balance in the weak gravity, and probably because of their spacesuits too. In lunar gravity they couldn't walk much like on Earth but hopped as we know. You say you were in 0.38g, what was your walk like in Martian gravity? $\endgroup$ – LoveForChrist Jun 26 at 15:42
  • $\begingroup$ It wasn't a recreational flight, so sadly I had no chance to test walking during lunar and marsian gravity parabolas. It involved sitting or kneeling on the floor and operating my experiment. But it's good enough that you can work normally in either case, though of course you more easily bounce at lower gravity (thus you are 'too strong' and push yourself too hard from the ground so that you make giant jumps, if you don't pay attention. The astronauts on moon were mainly hindered by their clunky space suit. $\endgroup$ – planetmaker Jun 26 at 23:40

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