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4

Checking for electron degeneracy is a matter of comparing the Fermi kinetic energy with $kT$. If $E_F/kT \gg 1$, then you may assume the electrons are degenerate. The central density of the Sun is around $\rho=1.6\times 10^5$ kg/m$^3$ and the number of atomic mass units per electron is around $\mu_e =1.5$. The number density of electrons is therefore ...


3

The first part is handled well by Chris Drost - the kinetic particle energies are a lot larger than their interaction energies, so the gas can be considered (approximately) ideal. The last part - yes, as long as the Coulomb energy is a lot lower than the thermal energy then the protons or He ions can be considered an ideal gas with the appropriate average ...


7

According to a NASA page, the density in the middle of the Sun is about 150 g/cm3. That's about 9 × 1025 protons in a 1cm3 box, or 450 million to a side, and using that spacing for a voltage calculation reveals a typical interaction energy of 65 eV or so. (If you've never seen this unit before, that is the energy used by a 1V battery to move an electron's ...


2

What emission lines? Unless you are looking at the chromosphere or corona the Sun does not have emission lines. Of course you can estimate photospheric abundances from a photospheric absorption line spectrum. That is how the solar abundances are estimated for most elements. However, you need a good spectrum to perform a detailed analysis. It should have ...


2

Using the basic equipment at your disposal, this will not be possible. A naive approach would be to take the measured line amplitudes, divide by the line-integrated absorption coefficients at the wavelength of each line for some nominal set of parameters characteristic of the solar atmosphere, then divide out the emission coefficients and compare the ...


5

Reflecting energy back into the photosphere using mirrors is equivalent to asking what happens if we artificially increase the opacity of the photosphere - akin to covering the star with large starspots - because by reflecting energy back, you are limiting how much (net) flux can actually escape from the photosphere The phenomenon could be treated in a ...


2

If you graph the temperature of your copper strip as a function of time you're going to get something like: This is because you have two effects. The light from the Sun heats the copper strip, but at the same time the strip cools. The equilibrium temperature (the dashed line) is the temperature at which the cooling balances the heating. If the intensity ...


0

The heat from Sun, comes in the form of radiation. To be specific, the infrared part of the EM light is responsible for heating. Infrared frequency is closest to the resonant frequency of most molecules. Thus, when these molecules absorb the infrared radiation i.e. the molecules' electron clouds oscillate at the infrared frequency, there's resonance, which ...


14

You probably are under the misconception that heat travels only via molecular interactions. (i.e, heat transfer by conduction, which needs a medium of sorts). Heat also transfers by radiation, which the sun is an enormous source of. Electromagnetic radiation does not need a 'medium' to travel through. All types of electromagnetic radiation carry energy, ...


4

The heat 'comes' as electromagnetic radiation, that is light. Light from the sun is electromagnetic radiation, that is a wave having energy and momentum or a very big amount of quantum particles, photons, that have energy and momentum. The interaction of this electromagnetic interaction is what heats the earth. Hope this helps.


0

Along w/ the info in the other answers, keep in mind that "white" is not a fixed item. In very dim conditions, your cones don't work, and the rods in your retina only report intensity, not color, so everything looks white/grey. In very bright conditions, your retina overloads, and all you sense is 'white.' Now, one can define "white light" analogous ...


1

"It is known" that each atom has a characteristic atomic emission spectrum, as long as the atoms are isolated from one another. Emission spectra are usually observed in gases at low pressure. But when the atoms are compressed into solids or liquids, the close proximity of the atoms distorts the environment in which the emission takes place, and shifts the ...


2

Yes and no. Fusion inside the sun produces light - but the atom are moving so fast that their electrons are not attached - it is a plasma. As such, you would be hard pushed to find emission lines in the sunlight. You will see some absorption lines - the colder hydrogen and helium further out will absorb little bits of the radiation. What you are left with ...


-1

The sun is a ball of gas, so "relative to its surface" and "walk upright" are a bit of a problem. You might fire a rocket engine hard enough to balance the force of gravity, and stand still. The acceleration of gravity gets weaker at large distance from the sun, but never is 0. $a = GM/r^2$. So you could say it is gone when it is too small for you to ...


10

Back before Copernicus (Or rather, before his view was accepted), we used to think the earth was the center of our solar system. Therefore, if you search for those models, you can find examples such as: This is, of course, based on observations rather than calculations, but it represents the complication of the solution nonetheless. (Image taken from ...


12

You certainly could define your origin of coordinates to be the center of the Earth. It would be a little tricky, because this would no longer be an inertial frame of reference, so there would be fictitious forces (or Coriolis forces). That is, your equations of motion would no longer look the same. One reason the standard barycenter frame of reference is ...



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