# Tag Info

85

Summary There's a 1 in 500 billion chance you're standing under a star outside the Milky Way, a 1 in 3.3 billion chance you're standing under a Milky Way star, and a 1 in 184 thousand chance you're standing under the Sun right now. Big, fat, stinking, Warning! I did my best to keep my math straight, but this is all stuff I just came up with. I make no ...

45

The sun isn't the same density all the way through. According to MSFC's solar interior page, the core density at the centre of the sun is a whopping 150,000 kg/m$^3$. Surrounding it the radiative zone is around 20,000 - 200 kg/m$^3$ (already less dense than water). Eventually at the edge is the convective zone - the density at the part that we see is much ...

37

The Pauli Exclusion Principle forbids two indistinguishable fermions occupying the same quantum state. It does not prevent them getting arbitrarily close together so long as they have very different momentum states. The big bang model relies on classical General Relativity. When we go back to scales where quantisation of space might become important (i.e. ...

28

Fusion inside of a star affects the sun's density (which does not happen with a planet). It produces an outward pressure that balances against the attraction of gravity, thereby reducing the density as long as the star is burning. Once a star the mass of the sun is no longer able to sustain fusion, what is left is a white dwarf which is in fact much denser ...

22

I feel it's a cheap answer but heavy Jupiters can get much denser than Earth because planets with Jupiter's mass stop adding size as they add more mass. A planet with Jupiter's size and 10-12 times Jupiter's mass would be over twice Earth's density. As far as Earth-like planets, there's a cool property of terrestrial planets, more mass means more tightly ...

21

The density of matter depends not only on its composition, but also on temperature and pressure. It's not meaningful to say that substance A is denser than substance B without specifying the conditions under which the comparison is being made. For a simple everyday example, at room temperature (and pressure) water is significantly denser than air. But ...

19

Let us define this as the largest observable density of a stable object, in order to exclude black holes which may have a very large (infinite) density at their centers or objects collapsing towards a black hole status. If we restrict the definition in this way, then the answer should be the core of the most massive neutron star that we know about. At ...

15

In short: no one knows for sure, but currently it looks that the probability is 1. Longer: On our current understanding, the Universe is probably infinite in space. This depends on the recent WMAP satellite results, which have shown a zero curvature of the Universe below measurement precision. The other two options were a positive curvature (thus, we would ...

11

I'd say the most important answer is because the volume of stars is counted differently than for (inner) planets.For the former, most of the gas surrounding the dense core is counted. The latter don't have significant enough amounts of it. This is even more pronounced with larger stars. VY Canis Majoris: "With an average density of 0.000005 to 0.000010 kg/...

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Straightforwardly no. For a start there are almost no free protons inside a white dwarf. They are all safely locked away in the nuclei of carbon and oxygen nuclei (which are bosonic). There are a few protons near the surface, but not in sufficient numbers to be degenerate. Let us assume though that you were able to build a hydrogen white dwarf that had ...

10

I assume you're asking about central supermassive black holes (SMBHs, one per galaxy), not stellar-mass black holes. The answer is yes, but what actually happens is the two SMBHs have to merge first, and then the resulting combined SMBH can sometimes be ejected from the combined (merged) galaxy. [Edited to add: Since you've updated the question with a ...

9

The vast majority of the particles in Saturn's rings are small, on the order of $\sim10^{-1}$ m or lower. The columnar number density, according to data from Voyager 1 and Earth-based observations, can be approximated as a function of particle radius by a power law for all particle radii $a$ in meters such that $0<a<1$, as can be seen on this log-log ...

8

The most-widely accepted hypothesis at the moment is that Mercury was struck by a large impactor that removed a significant fraction of its mantle (I believe this theory was originally proposed by Cameron & Benz in 1987, and the qualitative theory hasn't changed very much). For planets that are close to their parent stars (such as Mercury), the collision ...

8

As the surface of the earth is solid the crust, Correct. underneath the crust the mantle which is liquid, No the mantle is also solid, although more plastic than the crust. There is a liquid core below the mantle the liquid density is smaller than the crust. No the core is made of iron and considerably denser than the mantle rocks above it. ...

8

From the wikipedia page on Chthonian planet https://en.wikipedia.org/wiki/Chthonian_planet "Transit-timing variation measurements indicate for example that Kepler-52b, Kepler-52c and Kepler-57b have maximum-masses between 30 and 100 times the mass of Earth (although the actual masses could be much lower); with radii about 2 Earth radii, they might have ...

7

Icy objects, such as most in the Kuiper belt can reach an equilibrium if they are about 400km across, whereas the rocky asteroid Pallas, at 572km clearly has an irregular, non spherical shape. All rocky objects larger than Pallas (and there aren't many) are spherical. Rock tends to be stronger than ice. Rocky objects are able to withstand their own gravity ...

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Here is a plot I generated in 5 minutes at the site exoplanets.org To construct this I took planets discovered by the transit method and which had a $M \sin i$ measured using radial velocities. I divided the $M \sin i$ by the sine of the measured inclination angle (this is required to avoid using masses that have been estimated using an assumed mass-radius ...

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Proton degeneracy is not important, because its effect is much smaller -- much like nuclear particles in theory also are dictated by gravity, but the electromagnetic and nuclear forces are dominating, since they are much stronger. Proton degeneracy is weaker than electron degeneracy due to the far greater mass of the proton compared to the electron. The ...

6

The test to see whether degeneracy pressure is going to be significant is to compare $kT$ with the Fermi energy $E_F$ The Fermi energy is the energy level up to which all energy states would be occupied in a completely degenerate fermion gas. It is given by (for non-relativistic conditions) $$E_F = \frac{h^2}{2m}\left(\frac{3}{8\pi}\right)^{2/3} n^{2/3},$$...

6

There is no 1:1 mapping between density and composition/structure. You have to look at detailed planetary models. For example, some hot Jupiters are extremely dense ($\geq 10$ g/cm$^3$) but they are undoubtedly gas giants. The origins of this diversity are the source of much speculation and theory, but are certainly within the realms of known physics. An ...

6

Does "overhead" mean over the center of your head, or over some part of your head? If we assume the latter, it changes the problem! I don't want to recapitulate all MichaelS's lovely work above, so I'll do a quick back-of-the-envelope calculation borrowing from his numbers. The area of a human head as viewed from above (or below) is, umm, let's see, ...

5

The mean density of the star is really only defined by the formula $\bar\rho=M/V=3M/4\pi R^3$. The radius of a star is a generally a very complicated function of a star's other properties. When we determine the radius in stellar models, it's only because we've solved equations that describe the structure of the whole star, and read off the value at what we ...

5

Starting from the index you mentioned, I clicked through the links for some individual planets, which in turn link to discovery papers or other relevant observations. For planets around Kepler-23, -24, -25, -26, -27, and -28, the relevant papers are Ford et al. (2012) and Steffen et al. (2012), two out of a series of papers. Both papers used transit timing ...

5

This question is more complicated than it seems like it should be! There is no threshold mass or density beyond which an object becomes perfectly spherical; even supermassive stars are slightly oblong. The only exception is black holes, which are perfectly round up until you reach the quantum level. If we want a simple answer, most guesses are somewhere ...

5

Most simply, the density is determined by the number of molecules, the surface area of the planet, the temperature, and the gravity. But it sounds like what you really care about would be called the "column density", which depends only on the number of molecules and the surface area, and you are wondering why there is not some direct correlation between the ...

5

We can understand gravity as following a set of mathematical equations called "General Relativity" which were discovered by Einstein (and others) around the start of the 20th century. The same gravitational equations apply to black holes, stars, planets, people, apples etc. These equations are very hard to solve. Fortunately there is a very good ...

5

It's simply because you can "get closer" to it, that's all. No special sauce. You know how gravity is pretty weak far away, and gets stronger close by? The closer you get to the Sun - more specifically to the center of the Sun, because that's how you measure the distance - the greater the pull. However, once you reach the Sun's surface, there's a problem. ...

5

Yes, a planet can exist with the same mass as Earth, but have a different diameter - if it has a different density. The lower the density, the larger will be the diameter. In terms of an exact formula, we have: $$d(M, \rho) := 2 \cdot \sqrt[3]{\frac{3M}{4\pi \rho}}$$ The trick is that density is specifically a function of composition: hence, the extent to ...

5

Whether of not you include the mass of the atmosphere doesn't matter much. For instance, the mass of Earth's atmosphere is roughly $10^{-6}$ of Earth's mass, corresponding to the uncertainty in the gravitational constant $G$ used to calculate the mass in the first place. But the volume that you use for calculating the average density could potentially ...

4

No, being flat and being homogeneous is not equivalent$^\dagger\!\!\!$. Flatness refers to the geometry, which depends on the total energy density $\rho$; if it is above or below a certain critical threshold $\rho_\mathrm{cr}$, we call the Universe "closed" or "open", respectively, while if $\rho$ is exactly equal to $\rho_\mathrm{cr}$, we call it "flat". ...

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