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How much? Well how accurately do you need it? How do you want it quantifying? And in what wavelength range? Jupiter scatters a fraction of its incident sunlight. It also has its own luminosity (predominantly in the infrared). A quick calculation: The solar constant (flux at 1 au) is about 1370 W/m$^{2}$. Jupiter is situated about 5.2 au from the Sun (it ...


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The evidence is not only in Earth's climate but Earth's magnetic field too. For more evidence you can search for yourself too. Of cource there is the complete physical mechanism described, not just correlations. And the mathematical documentation is overwhelming. Pela: there is no such thing as self-tidal force, the only tidal forces to the Sun are those ...


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May I try to contribute to that conversation? Indeed the forces from planets are very weak BUT what is the sun? Gas collapsing because of gravity on one hand and on the other hand expanding due to nuclear reactions. All this in an equilibrium. So this small planetary power seems to affect the sun, in the absence of other stronger powers. An analogy is that ...


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The Sun does NOT revolve around an even bigger star. If it did, you'd see two stars, assuming the larger one didn't outshine the Sun. In that case, you'd only see one. Also, the earth would be a lot hotter, getting heat from two stars. We'd also be in danger of solar wind, more solar wind than we'd usually get from just the Sun. So, because we don't see two ...


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Does the Sun turn around a big star? No. Such a star, if it existed, would easily be the brightest star in the sky. You would have been taught about it early on in school if it existed. But it doesn't. For a while it was conjectured that the Sun had a small companion star to explain a perceived periodicity in mass extinction events. This too has been ...


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The Sun is not within the gravitational sphere of influence of any other star. The centre of mass of the solar system (which is very close to the Sun) instead orbits in the general Galactic gravitational potential. Because this has a roughly cylindrical symmetry (the Galaxy is basically a disk with a bulge in the middle), this means that it executes a ...


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To a first approximation when we calculate how fast an object is orbiting around some mass distribution we can assume that the gravitational attraction it experiences is only that due to the mass interior to its orbit. This approximation, known as the shell theorem should only be applied when the mass distribution is spherically symmetric, or most of the ...


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The answer to your simple question is that while a circular orbit could remain in the absence of any other gravitational fields, in reality you'll have an ellipse and it can be stable from the beginning, remember stable has a wide range of values. If you allow drag through tidal forces etc then the orbit will change, eg our Moon is slowly getting further ...


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The gravitational effect of the planets on the Sun can be calculated very accurately. The acceleration of a planet of mass $M$ at a distance from the Sun of $r$ is $$a = \frac{G M}{r^2}.$$ If you plug in the numbers, you'll find that the largest effect comes from Jupiter due to its enormous mass. The gravitational acceleration of Jupiter on the Sun is ...


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Both Mercury and Venus undergo retrograde motion, but are visually close to the Sun when this happens, so we don't notice it. Below is Mercury's right ascension from May 15 to July 15 of this year. Note that it starts off increasing, then decreases (retrograde motion) and then increases again. ...


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TL;DR Not for planets, but for other stuff yes. As explained in an earlier answer, planets are on circular orbits (quasi-circular actually but that's a detail). The closer to the star, the faster the planet is on its orbit. So, the retrograde motion observed for Mars cannot happen for Venus for instance. At most you could see Venus going retrograde when it ...


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You would need to have a planet orbit very slowly around the star. A planet moves retrograde because the planet is passing the other in its orbit so that the slower planet appears to be going backwards. So in order for this to happen you would need a planet having an orbit such that the outer planet passes the inner planet. But I think that this would ...


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The Wikipedia Precession article explains this. Precession of the Earth's axis is caused mainly by the Sun's gravity acting on the oblate spheriod of the Earth. The Earth bulges at the equator and the force of the Sun's gravity on the bulge makes the axis wobble like a child's spinning top - the axis traces out a cone shape in space. It has no effect on the ...


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Note: Answering from a comment posted on Space Exploration The classical stability analysis of these libration points assumes that we are examining the motion of a particle whose dynamics are perturbed by the gravitational impacts of a primary and secondary mass, so as a bottom-line-up-front type of answer, the mass of T is negligible - so any large ...


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Narrow answer: if all masses were equal, it would be a 3-body Klemperer rosette. It is known that such trivial KR configurations are not stable in the long term. http://en.wikipedia.org/wiki/Klemperer_rosette


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Kepler's account of how the third law came to be is as follows (Caspar p.286; emphasis mine): On the 8th of March of this year 1618, if exact information about the time is desired, it appeared in my head. But I was unlucky when I inserted it into the calculation, and rejected it as false. Finally, on May 15, it came again and with a new onset conquered ...


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Kepler's third law is trivial (in my opinion) compared to his first law. I am quite impressed that he was able to deduce that the orbits were ellipses. To get that, he had to go back and forth plotting Mars' direction from Earth and Earth's direction from Mars. He knew the length of both planets' years, so observations taken one Mars year apart would differ ...


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A circle is an ellipse with eccentricity zero. And in fact tidal evolution can drive orbital eccentricity to values negligibly close to zero. See Regarding the Putative Eccentricity of Charon's Orbit. From observations we already know Pluto and Charon move about each other in very nearly circular orbits. I expect when New Horizons flies by the system in ...


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You've been given an answer, and it's perfectly valid, but here's something from a different perspective (less strict). A circle is really just a particular case of an ellipse. Take an ellipse, and change it, by moving its focal points closer together. When those two points coincide, what you get is a circle. It's still an ellipse, technically - one that ...


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Why is it impossible for any planet (or moon, by the way) to orbit another body in a perfectly circular path? One way to look at it is from the perspective of probability and statistics. Think of position and velocity as random variables drawn from some continuous probability distributions. Given some position vector, the velocity vector has to have a ...


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You've asked a big question, too big perhaps for a Q&A forum such as this. Your question is the sole subject of graduate level aerospace engineering classes, e.g., University of Colorado ASEN 5070, Introduction to Statistical Orbit Determination, and is the subject of multiple graduate level texts, e.g., Statistical Orbit Determination by Bob Schutz, ...



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