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The orbit of the Sun around the Galaxy is quite complicated, because unlike the solar system, the mass is not completely concentrated at the centre. So, in addition to the roughly circular 230 million year orbit in the plane of the Galaxy, there are superimposed motions in and out of the plane and towards and away from the Galactic centre. These roughly ...

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Based on what I can access right now, the precession of Earth's axes are in a cone shape, so there isn't much wobble. But with all of those Near-Earth Objects, and nearby planets, there is probably some wobble (I'm no professional). There's also this thing called the Milankovitch Cycle, which shows how the combined effects of precession, external forces, ...

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There is quite simple formula that will give you the tidal-locking half-time $$T = C\, \frac{a^6 R \mu}{ M_s M_p^2}$$ $a$ - semi-major axis, or simply the radius of the planet circular orbital trajectory in meters $R$ - planet radius in meters $M_s$ - planet mass in kg $M_p$ - parent planet/star mass in kg $\mu$ - rigidity, approximately $3×10^{10}$ for ...

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The first part of your question has been asked before: Is Sun a part of a binary system? and the current (lack of) evidence for such a companion is discussed on the relevant wikipedia page about "Nemesis". To summarise: if it were a small companion star, or even a brown dwarf that had been cooling for 4.5 billion years and it had a 26 million year orbit, ...

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Revised: There are several reasons why the sun cannot be part of a binary system (even if the orbit is 26 million years). If the star has the mass of Sol, the binary system would orbit around the point equal distant from both masses. This would definitely show a change in parallax for some of the stars. A Sol size star in a 26 million year orbit (about ...

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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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