55

There are two forces that can cause the formation of a tail: the solar wind and radiation pressure. The first misconception in your question is "the dust [travels] slower than the nucleus". The tail is not left trailing behind the comet, it is pushed away from the comet by the sun. When the comet is moving away from the sun, the tail is in front of the ...


35

First, there is not just one tail, it is several, but when traveling far from a star, they are "aligned". When it gets closer the different materials behave differently, both depending on the temperature they start to vaporise and how they are affected by solar winds. I think this picture shows it in a good way. https://community.dur.ac.uk/physics....


17

Possibly you are labouring under the misapprehension that the number of photons is somehow a conserved quantity? That isn't true, there are more photons at any given wavelength when you are deeper into the star, because there is a temperature gradient. Cooler material further out is less emissive because fewer atoms are in excited states. The temperature ...


11

Radiative energy transport continues. The point is that the radiative flux, which is proportional to $dT/dr$ can be overtaken when the temperature gradient achieves the adiabatic value and convection starts. Once convection is started, it is very efficient and the majority of energy flux will be transported by convection. Details Broadly speaking, radiative ...


10

Radiation pressure is nothing but electromagnetic interaction. Imagine a hydrogen atom hit by a stream of photons coming from the same direction. Although the atom as a whole is neutral, the electron and the proton are physically displaced, forming a dipole, i.e. a positive-negative charge couple. Some of the photons therefore scatter against the dipole ...


10

Gas clouds with masses much higher than $10^3\,M_\odot$ are plentiful in galaxies; the typical star-forming cloud (the so-called molecular clouds) have masses of $10^3\,M_\odot$ to $10^7\,M_\odot$. When quasistars (hypothetical stars powered not by nuclear fusion, but by accretion onto a central black hole) cannot exist today, it is because all gas in the ...


9

There is a limit to how quickly a black hole can accrete matter. As a star gets close to a black hole it may be tidally disrupted and pulled apart. The material from the star will then fall toward the black hole and as it does so, the gravitational potential energy will go into heating it up. Now the stellar material would not normally be able to fall ...


9

the candidate star is an "average" 5 solar mass star, and the black hole is a 5 solar mass black hole Then their gravity is identical. Black holes don't have magic powers. A 5 M☉ star and a 5 M☉ black hole exert the same attraction from the same distance. The only difference is that the black hole would be much, MUCH smaller (about 30 km diameter in this ...


9

The electromagnetic energy density is dominated by the cosmic microwave background and the optical/IR background. This Physics SE answer, contains the plot below, showing the contribution of energy density at different frequencies. You can integrate under this "by eye" to see that the CMB contribution is the largest followed closely by an optical/IR bump. A ...


9

Just a short answer, and likely others will fill in more details. If there is ionization of some atoms, then generally there is recombination as well - you will have both processes going on, roughly in balance with each other. Typically when an electron recombines with an atom, it does so into some excited state. Then as it drops from that excited state ...


8

If you build a very massive protostar, more than a thousand solar masses, then it is possible for the core of the protostar to collapse directly to a black hole whilst it is still surrounded by a massive envelope. The collapse will happen "inside out", so that the envelope collapses at a slower rate. However, there is a maximum rate at which black holes can ...


7

Without any other information, you cannot distinguish between the two effects. $$ T = T_0 (1 + z) $$ A blackbody spectrum of temperature $T$ is identical to a blackbody spectrum of temperature $T_0$ with redshift $z$. For stellar/galactic radiation, we can use the fact that the radiation is not a perfect blackbody. For the CMB, we can use the fact that ...


6

There are a couple of relevant questions one would want to ask: 1) Do protons decay, and if so, what do they decay into? The answer appears to be no, or at least the theoretical lifetime of the proton must increase as a results of these experiments. If they do, eventually the universe could end up in a state of radiation (and dark energy, and dark matter, ...


6

Thermal radiation $\neq$ blackbody radiation. Thermal radiation is radiation that comes from a system where an equilibrium has been reached, where the various energy states are occupied according to the Boltzmann distribution and the particle velocity distributions are Maxwellian at some given temperature. That does not necessarily imply that the radiation ...


6

Radiation has energy, and energy exerts gravity. You could say that it "strengthens the force of gravity" (it definitely doesn't weaken it), but it would be more appropriate to say that it contributes to the total gravitational field. The effect it immensely small, however, and is in almost all circumstances completely insignificant compared to the gravity ...


6

It is not very clear what you mean. There are many cosmic sources of microwave radiation that emit so much energy and/or are so close to us that they confuse measurements of the cosmic microwave background. For example "compact sources" such as giant molecular clouds, supernova remnants, H II regions; "diffuse" sources due to dust in the galaxy and in our ...


6

Electric and magnetic fields can and do exist in a vacuum. Electromagnetic waves are just fluctuations in electric and magnetic fields. In a vacuum (meaning no charges or currents, which do require the presence of matter), then the solutions to Maxwell's equations are electromagnetic waves.


5

It is not possible to split a larger nucleus into hydrogen nuclei without expending a greater amount of energy that you receive back. This is because Hydrogen has (by far) the lowest nuclear binding energy per nucleon (protium has zero nuclear binding energy, though deuterium and tritium do have some). Therefore, such a process would decrease the entropy of ...


5

The Sun outputs several different kinds of things. Electromagnetic radiation The Sun is (partially) a black-body radiator at a temperature of near 6000 K, and therefore emits all sorts of electromagnetic energy, including UV and X rays. UV is stopped in the upper atmosphere. X rays are absorbed by the whole atmosphere, and are pretty weak anyway. The Sun'...


5

Imagine you have a solar collector measuring one square meter, which can directly measure the power (energy per unit time) striking it. If there are layers of atmosphere above the collector, some of the incoming light is absorbed or reflected by particles in the atmosphere, so you won't measure as much power as if you place the collector in space at the ...


5

We know that stellar radiation pressure balances the gravitational compressive forced of a star. We do not know that. Degeneracy pressure, thermal pressure, and radiation pressure are what collectively balance gravitation in a star. Degeneracy pressure dominates over the other two in white dwarfs and neutron stars. Thermal pressure dominates in low mass ...


5

Nothing that is known to exist Relativity does not allow for a massive particle to travel at the speed of light, but it doesn't prevent a particle from travelling faster than light. Such a particle has been called a Tachyon. No such particle has ever been observed. There are good reasons for believing that they don't exist. Such a particle would be ...


5

The Eddington limit represents the maximum luminosity that can be achieved by a body (such as the star) when there is hydrostatic equilibrium (http://astronomy.swin.edu.au/cosmos/H/Hydrostatic+Equilibrium). For luminosity greater than Eddington limit, the radiative force of the luminosity on matter exceeds the gravitational force on the matter. If the ...


4

Historically, two people (or groups of people) independently came up with different equations to model the blackbody equations in different parts of the spectrum. Rayleigh-Jeans law (classically derived) is valid for longer wavelengths and Wien's law (not Wien's displacement law) is valid for shorter wavelengths. The Planck Distribution approaches the two ...


4

Assume you have a spherical blackbody. The solar flux at the radius of the Earth is given to a good approximation by $L/4\pi d^2$, where $d = 1$ au. This is $f=1367.5$ W/m$^2$ (though note the distance between the Earth and the Sun has an average of 1 au). If it is a blackbody sphere it absorbs all radiation incident upon it. Assuming this is just the ...


4

There's a few parts to this question, I'll give it a shot. Also, since you're asking about the geothermal energy of the earth, I'm going to ignore radiation from space and radioactive van allen belts (Jupiter has a very radioactive van Allen belt). Also, as voyager discovered when it left the solar system, there's more radiation in the form of cosmic rays ...


4

The Sun's luminosity is $3.8\times 10^{26}$ W. Application of mass energy equivalence tells you it loses mass at a rate of 4.25 million tonnes per second as hydrogen turns into helium. This is practically nothing as far as the structure of the star goes. Over its lifetime, the Sun has lost about 0.03% of its mass in this way. Radiation pressure is a ...


4

Here's my approach to solving this problem. You provide some of the initial steps, but I'll still go through them just for completeness. Orbital Distance We're told the period is $T=5.5\:\mathrm{years}$. This means we can immediately calculate the orbital distance (or more precisely, the semi-major axis, $a$). Since we're talking about a comet orbiting the ...


4

The total energy of the universe consists of the mass energy of all the matter (both normal and dark), the mass-energy of the radiation and of the dark energy plus the gravitational potential energy of the universe (which is negative). The hypotheses is that this sums to zero. As far as I know, there is no experimental evidence for this. But there is good ...


4

Not noticeably darker. Assuming such a globule has a mass of 50 solar masses and a diameter of 1 light year, that would make it's average density about $2.2\times 10^{-16}\,kg\,m^{-3}$ which is fairly close to not being there at all in human terms. An imaginary tube of this stuff 1 AU tall and of area 1 meter squared would contain about $3\times 10^{-5}\,...


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