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## Hot answers tagged mathematics

20

There are several factors determining the inner limit to moons. Perhaps the simplest is that it needs to stay inside the Hill sphere, the region around the planet where the planet's gravity dominates over the sun's. If the planet's orbit has semi-major axis $a$ and eccentricity $e$ the farthest the moon can orbit is $$r_H \approx a(1-e)\sqrt[3]{\frac{m}{3M}}... 16 Both ellipticity f (also called flattening) and eccentricity e are measures of how elongated an ellipse is, based on the semi-major axis a and the semi-minor axis b (figure from wikipedia). They are defined respectively as$$f=\frac{a-b}{a}$$and$$e=\sqrt{1-\frac{b^2}{a^2}}$$For a circle, a=b, which implies that f=e=0. In modern orbital ... 16 The whole sphere has approximately 41,253 square degrees of solid angle.$$4\pi\left(\frac{180}{\pi}\right)^{2}\approx 41,253$$so for a hemisphere there should be half this number or about 20,627 deg2. I think you computation is missing the 4\pi steradians in a sphere term. This doesn't solve the disparity however. Perhaps the key is the term "... 14 The relationship between a, b and the eccentricity of an orbit e is$$ a = b\left(\frac{1+e}{1-e}\right)\ .$$It is clear therefore from a mathematical point of view that you cannot get a similar constant relationship by replacing a^3 by b^3 in Kepler's third law. You would have to use an expression involving both b and e. Another way of saying ... 13 Scale invariance and self-similarity Power laws basically mean that there is no preferred scale, i.e. that a physical property is scale invariant. Any deviation from a power law means that the Universe somehow thinks that the scale where it breaks down has some special significance. In other words, a power law describes self-similarity^\dagger. You can see ... 13 I am assuming that by adding 1 km/s, you mean increasing the tangential speed of the Earth by 1 km/s. This would increase both the kinetic energy and the angular momentum. This is a relatively small increase in both quantities and it would send the Earth into a slightly more eccentric orbit than it has now. It would not come close to Mars. Using the vis-viva ... 9 Ellipses have a "long radius" called the "semi-major-axis" which is the length from the centre to the ellipse measured along the long axis. And a "semi-minor-axis" which is measured along the short axis. Call the semi-major-axis "a" and the semi-minor-axis "b". Ellipses also have foci: which is where the ... 9 Consider this gif from wikipedia. All the orbits in the animation have the same orbital period T and the same semi-major axis a, but different semi-minor axes b_1,b_2...b_5. This shows that the orbital period is independent of the semi-minor axis. As an analogy, imagine twin figure skaters spinning with the same angular momentum. The red figure ... 9 Consider a child on a stationary swing. The fastest way to get them going is to push once every time they swing (a 1:1 resonance). If you push 581 times for every 137 swings, the pushes will mostly average out and the child won't get very high on the swing. Similarly, there are infinitely many orbital resonances possible, but when the integers in the ... 8 I have to admit that power-laws (in general) used to be my shtick so I am happy to shed some light on their general importance in physics which obviously also hold for astronomy. The main idea of a power law is nicely written in Wikipedia, but the essential part is the I highlighted in the following quote: [A Powerlaw is] a functional relationship between ... 8 GrapefruitIsAwesome has already explained why the sky is significantly larger than 3300 square degrees; I'd like to explain why the sky coverage is precisely the value it is. The wording is admittedly not ideal. The portion of the sky visible to the surveys are limited by 1) dust in the Galactic plane that prevents them from observing extragalactic sources ... 7 Kepler's third law is not that a^3/T^2 is a constant. It is that$$ \frac{a^3}{T^2} \propto (M_\odot + M_{\rm planet}) $$and so the left hand side depends on which planet you are looking at and should be larger for Jupiter than it is for the Mars for example. Does that account for your discrepancy? Almost. It should change the 4th significant figure, ... 6 Note that the Sun here is point-like and there is no refraction. What have I done wrong? Your supposition that the altitude of the Sun is directly related to the length of the day is wrong. Proof with counterexample: observe the altitude of the Sun on the equator and in the Moscow on the equinox; but still, the lengths are the same. And what is the correct ... 6 Using calculations from here: https://www.vanderbilt.edu/AnS/physics/astrocourses/ast201/orbitalvelocity.html New semi-major axis$$ a = \frac{150000000000 \cdot 0.0000000000667 \cdot 2\cdot 10^{30}}{2 \cdot 0.0000000000667 \cdot 2 \cdot 10^{30} - 150000000000 \cdot 30000 \cdot 30000)}\\ = 151.821\,\mathrm{million\,km} $$New orbital period$$ p = \frac{\...

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In a sense, even solving the two body problem as a function of time is unsolvable in terms of the elementary functions. The problem is that the solution involves the solving for the inverse of Kepler's problem, $M = E - e \sin E$. This inverse function is transcendental and cannot be expressed in terms of the elementary functions. That said, it's fairly easy ...

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I can show you an ESA Series exercise I did a few months ago for an astronomy class. Given the light curve of 12 cepheid variable stars in the galaxy M100 (which are very nice standard candles to measure large distances): We can find the distance between us and the M100, a spiral galaxy in the Virgo cluster, using the apparent magnitude m and the absolute ...

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There are a couple of oddities I see in this working. I'm not completely clear if $l_0$ is the length at zero celcius or at the equator, at 50 celcius. But I'll go with your interpretation, that 300 = length at 50 celcius. The temperatures seem extreme and unreasonable. It is warm at the equator and cold in Antarctia, but +50 to -90 seems rather unreal! ...

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From various sources such as Wikipedia, NASA, and various published papers, an orbital resonance is: when orbiting bodies exert regular, periodic gravitational influence on each other, usually because their orbital periods are related by a ratio of small integers So a ratio of $1:\sqrt{2}$ wouldn't count. What's more it wouldn't even make sense to talk ...

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When the planet is at one end of the major axis, that's an extremum with respect to time of distance from the sun (either aphelion or perihelion, depending which end of the major axis). That means that, at that point, the radial component of the planet's velocity is zero, and therefore there is no contribution of radial motion to the kinetic energy. Hence, ...

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I wrote the source you need some years ago: https://jumpjack.wixsite.com/progetti/sorgenti-ipsun The Arduino/Processing version was just a demo program to manually control a TENVIS camera by multiple buttons, it lacks the "astronomical algorithm". The Javascript version contains astronomical calculations and a demo page which connects to a local ...

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Short Answer: There is an inner limit to how clase a palnet could orbit to its star and keep a moonin orbit around it. But I don't know how to calculate it. As far as I know there is no outer limit to how far a planet could be from its star and have moons. A rogue planet in interstellar space far from a star could have moons. Long Answer: Every planet ...

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This example has less to do with scaling relations, and more to do with dimensional analysis - theoretical physicists like to express equations in terms of dimensionless ratios, because then it is really easy to plug in numbers and get a sense of how the equation behaves for certain parameter values. In this case, you have Kepler's third law, which is a well-...

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Yes. Assuming that the moon is distant enough to neglect parallax effects, the semi major axis is just the radius of the moon, $r$, and the semi minor axis is $|r\cos\theta|$ where $\theta$ is the Sun-Moon-Earth angle. So when the sun is behind the Earth, the angle is 0 and the ellipse is a circle (full moon) so the semiminor axis = $r$. When the sun is ...

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Answer to your question RA changes just a little bit through time (because of minor effects like parallax), but we can just say it is constant for a specific non-Sun star. Same applies to declination. Why is that? The declination and right ascension of vernal equinox (where the Sun is in March) is defined to be 0°. This point is fixed with respect to other ...

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The polytrope model of stars made early numerical calculations of stellar structure possible using anything from mechanical calculators to early electronic computers, and in certain cases, even analytically! In astrophysics, a polytrope refers to a solution of the Lane–Emden equation in which the pressure depends upon the density in the form $$P=K\rho ^{{(n+... 3 From the definitions of the Oort constants we know that$$A + B = -\frac{dv}{dr} $$at the solar radius. But A + B is small, implying that dv/dr \simeq 0^{\dagger} and we are dealing with a flat(ish) rotation curve. Assuming a spherically symmetric mass distribution with a density that is proportional to r^\alpha, we can write the centripetal ... 3 You need to do what is called “propagation of uncertainties”. You can search to get more information on that, but briefly if you have some function f(x) that depends on variable x, then the uncertainty \sigma_x on the quantity x is related to the uncertainty on f by$$ \sigma_f = \sqrt{\left(\frac{\partial f}{\partial x}\right)^2 \sigma_x^2} $$... 3 If you tilt a circular object, then along one axis the length will appear to be shortened, while along a perpendicular axis the apparent size will be unchanged. So you can use that to determine the tilt angle (inclination). If we define 0^\circ to be face-on, then when tilted at an angle i away from face-on, the shorter axis will appear to be decreased ... 3 The climate modelers' VERNAL is equivalent to this formula, where Y is an integer, ΔY = Y - 2000, and JD0 is epoch J2000 = JD 2451545.0 = 2000-01-01 12:00 TT:$$ \text{JD}_\unicode{x2648}(Y) = \text{JD}_0 + 78.813 + 365.24250~{\Delta Y} $$We can tune the coefficients to fit JPL DE431 for years -5000 to 9000:$$ \text{JD}_\unicode{x2648}(Y) = \text{JD}_0 ...

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A quick look at Stellarium (if we trust its accuracy) suggests longitudes ~90 east and west (Edit: it seems that each of these corridors get to enjoy both double sunrise and sunsets, not only one or another!). Yes, you can experience that effect near the poles, but take into account the local topography, as this is a timid effect we're talking about, and the ...

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