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The answer is simple: Yes, stars really do produce that many photons. This calculation is a solid (though very rough) approximation that a star the size of the sun might emit about $10^{45}$ visible photons per second (1 followed by 45 zeros, a billion billion billion billion billion photons). You can do the calculation: If you're 10 light-years away from ...

38

One thing to keep in mind is that objects that are bound gravitationally actually revolve around each other around a point called a barycenter. The fact that the earth looks like its revolving around the sun is because the sun is much more massive and its radius is large enough that it encompasses the barycenter. This is a similar situation with the Earth ...

32

The answer kind of depends on how old you are. At a very introductory level, say, maybe middle school or younger, it's "okay" to refer to Jupiter as a failed star to get the idea across that a gas giant planet is sort of similar to a star in composition. But around middle school and above (where "middle school" refers to around 6-8 grade, or age ~12-14), I ...

32

Anything the mass of a star is going to get hot like a star and fuse hydrogen like a star. In other words it will be a star not a planet! While it's technically possible to have a rocky planet the mass of a star, in practice when stellar systems form there aren't enough metals available to build such a large object. Large objects are invariably built from ...

25

Not quite like in the photo above, which shows more than what the naked eye can see, but yes, absolutely! Our galaxy (well, the chunk of it visible from these parts) is a naked-eye object. The fact that your question even exists shows how much time is now spent by people under light-polluted skies. It will not be visible from the city, however. You need to ...

25

Although I agree with all three of the above answers let me present a slightly different perspective on the problem. It's tempting to think of the light from the star as a flood of photons that behave like little bullets. However this is oversimplified because a photon is a localised object i.e. we observe a photon when something interacts with the light ...

21

Yes indeed to all your questions: mutually orbitting binaries do spin down, the system's orbital angular momentum thus decreases with time and the loss of energy and angular momentum is almost certainly owing to the emission of gravitational waves. Look up the Hulse-Taylor binary system: its spin-down has been carefully observed and measured since its ...

19

The estimates I've read are similar to yours: 200 to 400 billion stars. Counting the stars in the galaxy is inherently difficult because, well, we can't see all of them. We don't really count the stars, though. That would take ages: instead we measure the orbit of the stars we can see. By doing this, we find the angular velocity of the stars and can ...

17

It's because the value of the gravitational field at the center of a star is not the relevant quantity to describe gravitational collapse. The following argument is Newtonian. Let's assume for simplicity that the star is a sphere with uniform density $\rho$. Consider a small portion of the mass $m$ of the star that's not at its center but rather at a ...

17

Well, motion is relative so you can choose a frame of reference where one is stationary. If you do though, it makes the equations of motion quite complicated. Even in our solar system, the Sun isn't stationary. It orbits the center of mass of the whole solar system (barycenter), just as each planet orbits the center of mass. The center of mass of our ...

15

When astronomers started to get spectra of stars and began classifying them, the initial classification was based on the strength of the Balmer absorption lines in the spectra. The Balmer lines are created by electons in hydrogen atoms that are currently in the second energy level (N=2) absorbing energy and jumping up to higher levels. The stars with the ...

15

I work with stellar models, so I thought I'd chip in here. My instant reaction is that you shouldn't worry too much: determining the age of a star is difficult and different models will disagree (sometimes significantly!) on that age. How reliable is this research? I can't see an obvious reason to doubt the conclusion. What method do they use to ...

13

The analogy is facile. Helium fuses at a temperature ($10^8\ \text{K}$) roughly ten times higher than hydrogen ($10^7\ \text{K}$), so a better analogy would be alcohol and thermite. That higher temperature is achieved only by massive gravitational contraction after hydrogen fusion [EDIT: in the core] is exhausted. EDIT: To expand, different mass stars ...

12

Almost all exoplanets observed are near F, G, and K stars. In part, this is because astronomers are looking for earth-like planets, so they look at stars similar to our Sun, but there are also some physical reasons. Sahu et al (2006) have provided some evidence that red dwarfs (class M) are more likely to have planets than other spectral types, though it is ...

12

It seems that theoretically neutrino stars have been postulated. A google search came up with Supermassive neutrino stars and galactic nuclei. R.D. Viollier, D. Trautmann and G.B. Tupper. Phys. Lett. B 306 no. 1-2 (1993), pp. 79-85. The calculations have been done for you :) if you have access to a library: Abstract The characteristics of ...

12

The only stars you can reliably see are ones that are spewing enough photons at your eyeballs to appear stable. Any star which is so dim that photons entering your eye can literally be counted one by one, simply will not register in your vision, because your eye's retina is not sensitive enough. So your question is basically embroiled in observer bias; it ...

12

Well, you're right that a particle sitting at the centre of a star (or generally the centre of any spherical distribution of matter) feels no net gravitational force. So, in the absence of other forces, it will simply continue to sit at the centre. But every other particle in the spherical distribution will feel a gravitational force pulling it toward the ...

11

I think there are really three questions that need to be answered for this to make sense: is there a "normal" limit to how large a star can be? how can population III stars form with such large masses? how can population III stars retain their large masses? An answer to the first question is tricky. We expect large stars to be rare, and the largest ...

11

If they're sitting still, and are very bright, they are planets. Install Stellarium on a computer or a smartphone. First time you run it on a computer, enter your location in the settings (no need to do that again after the first time); on the smartphone, it deduces the location automatically each time. The program will show you what planets are visible at ...

11

Here is a nice answer, taken from http://www.enchantedlearning.com/subjects/astronomy/stars/twinkle.shtml The scientific name for the twinkling of stars is stellar scintillation (or astronomical scintillation). Stars twinkle when we see them from the Earth's surface because we are viewing them through thick layers of turbulent (moving) air in the Earth's ...

11

As others explained: When two masses interact gravitationally, it's not like the smaller mass is orbiting the larger mass. Both bodies orbit the common barycenter. When one of the two masses is extremely large compared to the other, the barycenter of the system is almost in the center of the larger mass, so the effect on the larger mass is negligible (like a ...

10

Given what we know about planetary formation (Link 1, Link 2, Link 3 and Link 4), and the theories around it, it would probably be a safe bet to say that ALL stars end up having some left over material that might become planets. I think the bigger question is how many of those planetary orbits stay stable enough throughout the life of the star? All these ...

10

For single stars (doubles also exhibit the spin-to-orbital angular momentum transfer), Rotation braking states: Stars slowly lose mass by the emission of a stellar wind from the photosphere. The star's magnetic field exerts a torque on the ejected matter, resulting in a steady transfer of angular momentum away from the star. A first order approximation ...

10

Y-dwarfs are a subtype of brown dwarfs, which don't produce energy like (or certainly don't produce as much as) normal stars. Brown dwarfs have an upper limit and a lower limit on their masses. Both these limits are informal and approximate. These limits aren't like planets, for which the IAU has an accepted definition, but they're reasonably well-defined. ...

10

The key as I understand it is metallicity. The Big Bang produced virtually nil above helium, so Pop III stars and their ancestral HI clouds had almost no metals. The forest of emission lines produced by even a tiny fraction of metal atoms acts to increase the cooling efficiency of the cloud enormously. At the extreme rate of cooling of modern clouds, ...

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Typically, a star (or stellar remnant, such as a neutron star, white/black dwarf, or black hole) will be the most massive thing in the area, by far. Planets, even gas giants, are a small fraction of the mass of a typical main sequence star. Now, as in Hal's answer, the relative mass of the planet and its star does make the center of mass, the barycenter, of ...

10

In most cases you are right, stars are pointlike. They are spread over multiple pixels not because of charge overflow (this can be overcome with shorter exposures and/or better equipment; there really should never be overflow in the image) but because the point spread function (PSF) of the telescope is larger than a single pixel (which is a good thing; you ...

10

Instead of the massive compact objects which could serve as a 'replacement' for the supermassive black hole inside the galactic center (which are discussed in the Viollier and Tupper paper from Anna's answer) I would like to point another possibility: halos of degenerate neutrino gas around galactic clusters. The order of magnitude calculations for the ...

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