Promoting astronomy discussion.
|
|
astronomy-chat.com Promoting astronomy discussion. |
mass. The force is equal to the Gravitional Constant times the first object's mass, times the second object's mass, and divided by the square of the distance separating them. F = (G * m1 * m2) / r^2 The orbit of a large planet around a star will cause a slight wobble (because they're both actually orbiting a common center of gravity), and this is detectable by noting tiny deviations in its spectrum (the Doppler shifts). That's the way astronomers have found most all of the extra-solar planet so far. You're not likely to notice much within a system though, because the effects are continuous and subtle. However, precise calculations of planetary positions and spacecraft trajectories always take into account the gravitational pulls of many other bodies. "IWJPoel" <Ihavenomailaccountatthemoment@no-e-mail.nope > wrote in message news:3fca1548$0$1498$e4fe514c@news.xs4all.nl... > Hello again! > > May I ask another question? > > I haven't found anything about this on the net. > > I read about the planets, comets, asteroids etc. being held in the gravity > of the Sun because the huge mass of the Sun pulls at these objects, so their > movement and behaviour is well predictable. > > But these other bodies in the solar system have a big mass as well. Do these > planets etc. also pull at the Sun as well, and if they do, is the force big > enough to make the sun move or to give it another shape? > > For example when all the planets are more or less lined up, would that give > a visible or even measurable reaction on the sun in any way? > > I admit that I have a very hard time putting planets, sun and all the other > solar system objects in the right perspective, especially when it comes to > weight and size since some things are more dense than others. > > Can anyone please tell me where to look for info on this or does any of you > know an answer to this? > > Thank you in advance, clear skies to you all! > > Greetings from the Netherlands, > Iko. > > > -- > "Odi et amo, quare id facere forasse requiris...Nescio, sed fieri sentio et > excrucior" > > |
<Ihavenomailaccountatthemoment@no-e-mail.nope > wrote: >I read about the planets, comets, asteroids etc. being held in the gravity >of the Sun because the huge mass of the Sun pulls at these objects, so their >movement and behaviour is well predictable. > >But these other bodies in the solar system have a big mass as well. Do these >planets etc. also pull at the Sun as well, and if they do, is the force big >enough to make the sun move or to give it another shape? Everything with mass in the Universe pulls on everything else with mass in the Universe. You shouldn't think of the Sun pulling harder than a planet because it has more mass. The Sun feels the same force from Jupiter that Jupiter feels from the Sun. Of course, that force has more effect on Jupiter with its lower mass. In fact, the Sun travels in a complicated path around the geometrical center of the Solar System because of the forces if feels from other bodies. But that path remains very close to the center of the Solar System. >For example when all the planets are more or less lined up, would that give >a visible or even measurable reaction on the sun in any way? Gravity is a very weak force, and the combined mass of all the planets is very small compared with that of the Sun. Certainly the positions of the planets perturb the position of the Sun slightly. The only gravitational effect that might affect the internal function of the Sun would be tidal forces, that is, the difference in gravitational forces across a body. In the case of the Sun, these are truly tiny forces. I'm not sure if the effects of tidal forces from the planets have been conclusively observed in the Sun or not. >I admit that I have a very hard time putting planets, sun and all the other >solar system objects in the right perspective, especially when it comes to >weight and size since some things are more dense than others. The Sun is 745 times more massive than the rest of the planets combined (and Jupiter alone is 2.5 times more massive than all the remaining planets combined). Jupiter has by far the most significant gravitational influence on the Sun. You might get a sense of the magnitude of its tidal forces when you consider that it is less than 1/1000 the mass and situated at a distance of over 1000 solar radii. Place a ball at one end of a soccer field, and a pecan shell at the other end. That is a fairly accurate representation of the masses and distances involved. _________________________________________________ Chris L Peterson Cloudbait Observatory http://www.cloudbait.com |
> > Hello again! > > May I ask another question? > > I haven't found anything about this on the net. > > I read about the planets, comets, asteroids etc. being held in the gravity > of the Sun because the huge mass of the Sun pulls at these objects, so their > movement and behaviour is well predictable. About 98% of the solar system's mass resides in the Sun. Although orbital mechanics is well understood, the complexity does produce chaotic behavior, especially for the minor planets (comets, asteroids, kuiper belt objects, etc.). > > But these other bodies in the solar system have a big mass as well. Do these > planets etc. also pull at the Sun as well, and if they do, is the force big > enough to make the sun move or to give it another shape? All bodies have gravitational pull on all other bodies. See: http://scienceworld.wolfram.com/physics/Gravity.html -Sam Wormley http://edu-observatory.org/eo/solar_system.html |
This fysics information is all rather new to me, I think I will need some time to get to my personal intellectual limits with this. Where are my old schoolbooks? This is a very cool group that you all have here, I have been lurking for a long time now and I like the huge amount of information and sources that you are sharing. I hope you don't mind my newbie questions every now and then. Greetings from the Netherlands, clear skies, Iko. |
If the sun vanished (this is just a thought experiment) would the earth fly off right a way or would it take 8 min. for us to fly off into the universe. "IWJPoel" <Ihavenomailaccountatthemoment@no-e-mail.nope > wrote in message news:3fca1548$0$1498$e4fe514c@news.xs4all.nl... > Hello again! > > May I ask another question? > > I haven't found anything about this on the net. > > I read about the planets, comets, asteroids etc. being held in the gravity > of the Sun because the huge mass of the Sun pulls at these objects, so their > movement and behaviour is well predictable. > > But these other bodies in the solar system have a big mass as well. Do these > planets etc. also pull at the Sun as well, and if they do, is the force big > enough to make the sun move or to give it another shape? > > For example when all the planets are more or less lined up, would that give > a visible or even measurable reaction on the sun in any way? > > I admit that I have a very hard time putting planets, sun and all the other > solar system objects in the right perspective, especially when it comes to > weight and size since some things are more dense than others. > > Can anyone please tell me where to look for info on this or does any of you > know an answer to this? > > Thank you in advance, clear skies to you all! > > Greetings from the Netherlands, > Iko. > > > -- > "Odi et amo, quare id facere forasse requiris...Nescio, sed fieri sentio et > excrucior" > > |
> If the sun vanished (this is just a thought experiment) would the earth fly > off right a way or would it take 8 min. for us to fly > off into the universe. Nobody knows. What you're asking about is the speed of gravity. In general relativity, the Earth travels in its orbit because the local curvature of space makes that orbit the straightest possible path in spacetime. If the Sun were to vanish, the consensus is that the effects of that vanishing could not propagate away from the Sun's position faster than the speed of light. As a result, the spacetime around the Earth wouldn't be affected for about 8 minutes. In addition, the tidal effects of the Sun would also vanish at the same time, probably causing all sorts of seismic events. But we probably wouldn't be too worried about that... A recent experiment purporting to measure the speed of gravity did indeed determine it to be roughly equal to the speed of light. However, the interpretation of the results is in dispute, with some leading specialists in relativity contending that all the experiment has done is to provide us with a measure of the speed of light. Brian Tung <brian@isi.edu > The Astronomy Corner at http://astro.isi.edu/ Unofficial C5+ Home Page at http://astro.isi.edu/c5plus/ The PleiadAtlas Home Page at http://astro.isi.edu/pleiadatlas/ My Own Personal FAQ (SAA) at http://astro.isi.edu/reference/faq.txt |
I always had hard time to understand why space objects have to "fly around" at all? What was the really first impulse invoking spinning of a planet and to orbit another celestial body when a matter collapsed? Brian Tung wrote: > In general relativity, > the Earth travels in its orbit because the local curvature of space > makes that orbit the straightest possible path in spacetime. Even though I know this definition very well I still can not understand why should it be considered as the explanation of circle trajectories? Why does not a planet simply fall down to its mother star in a seconds using "straight" trajectory (what is "straight" in the local curvature of space?) instead of circling for millions of years? Forgive my ingnorance, please. Roman If the Sun were > to vanish, the consensus is that the effects of that vanishing could not > propagate away from the Sun's position faster than the speed of light. > As a result, the spacetime around the Earth wouldn't be affected for about > 8 minutes. > > In addition, the tidal effects of the Sun would also vanish at the same > time, probably causing all sorts of seismic events. But we probably > wouldn't be too worried about that... > > A recent experiment purporting to measure the speed of gravity did indeed > determine it to be roughly equal to the speed of light. However, the > interpretation of the results is in dispute, with some leading specialists > in relativity contending that all the experiment has done is to provide > us with a measure of the speed of light. > > Brian Tung <brian@isi.edu> > The Astronomy Corner at http://astro.isi.edu/ > Unofficial C5+ Home Page at http://astro.isi.edu/c5plus/ > The PleiadAtlas Home Page at http://astro.isi.edu/pleiadatlas/ > My Own Personal FAQ (SAA) at http://astro.isi.edu/reference/faq.txt |
> > Brian, > > I always had hard time to understand why space objects have to "fly > around" at all? What was the really first impulse invoking spinning of a > planet and to orbit another celestial body when a matter collapsed? > > Brian Tung wrote: > > In general relativity, > > the Earth travels in its orbit because the local curvature of space > > makes that orbit the straightest possible path in spacetime. > > Even though I know this definition very well I still can not understand > why should it be considered as the explanation of circle trajectories? > Why does not a planet simply fall down to its mother star in a seconds > using "straight" trajectory (what is "straight" in the local curvature > of space?) instead of circling for millions of years? > Forgive my ingnorance, please. > Roman > > If the Sun were > > to vanish, the consensus is that the effects of that vanishing could not > > propagate away from the Sun's position faster than the speed of light. > > As a result, the spacetime around the Earth wouldn't be affected for about > > 8 minutes. > > > > In addition, the tidal effects of the Sun would also vanish at the same > > time, probably causing all sorts of seismic events. But we probably > > wouldn't be too worried about that... > > > > A recent experiment purporting to measure the speed of gravity did indeed > > determine it to be roughly equal to the speed of light. However, the > > interpretation of the results is in dispute, with some leading specialists > > in relativity contending that all the experiment has done is to provide > > us with a measure of the speed of light. > > > > Brian Tung <brian@isi.edu> > > The Astronomy Corner at http://astro.isi.edu/ > > Unofficial C5+ Home Page at http://astro.isi.edu/c5plus/ > > The PleiadAtlas Home Page at http://astro.isi.edu/pleiadatlas/ > > My Own Personal FAQ (SAA) at http://astro.isi.edu/reference/faq.txt Try to get the water to go down the drain without rotating. Conservation of Angular Momentum http://scienceworld.wolfram.com/physics/ConservationofAngularMomentum.html |
> I always had hard time to understand why space objects have to "fly > around" at all? What was the really first impulse invoking spinning of a > planet and to orbit another celestial body when a matter collapsed? Because the object starts out with some angular momentum. Our current best guess as to how the solar system evolved is that a cloud of gas and dust condensed. This cloud began with a certain amount of rotational momentum to it--angular momentum. You might think that on average in such a large body (probably a light-year or so in extent), the angular momentum would be zero, but in practice, the chances that the angular momentum would be precisely zero are tremendously slim. It's much more likely that there was a net rotation to the cloud, however slow. As this cloud condensed, the angular momentum remained constant. This angular momentum, however, is equal to the product of the mass of the cloud, its scale (its size, essentially), and its rotational rate (or angular velocity). Now the mass of the cloud remained more or less the same, but the scale decreased as the cloud condensed. To keep the momentum constant, the rate at which the cloud rotated had to increase, just as a ice skaters drawing in their arms in a spin begin to spin faster. Even after some particles in the cloud combined to form the planets, those planets retained the angular momentum in the cloud where they formed (and even stole some from the Sun), so they did not fall into the Sun. (Some planets may have, as a result of collisions, but since they aren't here, we don't speak of them, of course!) > Even though I know this definition very well I still can not understand > why should it be considered as the explanation of circle trajectories? > Why does not a planet simply fall down to its mother star in a seconds > using "straight" trajectory (what is "straight" in the local curvature > of space?) instead of circling for millions of years? Well, this is not an easy thing to understand without math. I'm going to try to explain it without a lot of math, but if you don't understand it, please realize that it's not your fault! (One result of trying to do this without math, by the way, is that I'm going to be using an old-fashioned way of thinking about general relativity that most people wouldn't use now.) Most of us understand how gravity and orbits work in Newtonian physics. There's a force, called gravity, which applies between any pair of objects with mass, such as the Earth and a ball (for example). This force is attractive, so that if you throw the ball straight up in the air, it doesn't continue up forever in a straight line, but instead slows to a stop, then falls back down to your hand. However, in general relativity, gravity is not a force. Rather, it's a curvature in space-time. We say that objects follow "geodesics," paths that are as "straight" as possible in the curved space-time. The problem is that this word "straight" doesn't mean what you think it does. This confusion in meaning is rooted in the difficulty of defining the term "straight" on a curved surface. So long as we constrain ourselves to flat surfaces, there's no confusion. If I draw two points on a flat sheet of paper, it's obvious what the straight line is between them. I just plop down a ruler and draw the straight line. But what if the two points are on a sphere? I can't simply plot down a ruler on the sphere, because the surface of the sphere is curved and the ruler is flat. We have to extend our definition of straight to apply to this curved surface, and the most common way to extend it is to recognize that not only is a straight line the most direct route between two points, it is the one that takes the shortest distance. On a sphere, that route is the segment of the great circle connecting two points. A great circle is a circle that cuts the sphere exactly in half, and for any two points (provided they aren't the same point or directly opposite one another on the sphere), there is exactly one great circle that connects them. Try it! For example, the shortest path between any point on the equator of the Earth and the north pole is the line of latitude that both points lie on. This line of latitude is a great circle; it cuts the Earth exactly in half. (Yes, I'm assuming that the Earth is a sphere for the time being.) And you can see that no other path between the two points is shorter than this path. It is the shortest path imaginable between those two points. What's not so obvious is the shortest path between, say, New York and Madrid, which both lie at about 40 degrees north latitude. You might think that the solution is to maintain a constant latitude of 40 degrees north. However, the circle on which that path lies--the circle at which the latitude is 40 degrees north--is not a great circle, so it cannot be the shortest path. Instead, the great circle connecting those two points is "tilted." Between New York and Madrid, that circle is north of 40 degrees latitude; at all other points, it is south of 40 degrees. That's why when you travel from New York to Madrid, the shortest path is not at a constant latitude, but travels in a gentle arc that is entirely north of 40 degrees. Here's another way to think of that. Both cities are at about 40 degrees north latitude, but Madrid has a longitude of 4 degrees west, and New York one of 74 degrees west. To get from New York to Madrid, you *must* traverse those 70 degrees of longitude. If you stay at 40 degrees north latitude, then each of those 70 degrees is 85 km, for a total travel distance of 70 times 85, or 5,950 km. However, you can shave some distance off if you recognize that degrees of longitude are shorter at higher latitudes. For example, at a latitude of 50 degrees north, each degree of longitude is more like 71 km. (Indeed, at the pole, each degree of longitude is of length zero!) The total path length can thus be made shorter by arcing toward higher latitudes, covering much of the longitude difference in those shorter degrees, then coming back down to 40 degrees north. However, going all the way to the pole to take advantage of those zero-length degrees of longitude makes no sense, because you use up so much distance just getting to the pole. An in-between solution is best, yielding a distance of about 5,800 km. On a map with a Mercator projection, this path *looks* curved, because the lines of longitude and latitude are drawn as straight lines. But on the Earth, which is curved, it is the great-circle path that looks as straight as possible, while the path along constant latitude is more curved. In the case of paths on the Earth, the distance is reckoned in three dimensions, because the surface of the Earth is curved into three dimensions. When dealing with general relativity, we must deal with at least one other dimension--the dimension of time. That means that any more or less complete treatment of the topic must incorporate four dimensions. However, since four dimensions are tough for us to visualize, and even three is a bit of a toughie, let's deal with just two dimensions: one time dimension, and one spatial dimension--up and down. Go back to the ball being tossed up and down. We'll say that the height of your hand is 1 m, and we'll start the clock at 12:00:00. The space-time coordinate of the ball resting in your hand is therefore (1, 12:00:00). If you throw the ball upward at a speed of 10 m/s, it comes back down to your hand in 2 seconds, having reached a maximum height of 6 m. At its peak, the coordinate of the ball was (6, 12:00:01), and when it lands back in your hand, it's (1, 12:00:02). Again, we understand this perfectly in terms of Newtonian mechanics, because in that framework, gravity is a force that creates a downward acceleration on the ball. We can compute how long it takes for that acceleration to slow the ball to a stop at the peak of its arc, and then how long it takes to fall back down to your hand. But in general relativity, we do it differently. The ball takes the path that follows the straightest line in a space-time curved by the Earth's mass. We don't start out with initial conditions and find the state at any point afterward, as we do in Newtonian physics; we start out with the initial and final states, and figure out all the states in between. Our question is, therefore: What is the straightest line between (1, 12:00:00) and (1, 12:00:02)? To answer this question, we need two further properties of general relativity: 1. Clocks run slower near a massive a object than they do further from it. In other words, a high-altitude atomic clock runs faster than one at low altitude. 2. The further the spatial distance between two points, the longer the space-time distance, as usual. But for time, it's different: the more an object advances in time between two points, the *shorter* the space-time distance. The naive answer to our question is that the straightest line between our two points is one where the spatial coordinate is constant at a height of 1 (meter). But after our discussion with great circles and shortest paths over the sphere, you should be wary of that answer. (Besides, it corresponds to a ball that hangs in mid-air at a height of 1 meter for 2 seconds. I don't think I've ever seen that--not without some other force acting on the ball, at least.) If we travel slightly higher up from the Earth, we should be able to take advantage of Property 1, and shorten the space-time distance. We can't go *too* far up, because the spatial distance incurred by going up will be more than can be made up for by the faster clock. Just as with the great circle route on the Earth, we have to balance out the benefit of going to a height where clocks run faster with the cost of getting up to that height in the first place. It turns out that the path which best balances those out is almost exactly the same arc as predicted by Newtonian physics. So why should we bother with general relativity if it gives basically the same answers as Newton, and Newton is much simpler? The reason is that these answers are only *almost* exactly the same. What's more, the answers are only that close because the mass of the Earth and the speed of the ball are so small by astronomical standards. With larger masses and faster speeds, the two theories give noticeably different results. For instance, many of us know about Einstein's prediction (around 1919) that distant starlight skimming by the Sun would be bent by the Sun's gravity through a certain small angle. What's not so well known is that Newtonian physics predicts a bending too! The difference is that Einstein predicted twice as large an angle as Newton did. The initial solar eclipse expedition by Eddington only marginally confirmed the case in favor of Einstein, but later experiments have more firmly established that where the two theories differ, general relativity is by far the more accurate. In fact, there are no well-established inaccuracies in it yet. Now, finally, let's get back to your question: How is an endless orbit a straighter line than a path directly into the Sun? The answer is, it isn't. They're both equally straight. In the case of the ball, general relativity predicts that the ball follows an arc in space-time that will take it into the center of the Earth after some tens of minutes. But that's only because we've only accounted for one spatial dimension--up and down. The path straight into the massive object only travels along that one spatial dimension. If we include the other spatial dimensions, then we find that there are many straight paths in space-time that avoid the Earth's center and instead end up at space-time point (1, 13:24:00). That's because the orbital period for a low-Earth orbit is about 84 minutes. The object then moves to the point (1, 14:48:00), then to point (1, 16:12:00), and so on. These paths are essentially ellipses and are followed by many artificial satellites (with suitable substitutes for "1 meter," of course!). They look curved because we're looking at them from the perspective that spatial lines are straight. But if we could see space-time as it really is, we would find that our spatial lines are a bit like the lines of latitude on a Mercator map--not really straight after all. Brian Tung <brian@isi.edu > The Astronomy Corner at http://astro.isi.edu/ Unofficial C5+ Home Page at http://astro.isi.edu/c5plus/ The PleiadAtlas Home Page at http://astro.isi.edu/pleiadatlas/ My Own Personal FAQ (SAA) at http://astro.isi.edu/reference/faq.txt |
> This cloud began with a certain amount of rotational > momentum to it--angular momentum. You might think that on average in > such a large body (probably a light-year or so in extent), the angular > momentum would be zero, but in practice, the chances that the angular > momentum would be precisely zero are tremendously slim. It's much more > likely that there was a net rotation to the cloud, however slow. Ah, this is the point! I see now. Your whole explanation is excellent! I admire you. Thanks. Roman P.S. Brian - brain, What a similarity! ;-) |
> Ah, this is the point! I see now. > > Your whole explanation is excellent! I admire you. Thanks. You're very welcome. > P.S. Brian - brain, What a similarity! ;-) Thanks--I've always thought so. :) Brian Tung <brian@isi.edu > The Astronomy Corner at http://astro.isi.edu/ Unofficial C5+ Home Page at http://astro.isi.edu/c5plus/ The PleiadAtlas Home Page at http://astro.isi.edu/pleiadatlas/ My Own Personal FAQ (SAA) at http://astro.isi.edu/reference/faq.txt |
|