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The paragraph notes previously that Mimas is not in hydrostatic equilibrium for its current rotation. A quick search for the moon and hydrostatic equilibrium turned up M. Burša (1984) "Secular Love Numbers and Hydrostatic Equilibrium of Planets". According to this paper the Moon, Mercury and Venus are all far from hydrostatic equilibrium. The discrepancy is ...


4

Based on observable data, I assume that there is a relationship between the size, rigidity and the passage of time; that will result in all objects subject to only their own gravitational influences (Given: no body is ever truly uninfluenced by others) becoming spherically shaped (round). That's called hydrostatic equilibrium. That's one of the main ...


3

For your second question, it depends upon whether you are talking about composition by number, or composition by mass (the numbers you give are roughly the composition by mass). If you want to get really specific, you can add in the other elements too. To answer the first, you add them up proportionally.


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As the comments already say, an object being a dwarf planet is a matter of convention. If the IAU says it's a dwarf planet, it's a dwarf planet. Otherwise, it's not. The requirements you are listing from Wikipedia are the IAU criteria for pronouncing objects as dwarf planets, but that does not mean that all objects fulfilling these criteria are dwarf planets....


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In terms of mean angular velocity, the distribution of rotation rates among main sequence stars is well known. Allen (1963) compiled data on mass, radius, and equatorial velocity, which was then expanded upon by McNally (1965), who focused on angular velocity and angular momentum. It became clear that angular velocity increases from low rates for spectral ...


2

The whole exercise illustrates the usefulness of vectors in addressing problems. $$\vec{F_g} = -\frac{GM(r)\delta m}{r^2}\ \hat{r}$$ $$\vec{F_P} = -\left(\frac{dP}{dr}\right) \delta r \delta A\ \hat{r}$$ $$\vec{F_P} + \vec{F_g} = 0$$ leads to the scalar equation of hydrostatic equilibrium $$ \frac{dP}{dr}=- \rho g,$$ where $\vec{g}= -g\hat{r}$.


2

Virial shocks appear when gas falls in toward a galaxy, and the shocked gas stays hot so the pressure in the post-shock gas can support the shock and keep it away from the galactic disk. The alternative is if the post-shock gas cools and loses pressure support, and the (non-virial) shock collapses right down to the galactic disk. The significance is that ...


1

Mass = $\text{atomic mass} \times \text{number of atoms} = 4.4400513610370694 \times 10^{30} \mathrm{\ kg} \approx 2.3M_☉$ Herein lies the problem: your units and exponents are messed up. The mass of your iron isotope is $55.934936 \mathrm{\ u}$, and there are $7.938⋅10^{31}$ atoms. Do the math, and you'd get a total mass of $4.4400513610370694 \times 10^{...


1

Seems to me that moons are regularly subjected to tidal forces that comets, asteroids and the like are not. Repeated stress, such as rocks in a rock tumbler, or two circles of glass on a mirror maker's table, tends to smooth things out. I'm sure there's a pile of academic papers on sub-disruptive levels of strain tending to increase the symmetry of many ...


1

I can only speculate but here goes nothing. 67P/Churyumov–Gerasimenko apparently came from the Kuiper Belt. There are a lot of KBO (Kuiper Belt Objects) in that region, so it is not unreasonable to expect some of these objects to have crashed into other objects of about the same mass, resulting in irregular shapes. Small moons (or other moons in the Solar ...


1

The Eddington luminosity is defined in this way - but I guess that is not the answer you are looking for! However, if we ask what isotropic luminosity is required in order to balance the inward gravitational force on a lump of gas then yes we could consider gas pressure as well. The correct formulation is that the pressure gradient due to gas plus ...


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Yes, a droplet of water would do that. However, at the planetary scale, the contribution of superficial tension is negligible. At that scale, gravity dominates all other forces. In fact, you could model a planet as a loose mass or rubble, kept together by gravity alone, and the model would be very close to reality.


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Video: Space Station Astronauts Grow a Water Bubble in Space Surface tension tends to draw the water into a nice sphere.


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Strictly speaking (as far as I know), hydrostatic equilibrium applies whenever a fluid balances external body forces with the pressure gradient. From Wikipedia: In continuum mechanics, a fluid is said to be in hydrostatic equilibrium or hydrostatic balance when it is at rest, or when the flow velocity at each point is constant over time. This occurs when ...


1

I think you're asking, "If we know an object's shape, can we determine if it is in in hydrostatic equilibrium?" If so, one might wonder if astronomers classify basketballs or ball bearings as being in hydrostatic equilibrium, since they're so spherical. Below about 100 km in radius, the answer is in general no. Given some population of randomly lumpy ...


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To get an order of magnitude estimate you can just use the total mass $M$ and luminosity $L$ of the star and an assumption of your fusion process. Main sequence stars fuse Hydrogen in to Helium through the proton-proton chain, which converts 0.7% of mass into energy. So the estimated lifetime of the star would just be: $0.007\frac{Mc^2}{L}$ ($c$ is the speed ...


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