The Yarkovsky effect is responsible for changes in the rotation and orbit of some celestial bodies, most notably asteroids. It has been measured on asteroids, such as 6489 Golevka and 1999 RQ36.
What is the largest object on which the Yarkovsky effect has been observed? Scholarpedia has an excellent article and list of asteroids, but the list isn't necessarily complete, and I don't know if the effect has been detected on other (non-asteroid) bodies.
Infinities are hard to bend the mind around, but in this case, the merge is not impossible. Yes, the distance between them must reach zero in order to make the black holes merge, but the rate of which energy is lost to gravitational waves also increases when they get close to each other.
We are therefore dealing with a $frac{mathrm{infinity}}{mathrm{infinity}}$ problem, where a finite limit may, and in this case does exist.
Do not forget your $lim$'s!
I've been searching on Internet about how to calculate which RA is in zenith at a given location in a given time but I haven't find anything (or maybe I don't know how to search).
Using my star finder I have not draw any conclusion.
How can I calculate which RA is on zenith on a given location in a given time?
I bought this binocular because I've read countless times that one should start with a pair of binoculars before diving into telescopes. I've seen details of the moon and I can notice Venus round shape on some nights. I also saw Jupiter and it's two larger moons, as faint as the faintest stars, but this was on a beach trip so the sky was clearer than what I'm used to.
I live on a place with a 7 or 6 on the Bortle Scale, is there something else I should try to see?
The main factor is the velocity of the encounter. The higher the relative velocity between the two galaxies, the easier it is for them to pass through each other without being slowed down enough for a proper merger to take place, and without being strongly distorted by the encounter. (In a very high speed encounter, the two galaxies will spend almost no time close enough for tidal forces to be effective.)
In clusters of galaxies, where the typical velocities are around 1000 km/s, mergers are rare; in small groups, where the velocities are around 100 km/s, mergers are more common. So the Milky Way and Andromeda are pretty much destined to merge.
During a merger it's possible for stars in the outer parts of the galaxies to be ejected by tidal forces, but the central regions of both galaxies should definitely merge. Thus, the supermassive black holes (one in each central region of the original galaxies) will end up in the center, too. (Unless you have something exotic like a merger where one of the galaxies already has an unmarked binary SMBH; then you might get one of the three SMBHs ejected via a 3-body interaction. But I suspect that's pretty unlikely.)
Not at all a dumb question, but actually you can see distant galaxies with the naked eye. From the northern hemisphere, the Andromeda Galaxy, our biggest neighboring galaxy, is visible if you know where to look, and is at a reasonably dark place. From the southern hemisphere, the two smaller, but nearer, irregular galaxies called the Small and Large Magellanic Clouds are visible.
The reason that more distant galaxies are not visible, is due to the inverse-square law: As the light particles (photons) recede from the galaxy (or any other light source), they are distributed over an ever-increasing surface. That means that a detector (e.g. your eye) of a given area will catch less photons, the farther it is placed from the galaxy. The law says that if in a time interval Δt on average it detects, say, 8 photons at a distance D, then in the same time interval, at a distance 2D it will detect 8/22 = 2 photons. At a distance of 4D, it will detect 8/42 = 0.5 photons. Or, equivalently, it will need twice the time to detect a single photon.
The bottom line is that in principle you can see the very distant galaxies, but the photons are so few and arrive so rarely, that your eye is not a good enough detector. The benefit of a telescope is that 1) it has a larger area than your eye, and 2) you can put a camera at its focal point instead of your eye and take a picture with a large exposure time, i.e. increasing the Δt.
3 parsecs is the distance from us (Earth, Sun etc) to the star.
At a distance of 1 parsec, the apparent movement of a star wrt the background, as the Earth orbits on its 1 AU orbit is 1 arc second.
So, from 1 parsec distant, the earth would move by 1 arcsec each 6 months. It woukd have an apparevt angular radiusof 0.5 arc sec.
Now this star is 3 parsecs distant, so an orbit with radius 1au would appear to have a angular radius 3 times smaller (here I am approximating, it is valid because the angle is small) or 1/6 arc sec. The planet in the question has a smaller orbit, 6/10 of the Earth, or 0.6 au
The convenient units, and tge the small angles mean that we can avoid trionometry in the solution.