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2

For the eye, draw a picture. You have an isosceles triangle with the small angle one arc minute and base 10 AU. What is the altitude? Bisect the small angle and you have two right triangles. For the second, you need to compute the small angle of the same triangle from the diffraction limit on a diameter of 2.4 meters. Can you find that equation in your ...


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Um, a snarky answer is that now neither is proper. As Anna notes, the $r=0.20$ value assumes no contribution from galactic dust polarized emissions ("foregrounds"), while $r=0.16$ results from BICEP2's indirect estimate of such foregrounds. However, now Planck has published a much more direct measurement of these dust emissions, and they're much higher ...


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So first of all, $\Phi(L)dL$ is, as you say, the number density (per unit volume) of galaxies with a luminosity between $L$ and $L+dL$. The constant $\Phi_0$ is a normalisation constant, and defines the overall number of galaxies per unit volume. To calculate the average luminosity we simply add up the luminosity from each galaxy and divide by the number ...


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Answer to your old question: Right now, the perihelion currently occurs around the same time as (2 weeks after) the northern winter solstice, but 10000 years from now it will occur around the northern summer solstice. So, like other people have said, it's just coincidence that right now they both happen at about the same time. Answer to your new question: ...


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Is this not a question about human physiology rather than physics? We are discussing the variations in observed response of the human eye to light with different qualities: overall intensity, and distributions over the visible spectrum. This question seems to ascribe all of the observed differences (or lack thereof) to the quality of the light; "the Moon ...


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But I've never seen that happen. You haven't looked then. The rising or setting Moon is rather reddish, just as is the rising or setting Sun. However, there is a difference between the Moon and the Sun. You can look directly at the Moon, even a full Moon, regardless of where it is in the sky. On the other hand, you can only glance at the Sun when it is ...


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This is just an opinion, but the moon on the horizon is simply less visible than the sun is. I suspect that color changes it makes are more subtle and less easily noticed. However full moons are often noticeably orange. Here is a page with a wonderful time lapse view. http://www.pikespeakphoto.com/moon-rising.html


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Neither the antumbra nor pentumbra have a fixed brightness. It varies across the region based on the amount of the solar disk that is visible. Each has regions where it is brighter than places in the other. It does not make sense to call one "brighter" than the other.


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I'm going do do the annoying teacher thing where I answer a question with a question. Which is darker, a partial solar eclipse, in the penumbra of the moon, or a planetary transit, in the antumbra of an interior planet?


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In the penumbral area, part of the disc of the moon obscures a section of the sun. In the antumbral area, the entire disc of the moon obscures a, logically, larger section of the sun. Therefore the antumbra cannot be brighter than the penumbra during the same event.


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I did some research and I think I can answer my question myself now, after all. I hope you find it interesting. As it turns out, the technology of the Kepler space telescope would indeed allow detection of all Solar system planets except Mercury and probably Mars, i.e. all of them are big enough to be seen by it from a distance of about 2,000 ly. However, ...


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Well the light which is reaching there now, the light which is being detected by a telescope set up in the Kepler planets was emitted from the Sun's solar system 2000 years ago since light took 2000 years from here to reach Kepler planets. So any event which occurred 2000 years ago in our solar system will be visible now in the Keplers. If we are (somehow) ...


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Have a look at this article. It gives the number as $10^{24}$ rather than $10^{23}$, but it's such a vague estimate that a factor of ten is within the expected error. The number is the number of stars in the observable universe i.e. within 13.7 billion light years of Earth at the time the light we see today was emitted. Note that visible means visible to a ...


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Here are the systems I found: 6: ADS 9731 Beta Tucanae Gamma Velorum Kappa Tauri Mu Sagittarii 7: AR Cassiopeiae Nu Scorpii ... no physical multiple stars of greater multiplicity yet found.


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Even without setting $\sin \theta=1$ you have a cubic equation for $m_2$. Such things can be solved, like the quadratic formula but messier. Various computer algebra packages incorporate the solution. For graphing, you can just pick a bunch of $m_2$ points, solve for $m_1$, and plot the points.


3

The first definition of $\mu=GM$ is the standard definition of the SGP. The second one comes from the velocity of a circular orbit. If you have an object in a circular orbit of radius $r$ and velocity $v$ around a body of mass $M$, then the velocity is given by $$v=\sqrt{\frac{GM}{r}}$$ From this you can see that $rv^2=GM$ for circularly orbiting objects. ...


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The Hunt for Exomoons with Kepler Project has so far failed to find any exomoons. This is a negative finding (so far). Negative findings are always a bit trickier to explain than are positive findings. This negative finding might mean something very significant, or it might have very little significance: Maybe exoplanets are much less likely to have ...


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You are looking for the initial mass function (IMF). This tells the probability of finding a star of mass $m$ (in solar mass units). The prototypical IMF is the Salpeter IMF, $$ \phi(m)\sim m^{-2.35} $$ This gives a decent and quick approximation, though a multi-power-law fit seems to be better; thus we have the Kroupa 2001 model: $$ \phi(m)\sim ...



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