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In the chapter The Ultraviolet Radiation Environment of Earth and Mars: Past and Present, the author describes that, despite being closer to the UV source (the sun), Venus's atmosphere is so dense that no UV-C or UV-B radiation reaches the surface of the planet and only very small amounts of UV-A penetrate. As for Mars, the very thin $CO_2$ ...


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The Earth's atmosphere stops most Gamma Rays. It is "as thick to gamma-rays as a twelve-foot thick plate of aluminum". The Gamma Rays that make it to our atmosphere and impact another particle are absorbed. Secondary particles are produced in this interaction, and these particles can be more penetrating and damaging. Source: Gamma Rays and Our Atmosphere ...


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That's a good question and I think the answer may surprise you. It turns out that indeed, there's a lot of gamma ray radiation being produced in the sun's core from fusion reactions, so why are we not bombarded by gamma ray radiation? Those gamma ray photons need to escape from the sun's core, into the outer edge, and then finally from the surface. These ...


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Actually, we don't need a lot of protection from solar gamma rays because they never reach us. Here's Wikipedia's take on the matter: Although the Sun produces Gamma rays as a result of the nuclear fusion process, these super-high-energy photons are converted to lower-energy photons before they reach the Sun's surface and are emitted out into space. As a ...


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As you can read on wikipedia Sun produces Gamma rays as a result of the nuclear fusion process, these super-high-energy photons are converted to lower-energy photons before they reach the Sun's surface and are emitted out into space. As a result, the Sun does not emit gamma rays. The Sun does, however, emit X-rays, ultraviolet, visible light, ...


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All celestial bodies lose atmosphere due to a portion of the gas "near space" exceeding escape velocity. The velocity distribution of an ideal gas can be found using the Maxwell-Boltzmann distribution. So an easy approximation for this problem is to say we only want $10^{-6}$ of the molecules to have escape velocity. Using oxygen at 300K, results in an ...


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I believe the confusion is that you believe pressure will always increase as temperature increases. This is only the case in a closed environment such as inside the tire. In an environment such as the atmosphere which is, essentially, in an unconfined environment, the density will decrease with temperature as well. This does not happen inside of a closed ...


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Small elaboration on t.c's answer: If you say that $$F = \frac{G\ M_e}{R_e^2}m = g\ m$$ at the surface of the earth $R_e$, then at other heights you can write (for $h << R_e$): $$\begin{align} \\ F &= \frac{G\ M_e}{(R_e+h)^2}m\\ &= \frac{G\ M_e}{R_e^2(1+\frac{h}{R_e})^2}m\\ &\approx\frac{G\ M_e}{R_e^2}(1-2\frac{h}{R_e})m\tag1\\ &= g ...


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You can use the Newton's law of gravitation which is given by: $$F = G \frac{m_1 m_2}{r^2}=\left(G \frac{m_1}{r^2}\right) m_2$$ The gravitational constant, $G$, the weight of Earth, $m_1$, and the radius are constants, so: $$G \frac {m_1}{r^2}=(6.6742 \times 10^{-11}) \frac{5.9736 \times 10^{24}}{(6.37101 \times 10^6)^2}=9.822 = g$$ You can adjust $r^2$ ...


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Here's a simple, non-mathematical, answer. Although the pressure at the surface does depend on the velocity of air molecules that's not the whole picture. It is more precise to say that it depends on the rate of collisions. The collision rate depends on the velocity of the molecules, i.e. the temperature. But it also depends on density of molecules. Higher ...


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According to the MTU webpage Speed of Sound in Air, some things to consider: if the ideal gas model is a good model for a real gas, then you can expect, for any specific gas, that there will be no pressure dependence for the speed of sound. This is because as you change the pressure of the gas, you will also change its density by the same factor. ...


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Why does the air pressure at the surface of the earth (resulting from collisions of molecules on the surface of the earth which has to do with the velocity of the particles) exactly equal the weight of the entire air column above it (which just has to do with the number and mass of the molecules in the air column)? That's not exactly true. Deviations ...


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Suppose the pressure at the Earth's surface is $P$. Consider an air column of cross-sectional area $A$. The upward force on the column is $F_{\text{up}}=PA$. Denote the weight of the column as $W$. By definition of "weight", the downward force on the column is $F_{\text{down}}=W$. Suppose the pressure is too low, such that ...


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This is a very old question, but none of the answers fully address the question. I'll frame my answer in terms of answers to a series of questions: How much does temperature vary with altitude? Why does pressure vary with altitude? Why does temperature vary with altitude? What about the second law of thermodynamics? Why is the Tibetan Plateau so cold so ...


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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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Molecules in the outermost layers of the atmosphere are always reaching escape velocity - but there is sufficient statistical fluctuation that you will never, ever be able to demonstrate that your shout made a particular molecule escape. Let's do some math. Assuming that your sound wave is still a sound wave (rather than a shock wave) when it leaves your ...


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I'll just give a short outline (many caveats though): Energy in sound waves drops off as the square of the distance (a sound wave spreads out as a sphere from your mouth). If we do not take dissipation into account, you need to compare the maximum energy of your shout and divide it by $R^2$ with $R$ the distance you want to consider. Compare the kinetic ...



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