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27

Jupiters moon Io is heated through the gravitational pull of Jupiter, but when Io is heated because of this, where does that energy come from? How does conservation of energy work for this effect, where is energy "lost"? TL;DR: The energy ultimately comes from Jupiter's rotation. Io is tidally locked; it has the same orbital and rotation rates. If Io ...


16

In astronomy parlance, the Sun has a "metal"$^{1}$ mass fraction of about $0.02$. A solar mass is $\sim2\times10^{30}\;\rm{kg}$, so the sun contains about $4\times10^{28}\;{\rm kg}$ of "metals". That's about $20$ times the mass of Jupiter. A lot of that metal mass will be ${\rm C}$ and ${\rm O}$ and other elements a chemist would call non-metals, but I think ...


11

Estimating the mass of a "single star" can be a very difficult task, though perhaps your question is too pessimistic. There are a number of suggested relationships linking the mass of a star to its luminosity. These can of course come from stellar evolution models, but they can also be empirically calibrated using stars in resolved binary systems of known ...


9

To add to Dancrumb's answer and deal more specifically with energy, page 21 of the MIT thesis says the following: As a secondary object orbits around a primary body, the gravitational force of the primary causes a distortion in the shape of the secondary, and vice versa. We can refer to one body as the perturber and the other body as the extended ...


7

A binary star system, which is quite common, will allow us to determine mass with great accuracy, using Kepler's Third Law of Planetary Motion which is as follows: $$ \frac{T^2}{r^3} = \frac{4 \pi^2}{GM_{sum}} $$ Using the orbital period, $T$, we can determine the acceleration and affect that stars have on one another to determine mass. Once we have the ...


7

Our primary method of understanding stars is using optical telescopes. We can directly measure the temperature of the "surface" of a star by fitting its spectrum to a blackbody. However, this gives us the temperature only at the "surface" or the optical depth of a visible photon. We know that stars are extremely hot through the hydrodynamic/thermodynamic ...


5

"Storing" something implies the purpose is to put is somewhere safe so that it can eventually be retrieved. Heavy metals should eventually sink to the center of a star, but how are you going to retrieve it? Even in a science fiction context, it's hard to imagine a plausible means or retreiving a large pile of heavy metal from the center of a star. ...


5

Io undergoes synchronous rotation as it orbits Jupiter. As a result, the same part of Io faces Jupiter all the time. Since Jupiter is so massive and Io orbits so closely, Io's shape is distorted due to tidal forces, that is to say the difference between the gravitational pull on Io's closest side to Jupiter and its furthest side to Jupiter are different and ...


5

So a while ago I did a little project where I grabbed a "standard solar model" from this paper, which gives me some information that's useful for actually making an estimate. (Unsurprisingly the link given to download the data has changed in the last ten years; I haven't sleuthed to see whether the data is still publicly available.) Only about 1.5% of the ...


3

What caused the transition,[...] As the universe expands, the number of particles per unit volume goes down. But in addition to this, photons suffer a cosmological red-shift. So the density of mass-energy due to nonrelativistic particles goes down, and the density due to photons goes down, the latter goes down faster. By extrapolation, we predict that ...


3

I actually answered a related question a couple of days ago on Astronomy. Small world! One of the important properties of the Oort Cloud is that objects in it are not strongly influenced by the Sun. After all, its inner edge is roughly 2,000 AU away - 300 billion kilometers from the Sun. The Sun's gravitational influence in that region is rather weak, so ...


3

I had a quick look at the paper - its mostly nonsense. The intrinsic light from a quasar is completely dominated by its emission line spectrum and a mostly featureless continuum. The emission lines give the true redshift of the quasar. Absorption lines in quasar spectra are predominantly due to foreground gas clouds at lower redshifts than the more distant ...


3

The resolution is much smaller than the granulation pattern. i.e. The granules are well resolved and are of order 500-1000 km in diameter, not 50 km. The sunspot is of order the diameter of the Earth. Granulation is caused by convective cells rising and falling such that they just poke up into the bottom of the photosphere - i.e. that region where most of ...


3

This is due to spinning forces, such as centrifuged force. Remember that galaxies form (at least the regular ones) from the spinning of matter around a galaxy nucleus. So, you're not wrong. Gravity is isotropic, but in this cases the spinning forces are the definition to the galaxy form.


2

According to Wikipedia The Chandrasekhar limit is the maximum mass of a stable white dwarf star. The limit was first published by Wilhelm Anderson and E. C. Stoner, and was named after Subrahmanyan Chandrasekhar, the Indian-American astrophysicist who improved upon the accuracy of the calculation in 1930, at the age of 19. White dwarfs with masses greater ...


2

Elements heavier than iron are produced mainly by neutron-capture inside stars, although there are other more minor contributors (cosmic ray spallation, radioactive decay or even the collision of neutron stars). Neutron capture can occur rapidly (the r-process) and occurs mostly inside supernova explosions. The free neutrons are created by electron capture ...


2

Your first paragraph is not quite right. Gas pressure does not "stop" upon formation of an iron core, it is merely that the star cannot generate further heat from nuclear reactions and becomes unstable to collapse. i.e. The star does collapse! Perhaps what you mean is what halts the collapse (sometimes) before the star disappears inside its own event horizon ...


2

Consider a satellite in orbit about the Earth and moving at some velocity $v$. The orbital velocity is related to the distance from the centre of the Earth, $r$, by: $$ v = \sqrt{\frac{GM}{r}} $$ If we take energy away from the satellite then it descends into a lower orbit, so $r$ decreases and therefore it's orbital velocity $v$ increases. Likewise if we ...


1

For stars (which have huge amount of mass and density), gravity is taken to be responsible for the heat increase. because heat and volume (thus density) thus gravitation of a (massive) star, are related. This is exactly one of the factors that make nuclear fusion (in stars) possible. The two effects thermodynamics (and kinetic energy) and gravity are ...


1

If a supernova occurred within 100 light years of Earth it would wreck the atmosphere. However, on a more mundane level it has been suggested that cosmic rays cause water droplets to form in the atmosphere and affect cloud cover and hence climate. The amount of cosmic rays also depends to some extent on the changes in the magnetic field of the sun. Reference ...


1

I'll tackle Gamma Ray Bursts (GRBs) since neither of the current answers have addressed that yet. This section of the Wikipedia article on Gamma Ray Bursts state that a GRB from a Wolf-Rayet star within the Milky Way Galaxy could have a devestating effect on Earth. Specifically, A GRB at a distance of 8,000 light-years from Earth could deplete 25% of the ...


1

The core-collapse phase of a supernova commences once the final stages of nuclear burning are complete. This final phase of fusion reactions involving silicon, produce a core composed of iron-peak elements (not just iron). The cessation of nuclear burning leads to the contraction of the core. This happens relatively slowly at first, on a timescale given by ...


1

Fun, So you are asking about the thermal energy content of the sun? If we assume that all the hydrogen is dissociated. (single atoms) Then each atom has three degrees of freedom and carriers 3/2 kT of energy. So count up the number of atoms at each temperature.... That will work until the atoms ionize. Then there will be equal energy in all the ...


1

Here are the systems I found: 6: ADS 9731 Beta Tucanae Gamma Velorum Kappa Tauri Mu Sagittarii 7: AR Cassiopeiae Nu Scorpii ... no physical multiple stars of greater multiplicity yet found.


1

A "collision course" is a very fuzzy concept: if you are "barely going to hit" you are on a collision course but don't need a lot of deflection. However, let's assume for a moment a stationary earth, a meteorite of mass $m$ at distance $D$, heading for earth of radius $R$ with velocity $v$. The equations you need are conservation of angular momentum and ...


1

The apparent radial accelleration due to moving in a circle is ω2R. For earth, ω = 2π rad/24 h = 73 x 10-6 rad/s. Earth's radius is about 6.37 Mm. Therefore the upwards accelleration at the equator is 34 mm/s2. That's pretty small compared to the 9.8 m/s2 downwards accelleration due to gravity, so we generally ignore it. Then there is ...



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