piff

I recently purchased a Skywatcher mak with their fancy new "Avant" mount which comes with a very smart looking tripod. However I am finding considerable difficulty following the instructions for fitting the accessory tray to the tripod - not something I would normally anticipate being a major problem! The instruction manual says to "align the accessory tray" and then "rotate the tray to lock it into place". Now the problem is that the accessory tray is specifically designed so that the lug-holes in its centre are not in alignment with the triangular corners of the tray. When fitted onto the lugs on the "spider" attachment which is fixed between the legs of the tripod, the corners of the tray stick out into the spaces in between the legs (see photo attached). It is not however possible then to "rotate" the tray as per the directions because, one the lug holes on the tray are fitted onto the lugs attached to the tripod, it does not rotate. The attachment on which the tray fits does not move or turn at all. It is all moulded as one piece, fixed in position, and does not twist. Where am I going wrong?

Thanks, everyone who has contributed. I think I have grasped the idea now.

HI Marvin - thanks for your input. But doesn't the degree of "warping" depend purely on the mass involved, rather than the volume? This is where I got confused and why I posed the original question - the volume is the only thing that has changed. The mass remains the same.

I understand - I think! - that the gravity field of a black hole is so great that light photons from within the event horizon (or Schwarzschild limit) cannot leave it - hence the name. However I recently read that the mass of a black hole - which is after all just the crushed core of its parent star - is the same mass that the core had before the star's collapse. That seems to make sense. No additional mass has been acquired in the meantime, I assume. It would follow, it seems to me, that its gravity cannot be any greater than the gravity the core had beforehand. Yet we never hear about the star, before its death, having a gravitational field so strong that light could not escape from it. Why and how is the black hole's gravity so much stronger after the core collapse, that even light cannot escape, when its mass is no greater than it was before when light could escape it perfectly easily?

I have been reading that whereas the fusion of hydrogen into helium in the core of a star requires temperatures of 15 million or so, the next stage of fusing helium into carbon requires temperatures of something like 100 million degrees. How does the star achieve such a temperature when it has no more mass than it did before, when it only achieved 15 million degrees?

I have a go-to Nexstar SCT. Having done the star-alignment, if I now turn off the power will its "onboard computer" still remember this when I turn it back on next time, or will I have to go through the whole procedure again every time I use it? If it can "remember" the initial alignment, will it be necessary to place the telescope back in exactly the same spot again, or is there a margin for approximation or "wriggle room" in that respect?

Thanks guys. I am, actually tossing up between two Skywatchers, both 102 mm aperture. One is a refractor with a focal length of 500 mm and the other is a Mak with a focal length of 1300 mm - the latter being only a fraction of the size and looks very convenient as a "gab and go" scope but I am concerned whether this would have equal viewing quality

Given a choice between two telescopes of equal aperture - and therefore presumably capable of supporting equal maximum useful magnification - what are the advantages/disadvantages of choosing one with a longer or shorter focal length?

Millimetre for millimetre, is it considered that you get clearer images or greater magnification or both from a refractor, or from a Schmidt-Cassegrain of equal aperture?

But presumably the zodiacal system was devised long before "apps" or "software" existed? What is the reality behind the curious assertion that the sun moves against a background of stars when no such background is ever visible?

I have just been watching a DVD "explaining" how the sun moves through the constellations of the zodiac during the year, starting with the statement "in January we see the sun against the background of the stars in Sagittarius" and continuing in this vein. I have read numerous similar accounts in various books so it is obviously fairly standard patter, but I have never been able to understood this. I never see the sun against the background of any stars at all, because when the sun is in the sky they are all invisible. I know the stars are still there of course, but the point is, we cannot see them. So what is the point of saying "we see the sun against the background of the stars in such-and-such constellation? How are we supposed to be able to tell which stars are in the background when this background is wiped out by the glare of the sun?

I have been confused for some while by what seems to be a contradiction between the principles of gravitation expounded by Galileo and those expounded by Newton - both of which we know to be correct, so I must be missing something here! Galileo established that two objects - regardless of differences in size, mass or shape - will, discounting air resistance , fall to the ground at the same rate. We had this principle impressively demonstrated by astronaut Dave Scott during his moon walk on the Apollo 15 mission when, taking advantage of the absence of any atmosphere on the moon, he simultaneously dropped a feather and a hammer onto the moon's surface. On the other hand Newton's formulation of the force of gravitational attraction between any two objects states that this is proportional to the product of the mass of BOTH objects. SO it seems to me that if the strength of gravitational attraction between the ground and a hammer on the one hand and of a feather on the other, depends at least partly on the mass of those two objects, as well as the mass of the earth (or the moon) then the calculation of those forces ought to produce very different figures in each case. Where am I going wrong here?