# Tag Info

90

There are plenty of satellite galaxies orbiting larger galaxies. The question is how long are you willing to wait for an orbit? The Milky Way has a mass $M$ of something like $6\times10^{11}$ solar masses, or $10^{42}\ \mathrm{kg}$. The small Magellanic Cloud is at a distance $R$ of $2\times10^5$ light years, or $2\times10^{21}\ \mathrm{m}$. A test mass ...

36

The answer lies in something called the virial theorem. You are correct, a cloud that is in equilibrium will have a relationship between the temperature and pressure in its interior and the gravitational "weight" pressing inwards. This relationship is encapsulated in the virial theorem, which says (ignoring complications like rotation and magnetic fields) ...

27

They do! There's an entire class of galaxy, called a 'satellite galaxy' which is defined entirely based on them orbiting a larger galaxy (which would be called a 'central galaxy'). Our own milky-way is known to have many orbiting satellite galaxies, or at least 'dwarf-galaxies'. If dwarf-galaxies aren't enough, the milky-way itself is gravitationally ...

13

Elements up to and including iron can be produced exothermically by fusion reactions in stars. Producing heavier elements is then endothermic. The reason for this is that the binding energy per nucleon is maximised in nuclei around the "iron peak". This means that if you tried to add something to an iron nucleus, the resulting nucleus would have a smaller ...

13

As gas clouds collapse, they increase in internal energy (measured by temperature). This is part of what causes their pressure to increase. As they increase in temperature, though, they also increase the amount of radiation they emit. As they emit radiation, their internal energy decreases and thus their pressure also decreases, allowing for further ...

4

Do you mean anything in the real universe or just theoretically? If the latter, then I can think of a few phenomena: Heat: Just heat it up until the thermal velocity at the surface is greater than the escape volcity. Then neutrons will just fly off and it will evaporate (sublimate?). Spin: Wind it up until the tangential velocity at the equator reaches ...

4

Although not a complete answer, one place to start is with the coldest naturally occurring place in the universe, which is the Boomerang Nebula, a planetary nebula that is around 1 K. As best as I can tell, this cooled below the CMB temperature simply by adiabatic expansion, and is insulated in its interior from CMB heating. Is this a feasible way to get to ...

3

That the numbers of photons far, far outweighs the number of neutrinos can also be determined as follows. One of the main contributors of energy production in the Sun's core is the proton-proton chain reaction, which produces one neutrino and about $26.7\:\mathrm{MeV}$ of energy per helium atom produced. This colossal amount of energy is gradually ...

3

For a uniform, spherical distribution of mass (cloud of gas and dust) of radius $R$ and mass $M$ in absence of magnetic, radiation fields etc, we have $dm = 4\pi \rho r^2 dr$ and the potential energy of a spherical shell of inner radius $r$ and outer $r + d r$ is $dU = -G\frac{m(r)dm}{r}$, $m(r) = \frac{4}{3}\rho r^3$, and a simple integration yields, ...

2

You can get a rough idea from the virial theorem. This tells us that for a gravitationally bound system the kinetic energy $T$ and the potential energy $V$ are related by: $$2T = -V$$ or obviously: $$T = -\tfrac{1}{2}V$$ Suppose we start with our dust cloud particles at infinity with $T = V = 0$ and let the system collapse until the potential energy ...

2

The neutron star crust is separated into outer and inner regions. The outer is a crust of neutron-rich nuclei surrounded by degenerate electrons. The inner is similar, but the nuclei are even more neutron-rich and there are degenerate neutrons too. The (qualitative) answer to your question looks at the ratio of electrostatic (Coulomb) energy to the thermal ...

1

The probability of light getting to an optical depth $\tau$ is $\exp(-\tau)$. So the probability of it being (singly) scattered would $1 - \exp(-\tau)$.

1

I'd like to know what would happen if Venus was flung into a highly eccentric orbit like Sedna (except maybe with its current perihelion) with an orbital period measured in thousands of years. It's kind of a weird question but the first thing to consider is whether the orbit crosses any other planetary orbits, cause if it does, the biggest effect of ...

1

I'm going to add another way to break up a neutron star. Shoot antimatter at it. The difficulty with breaking up a neutron stars is that, once they undergo the compression to become Neutron stars, their gravitation tends to keep them there. The minimum size for a Neutron star to form is about 1.2-1.5 solar masses, but once it's shrunk down, the mass it ...

1

The neutron star is still "regular enough mater" for that it would react to anything a normal object would react. To me the point is more "since its center is not far to collapsing to blackhole, is it possible to shake (or breakup) a neutron star without making it collapse".

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The secret is to evacuate the heat, mainly by radiation. But for this you need dust or "metals", since H and He alone radiates very unefficiently. Paradoxically it is not so easy to collapse completely enough. ( BTW for dark mater there is no possible radiation to dissipate energy, which keeps it fuzzy and a lot less concentrated than ordinary mater.)

1

Iron fusion can take place in stars - what you need is lots of iron and very high temperatures. These conditions exist in the cores of massive stars near the ends of their lives. For example alpha particles can fuse with an iron-56 nucleus to produce nickel-60 and then zinc-64; these reactions are barely endothermic. The problem is that there are competing ...

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