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1

I don't think you can draw such conclusions from a simple graphic. They've probably just cut it off at somewhere around 200nm because the power output at shorter wavelengths is almost irrelevant. If you want a more accurate graph for shorter wavelengths, try this one. You can set it up for 5900K and graph values below 240nm.


2

It seems to me that the person drawing the graph was a bit sloppy - the ideal black body radiation ("idealer Schwarzer K├Ârper" - Temperatur 5900 K) does not cut off sharply at 240 nm as shown. Instead, it should look like this: when calculated from Planck's Law. I suspect some bug in the method used to calculate the values in the plot you reproduced - ...


1

It will simply be swallowed up. There's no fall involved. All the material that once made up Mercury will become part of the mass of Sun and presumably over time diffuse completely into that mass, so that no trace of Mercury will be found, even though its actual matter hasn't been destroyed.


0

The core of the Sun is radiative. That means that energy is transported outwards primarily by photons and is stable to convection. This means, to first-order, that the centre of the Sun is not mixed up by convection. As the hydrogen in the core is burned, it forms helium, which has a larger atomic mass. The helium sinks towards the middle and the core fills ...


8

The image below represents the Sun's density gradient, which shows how the density changes with the radius. The ground we stand on should have a density between 2 to 3 $g/cm^{3}$. That should put you just above the water point on the vertical axis. The corresponding radius is then about 0.45 of the solar radius. Note that the vertical axis is in a ...


2

As @KyleKanos point out, "the answer is a google search away", but it's not quite as simple as he suggests. The mean density doesn't answer your question (the mean density turns out to be about 1.4 times the density of water by the way). An ill-defined idea in your question is the "surface" of the sun. Where is the "surface" of the sun given that none ...


0

In about four billion years the sun will have used up most of the hydrogen in its core and fuse helium into heavier elements. The fusion of helium releases more energy than the fusion of hydrogen. This will cause the sun to expand to, at least, the orbit of Venus and likely the Earth. Both planets will, of course, be vaporized.There was a similar question ...


0

In a dilute gas, the photon density should be the same as inside an evacuated black box of the same temperature (independent of the gas density). Edit: In other words, the massive particles in the Sun have a temperature based on their kinetic energies. The photons must be distributed as black-body radiation at the same temperature. The density of photons in ...


2

They do radiate light randomly in all directions -- in the object's own reference frame. But not from the Sun's reference frame. The effect in general is called "relativistic beaming." Here's the clearest derivation that I know. Take the Pauli matrices $\sigma_i$ and adjoin the identity matrix to them as $\sigma_0 = I$. Now take a four-vector $v^\mu$, and ...


7

By "consume" we mean "convert into helium." That $6\times10^{11}\ \mathrm{kg}$ of hydrogen is part of the Sun (specifically it is found in the core of the Sun), and it is converted into $6\times10^{11}\ \mathrm{kg}$ of helium. The Sun doesn't need to suck up material from space. Note that this amount of material is miniscule compared to the $2\times10^{30}\ ...


3

It's a bit of a puzzling question. I'll try to work it out, but one of the tricky parts is that atoms absorb and emit photons all the time, higher temperatures emit higher wavelengths. Photons created in the sun (per second) can be estimated, but those are fusion gamma rays. The sun burns about $564$ million tons of hydrogen per second. (Source), and 1 ...


-1

The number of atoms in the sun is on the order of $10^{57}$, see here. The number of photons emitted per second is on the order of $10^{44}$, see here. The difference is on the order of $10^{13}$. So if photons emitted for $10^{13}$ seconds, 315,00 years, the number of photons would begin to overtake the number of atoms in the sun. The sun is much older ...



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