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Gravitational constant: 6e-11 (meters, seconds, and kg).

Jupiter radius: ~ 69,000,000 meters (google)

Jupiter mass: $2*10^{27}$.

1 kg, at Jupiter's radius, experiences ~100N of force:

$$ F = G * \frac{J_m * 1.0}{r^2} \approx 6*10^{-11}\frac{2*10^{27}}{5*10^{15}}\approx 20_N $$

However, if one just moves out to 100,000 kilometers, you get down to $\approx 10_N$, the same as earth.


What explanation is there for why there is no atmosphere at the earth-equivalent level of gravity?

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  • $\begingroup$ On Earth, the atmospheric pressure at an altitude of 70km is about 1/9600 of the air pressure at sea level, but the strength of the gravitational acceleration at that altitude is only slightly less than that at sea level. See en.wikipedia.org/wiki/Scale_height $\endgroup$
    – PM 2Ring
    Feb 8, 2019 at 8:20

3 Answers 3

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Perhaps it is simpler not to compare Jupiter to Earth, but to think about what happens as you add more and more gas to a gas planet. Even if you make a planet out of incompressible stuff, the gravitational field at the surface increases as the planet grows. This effect is magnified when making a planet out of compressible stuff (gas). The pressure inside the planet goes up for two reasons: there is more mass pressing down from above, and more gravity pulling on it. So the density increases, meaning that the volume of the planet grows more slowly than its mass. The radius of a planet made of incompressible stuff would grow like $M^{1/3}$. We've just decided that the radius of a gas planet grows more slowly. Meanwhile, the radius at which the gravitational field of the planet reaches some small value (small so as to ensure it is outside the planet) grows like $M^{1/2}$ (that is, faster than the radius of an incompressible planet, and even more faster than that of a compressible one). Thus it is clear that as you add gas, you steadily increase the gravitational field at the "outer edge" of the planet, however you choose to define that.

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  • $\begingroup$ This is a great answer $\endgroup$
    – Chris
    Feb 8, 2019 at 6:41
  • $\begingroup$ However, the mass of the planet is an invariant in the question. If I add gas, I violate the parameters. Instead, I am talking about adding energy. $\endgroup$
    – Chris
    Feb 11, 2019 at 16:49
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the fact that there is no atmosphere near jupiter at the equivalent g level radius for earth is because any parcel of gas at that radius has no solid surface to stop it from falling closer to jupiter. so it falls, and hence there is no appreciable amount of gas at that distance from jupiter.

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  • $\begingroup$ Right, so what you are saying is: if the earth had no crust, and it's mass held the same number of joules we enjoy now, and was just a point source of 1g gravity, we would have no atmosphere at our current distance from the center; instead, we would observe merely a denser outer atmosphere that was substantially closer to point of origin of Earth's gravitational pull, on a statistical basis. $\endgroup$
    – Chris
    Jan 8, 2018 at 19:51
  • $\begingroup$ what I am getting at is: suppose that Jupiter isn't hot...but rather that it is very cold (much colder than it used to be), and that it's outer atmosphere is no longer at the same distance from the center as it once was... (thus explaining the ice rings). $\endgroup$
    – Chris
    Jan 8, 2018 at 19:53
  • $\begingroup$ if it were very cold, its gas would contract and its apparent radius would shrink. however its MASS would remain the same, as would its gravitational pull. the argument still holds. $\endgroup$ Jan 8, 2018 at 22:24
  • $\begingroup$ That was my point: the current radius of jupiter is smaller than it used to be because it is colder now than it used to be, but it’s mass is still the same, so the gravity is still the same at the same distances... $\endgroup$
    – Chris
    Feb 8, 2019 at 4:40
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The strength of gravity at a point is not directly related to the pressure of the atmosphere at that point. Instead it is related to the pressure gradient. To a decent approximation

$$ g = \frac{1}{\rho}\frac{dP}{dz} \sim \frac{T}{P}\frac{dP}{dz} $$

as this is the pressure gradient that provides an upward force to cancel out the downward force of gravity.

Even ignoring the solid surface of the earth or temperature differences, solving this gives different surface pressures for planets of the same surface gravity but different sizes.

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  • $\begingroup$ No. Actually, solving the equation I gave shows that higher temperature gives a lower surface pressure (assuming the planet is relatively fixed in size). This makes sense- it is obvious in the limit as $T\rightarrow\infty$ that all the atmosphere escapes. Cold is better for holding an atmosphere. $\endgroup$
    – Chris
    Jan 8, 2018 at 22:30
  • $\begingroup$ Using your definition for surface would always give a surface pressure of zero. Heating up the planet to make it ten times larger would give you an atmosphere at $100,000~\rm km $, but it would also change the gravitational profile of the planet, so you still wouldn't have an atmosphere at the distance of earth equivalent gravity. $\endgroup$
    – Chris
    Jan 8, 2018 at 22:38
  • $\begingroup$ @bordeo Sure, there are probably conditions that would make it happen. There's no reason to believe it would be at a survivable temperature, though. Giving it a temperature of $300~\rm K$ does not give you one atmosphere at a level where acceleration is $1~\rm g$. I can update my post to show that, if you like. $\endgroup$
    – Chris
    Jan 8, 2018 at 23:01
  • $\begingroup$ @bordeo To float a platform, what you need is a large pressure gradient. There's no way you could float a platform at roughly earth temperature, atmospheric pressure, and earth gravity, since the pressure gradient at that point is identical to earth's surface. So floating a platform there would be the equivalent of making a platform that floats just above the surface of earth. $\endgroup$
    – Chris
    Jan 8, 2018 at 23:12

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