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1

This depends entirely on the kinematics. Imagine I'm throwing an electron at a positron. If the two interact, then the resulting photons (two or more, although the more you add, the less likely the process is of occurring) will have energies entirely dependent on how fast you threw them together. If you throw them together hard enough, you can create heavier ...


0

Bad question- the speed of light should be rephrased as "Speed of electromagnetic radiation" and then specify what wavelength of radiation we're talking about and which part of the neutron star is being discussed. Nobody has determined the EM properties of the inside of neutron stars so the question is not on target. Different parts of a neutron star have ...


1

I'm not entirely sure what you mean about 'pulling hydrogen', all bodies, whether they be planets or literally human bodies, will pull hydrogen via gravity. Earth can lose the H it attracts as H is so light that it can have speeds greater than the escape velocity (just due to random thermal motion). Perhaps Jupiter is sufficiently massive that this happens ...


2

Jupiter will never (not on any timescale like the lifetime of the Sun anyway) accrete enough mass to begin hydrogen fusion. It would need to accrete 12 times its current mass to undergo a brief period of fusing its interior deuterium and to accrete more than 70 times its current mass to attain a central temperature high enough to sustain hydrogen (pp chain) ...


0

Not all galaxies are spiral, there are elliptic and irregular shaped ones also. Search for them on Google and images abound of non spiral shaped ones. A more interesting question is: why are galaxies, in general, laid out in a flat plane, like a dinner plate? According to studies dating back over 40 years ago, spiral galaxies are gravitationally unstable ...


0

Nothing wrong with your reasoning. Planets aren't plane mirrors, they are more like the first case that you discuss. The reflected light from the Sun does not come straight back to the observer, it is reflected over a range of angles. The luminosity of the planet will actually be $$L \simeq a \pi R_p^{2} \frac{L_{\odot}}{4\pi d^2}$$ where $R_p$ is the ...


1

Yes. There are a few ways this happens. Ultimately, it's an energy budget issue. There is less "mass" being radiated, and more "energy" being radiated. Of course, they're fundamentally related. Here are some of the ways this can happen: Loss of energy due to tidal dissipation Radiative Heat Transfer / Radiant heat loss Atmospheric particle escape ...


1

The formula you have quoted is an intrinsically unreliable age estimator for a number of reasons. (i) Yes, you may have to assume $n=3$ for magnetic dipole radiation (or at least that an $n$ measured now has been constant); (ii) you may have to assume $P_0/P=0$, which will be unrealistic for young pulsars and (iii) it assumes that the geometry of the ...


5

All bodies with temperatures higher than 0Kelvin radiate away electromagnetic energy according to black body radiation. The power emitted goes according to the Stefan-Boltzmann law. The law states that the total energy radiated per unit surface area of a black body across all wavelengths per unit time (also known as the black-body radiant exitance or ...


1

Black holes need to conserve the angular momentum of all the matter that collapses into them from a spinning star. As the radius of a black hole is considerably less than the radius of the star, it needs a really fast rate of rotation to conserve the angular momentum of all that matter collapsed into a much smaller radius. Here is an account of the ...


1

From the question you've linked to, I assume you're asking what would happen if a dense and insanely huge water/ice body was to undergo strong gravitational collapse. Stars are made up of Plasma, and Plasma is extremely high energy stuff. The pressure energy density on the molecules during the gravitational collapse process is more than enough to rip ...


0

I believe the first formula you give is obtained by approximating the integrand $ S(E) e^{-(E/k_BT-\sqrt{E_G/E})} $, where $E_G$ is the Gamow energy, by a Gaussian centered about the maximum value $E_0$. However, if you take $T\rightarrow \infty$, then the integrand becomes $$ e^{-\sqrt{E_G/E}}\rightarrow 1,$$ and the integrated cross section blows up, as ...


0

The iron-peak elements are mostly the product of alpha capture reactions onto nuclei that begin with a similar number of neutrons and protons ($Z = N$). The nuclear burning associated with carbon and oxygen (in type Ia supernovae) or silicon (in the cores of massive stars at the ends of their lives) is very fast or even explosive. The important reactions in ...


1

As the Earth isn't a closed system with regard to other objects in the Solar System, including atomic particles, its momentum is affected by collisions. But if the collisions were inelastic (if the atomic particles were absorbed into elements of the crust and the atmosphere), the momentum of the atomic particles and the momentum of the Earth involved in the ...


0

It does affect the Earth, but at a rate so slow that the sun will expand into a red giant before that happens.


4

The final stage of nucleosynthesis at the core of a massive star involves the production of iron-peak elements, mostly determined by competition between alpha capture and photodisintegration. The starting material is mostly Si28 and weak processes are unable to significantly alter the n/p ratio from unity on short enough timescales. Thus the expected outcome ...


3

A very nice question about a common misconception in books on astrophysics (I've made the same mistake in a comment here). According to M.P. Fewell, the origin of this misconception lies in the theory of stellar nucleosynthesis and the abundance of the elements. While other nuclei have higher binding energy per nucleon, $^{56}\mathrm{Fe}$ is more abundant ...


1

The materials that made up the solar system can be studied through the analysis of pre-solar grains and the abundances of various isotopes in primitive meteoritic material. Pre-solar grains were formed in the photospheres of stars pre-dating the Sun. These grains were then expelled into the interstellar medium (ISM) in stellar winds and also in supernova ...


2

This is a complicated problem but if we make several assumptions, we can an order of magnitude estimate that should address your question. Power Source First, the sun is the source of power/energy, and we know its luminosity is ~ $3.846 \times 10^{26}$ W. Therefore, the power per unit area at various distances can be determined by dividing this result by ...


3

It all depends on the closest approach of any stars to the Sun. When galaxies collide it is not that their stars crash into each other, because their individual cross-sections are extremely small when compared to the space between them. This is dealt with in qualitative terms on the wikipedia page on the likely collision. The Milky Way disk at the ...


1

When we assume hydrostatic equilibrium, the pressure gradient is taken. Maybe the pressure should be 0 when $\xi>\xi_0$, so that constraint should be explicitly applied. This is generally understood as an implicit restriction/constraint of the hydrostatic equilibrium (HSE) condition for starting the Lane-Emden equation: HSE is for the star and not ...


0

The assumption $$ P=K\rho^{1+1/n} $$ is a very simplistic mathematical model and should be applied with care. When the model predicts matter has negative density at some point, it is a sign that it is too simple to get it right there. The model neglects lots of things - radiation, magnetic field, solar wind, etc.


1

I would go with the hydrostatic equilibrium condition being the source of the issue here. When there is no polytropic gas when $\xi \gt \xi_0$, there is no need for any hydrostatic equilibrium to be in place - the gravitational potential is free to take on any value, determined (together with boundary conditions) by Laplace's equation: $$\nabla^2\Phi = 0$$ ...



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