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Hot answers tagged stellar-astrophysics

37

It depends on what object it's acting on. There are many objects, including stars, that have magnetic fields where Lorentz forces on charged particles like electrons and protons are stronger than the gravitational force on them. Also remember that the strength of the Lorentz force depends on the speed of the particle moving through it, so a fast enough ...

22

Am I correct in saying that the fusion process of the sun is constant, i.e. X amount of fusion happens per day, more or less? Yes, at least over human timescales. You could reasonably expect the fusion rate within the sun to be the same today as a few thousand years ago or into the future, give or take some small fraction. Why does this not speed up, i....

17

Let's look at the proper magnetic force (as opposed to the Lorentz force on a moving, charged object described in @KenG's answer) on a specimen $S$ of magnetized material with mass $M_S$ as a way to try to compare. Let's arbitrarily assume it has a fixed, permanent magnetic moment $m_S$. We can't use iron because it will saturate too easily. Then let's ...

14

The total entropy actually increases, as the molecular cloud shrinks under gravity. It may seem that as the molecules are getting closer, they are more ordered, which means less entropy. That is however only one part of the process. The second (important) part is: when the molecules are closer, they also have higher kinetic energy (since they descended into ...

13

No, the fusion rate of the Sun is not absolutely constant in time. The Sun is gradually becoming more luminous and that luminosity is provided for almost exclusively by fusion in the core. However, the rate of increase is not large, of order 10% per billion years. The fusion process is extremely slow (and inefficient in terms of energy release per unit ...

12

Short answer: Without tunnelling, stars like the Sun would never reach nuclear fusion temperatures; stars less massive than around $5M_{\odot}$ would become "hydrogen white dwarfs" supported by electron degeneracy pressure. More massive objects would contract to around a tenth of a solar radius and commence nuclear fusion. They would be hotter than "normal" ...

9

No. There is no consensus. The discrepancy between the predicted big bang nucleosynthetic abundance of Lithium 7 and the measured value can be summarised as follows. If we take what we know about the the baryonic mass density of the universe and the Hubble constant, we get a self-consistent picture between the cosmic microwave background, observations of ...

9

It isn't impossible, but the short answer is "no". A gravitational field will accelerate all matter and energy equally while a magnetic field will only accelerate moving electric charges (other magnets). The force due to gravity is proportional to the inverse square of the distance, and the force due to magnetism asymptotically approaches the inverse cube ...

8

Let me try to add some numbers to Steve's answer. The Sun's luminosity is about $L_{\odot}=4\times10^{26}\text{ J/s}$. Now, if we assume that the majority of that energy comes from nuclear fusion, we have $$L_{\odot}=\frac{dE}{dt}=\frac{d(mc^2)}{dt}=c^2\frac{dm}{dt}=c^2\dot{M}_{\text{nuc}}$$ Therefore, we can write the mass-loss due to nuclear fusion as $\... 7 There is no specific answer to this -- anything from "just sits there" to flys away at high speed is possible. It all depends on the symmetry of the supernova (SN) explosion. Extensive modelling shows that the explosions can be quite asymmetrical, and if they are the gravitational waves created can give the new black hole (BH) quite a kick. If the ... 7 The Earth is a moving (actually, accelerating) platform from which we make our observations. If you want to describe the motion of a distant celestial body, then it does not make much sense to provide a geocentric velocity, because this will depend on exactly when the observations were taken (because the Earth orbits with a speed of about 30 km/s, but the ... 7 So if I read your question correctly, you're asking why as stars get dimmer, they are given higher magnitudes? The reason is purely historical. The ancient greeks assigned stars with 6 brightness levels or magnitudes. The brightest stars were of first magnitude and the least bright stars (to the naked eye) were of 6th magnitude. In the 1800's this system ... 6 The slowest reaction rate in the pp chain determines how quickly hydrogen can "burn" in the core of a sun-like star. That rate-determining step is actually the fusion of two protons to form deuterium via the diproton and a weak interaction decay. The fusion of lithium, whereby it fuses with a proton and then splits into two Helium nuclei is actually part of ... 6 Plasma consists of ionised particles. The electrons are not bound to their nuclei and are free to move around within the star. However, they are not free to leave the star. Without the electrons to maintain a neutral charge, there would be massive electrical repulsion from all the positive protons and helium nuclei, the result would probably be something ... 6 What is it? An IMF,$\Phi(m)$, is defined such as$\Phi(m){\rm d}m$gives the fractions of stars with a mass between$m - {\rm d}m/2$and$m + {\rm d}m/2$, and with a normalized distribution $$\int_{m_{\rm min}}^{m_{\rm max}}m\Phi(m){\rm d}m = 1\ M_{\odot}.$$ Note that these boundaries ($m_{\rm min}$and$m_{\rm max}$) are ill-defined, but typically of ... 6 This is basic thermodynamics. When you compress a gas, you inject energy into it. Think of the pump you use to inflate the tires on your bike. It takes some force to move the piston, right? That effort is not wasted, but goes directly into the air in the pump. Now the air has more energy. But what happens to a gas when you put energy into it? It's ... 6 I think that there isn't a strict answer to this question. However, I believe the answer is that there's a difference between the core of a hydrogen-burning star and the core of a protostar or star-forming, gas cloud. For a hydrogen-burning star, the core, as you say, is the region of the star where fusion is taking place. This is surrounded by the ... 6 The energy derives from gravitational potential energy. The core of a bit more than a solar mass collapses from the size of the Earth to a 10km radius. Some of the gravitational energy (a tiny percentage) released is transferred to the overlying envelope and blasts it into space. Further energisation takes place due to radioactive decay. A ball of neutrons ... 6 Neutron stars are one of the possible end products of the evolution of stars greater than around 8 solar masses. If you start out with a close binary pair of these fairly massive stars -- not common, but not rare, either -- the more massive star will evolve to a red giant and tides (or even friction) in the extended envelope will pull them closer together.... 6 We use the position of the sun (or more accurately the centre of mass of the solar system) as this gives us a very nearly inertial frame of reference. An inertial frame is one which is not changing its velocity. The surface of the Earth does not define an inertial frame. The Earth orbits the sun and so it is moving in opposite directions in summer and ... 5 Main sequence stars are defined by being hot enough in the core to fuse hydrogen, so their core is at least about 10 million K, and can get up to 20 million K for the more massive ones (because they are more luminous, so their nuclear fusion has to crank itself up a bit more). The way they get their cores hot enough to fuse hydrogen is via gravitational ... 5 In general, you can't. If obtaining spectra in regions where there is expected to be a spatially varying background then you either need to do long-slit spectroscopy so that you have a good measurement of the ISM contribution either side of your source, or you do integral field spectroscopy with the same idea. The problem is that the line strengths for the ... 4 This comes from a misunderstanding of local and absolute. There is nothing to prevent a local increase in order - overall, order still decreases (or in common terminology, entropy increases) From Wikipedia: According to the second law of thermodynamics the entropy of an isolated system never decreases, because isolated systems spontaneously evolve ... 4 This question is very broad - there are very many techniques for estimating temperatures, so I will stick to a few principles and examples. When we talk about measuring the temperature of a star, the only stars we can actually resolve and measure are in the local universe; they do not have appreciable redshifts and so this is rarely of any concern. Stars do ... 4 The Kaggle galaxy zoo challenge is an example of a problem begging for ideas from outside the field. Sander Dieleman, with a background in deep learning and feature learning, bravely stepped forward, creating an image classifier utilising convolutional neural networks; his full solution is described fluently here. These kinds of techniques could be applied ... 4 If you have good knowledge of software development and pattern recognition, there are several problems that you could assist in solving. Much of observational astronomy requires long time series data, and removing the noise from this data. I have just left the field where some colleagues are trying to develop some software to use image subtraction techniques ... 4 The wording provides a contrast with the Hayashi track phase that immediately precedes it, where the luminosity decreases by orders of magnitude with little change in effective temperature. You are correct though, the Henyey track for low-mass stars is not horizontal in the HR diagram, however it is a critical point that the luminosity does not change very ... 4 Yes, there are electrons inside stars, they have not gone anywhere. I must say your new theory of electromagnetic force does not make any sense to me, but there are actually a few cases where electrons "disappears" from a star: First off, the nuclear fusion of 4 hydrogen atoms into one helium atom actually consumes electrons. Let us count the particles ... 4 The effective temperature$T_\mathrm{eff}$of a star, which is presumably what's been plotted, is defined through its relationship with the star's radius$R$and luminosity$L$by $$L=4\pi R^2\sigma T_\mathrm{eff}^4$$ This comes from the assumption that the star radiates like a black body at the photosphere. While this isn't strictly true, it's quite ... 4 Your feeling is right: You shouldn't convolve the spectrum and the filter, you should only multiply so that flux outside the bandpass is suppressed. Subsequently you integrate the resulting function over wavelength, so that flux density (in energy/time/area/wavelength) becomes flux (in energy/time/area). Simply setting the flux to 0 outside$\lambda_1$and$...

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