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piff

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About piff

  • Rank
    Nebula

Profile Information

  • Gender
    Male
  • Interests
    Astronomy (obviously); Esperanto; History; Music (both performing and listening)
  • Location
    Hastings, Sussex
  1. Thanks, everyone who has contributed. I think I have grasped the idea now.
  2. 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.
  3. 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?
  4. 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?
  5. 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?
  6. 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
  7. 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?
  8. 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?
  9. 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?
  10. 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?
  11. 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?
  12. Can anyone explain why the Herzsprung-Russell diagram is always shown with the x-axis going in reverse - i.e. decreasing as you go along it rather than increasing as is normal in just about every other such graph or diagram ever conceived?
  13. I have a little 70 mm Celestron travel-scope. Not a very serious instrument of course but handy for taking around with me and showing some of the larger solar system objects to people who are new to it. However it seems extraordinarily difficult to find objects while looking through it. I guess I really need a finderscope but there are no screw holes to attach one and I am reluctant to bore holes in the tube myself. Are there such things as finderscopes that can be attached using some sort of adhesive, rather than being screwed on?
  14. I recently bought a second-hand Celestron Nexstar 8 SE Schmidt-Cassegrain with go-to capability. In the manual it states "The Nexstar can be powered by eight AA batteries (not included) an optional 12 volt AC adaptor, or an optional car battery adaptor." The next few paragraphs describe exactly how to instal the batteries in the base, but rather unhelpfully there is not another word in the manual that I can find about how to provide power tot he telescope using a car battery charger, which is the method I would much prefer, rather than running through innumerable sets of batteries. I purchased a car battery charger especially to use with the telescope, but I cannot see for the life of me how or where to plug it in. The only input plugs I can see on the mount are of the "telephony" type, square in shape. The cable that came with the charger has a "car cigarette lighter" type of plug at one end and a tiny circular "mobile phone" type plug at the other end. Neither looks likely to fit into any visible orifice on the telescope mount. How exactly is one meant to use a car charger to power the telescope?
  15. Hi Guys Very grateful for your input on this. You all seem to agree that radiation is the answer but I would be grateful for a little further clarity on this. Sorry to sound dim, but I had the impression that radiation means that particles are actually ejected from the "parent" body. Are you saying that molecules of rock get so excited by the heat energy they have absorbed that they leap off the surface into space? That would explain it, certainly. It just sounds a little counter-intuitive, if you know what I mean. If that is what happens, presumably the molecules of rock would not reach escape velocity and would fall to the surface again, adding to the thin layer of surface dust?
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