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csmyth3025

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Everything posted by csmyth3025

  1. The formula for calculation the centripetal (inward) force exerted on an orbiting body is: Essentially the acceleration is v2/r. The equatorial rotational velocity for 87 Sylvia is given as: (ref. http://en.wikipedia....characteristics ) In this case, the equatorial velocty would be (230,000 m/hr)/(3600) = ~63.89 m/s. The radius used for this calculation would be: (63.89 m/s)/(2*pi/18,648 s) = 189,620 m = 189.62 km radius = ~380 km diameter. This is pretty close to the 384 km diameter long axis given in the Wikipedia article. The mass of 87 Sylvia is given as 1.478x1019 kg in this same article. Using round numbers, the acceleration of gravity at ~190,000 meter radius = (G*m)/r2 = [(6.674x10-11m3/kg*s2)(1.478x1019 kg)]/(1.9x105 m)2 = ~(9.86x108 m3/s2)/(3.61x1010 m2) = ~0.027 m/s2 Since this number isn't even close to the 0.0729 m/s2 equatorial surface gravity givenin the article (not even counting the effect of rotation), I've obviously done something wrong. Help! Chris Edited to correct spelling errors
  2. I think what Elen Sila is saying is that if Earth were suddenly taken out of the picture, Sam would be traveling in some direction with a linear velocity of ~465 m/s (assuming he is at the equator). Mary, who is in geostationary orbit directly above Sam, would be travelling in the same direction with a linear velocity of ~3,070 m/s. In this scenario the difference in (linear) velocity between Sam and Mary would be ~2,605 m/s. If one were to ignore the gravity of the Sun and the galaxy as a whole, they could both be considered to be in their own inertial frame of reference. Sam would see that Mary is moving away from him at 2,605 m/s, and Mary would see that Sam is moving away from her at 2,605 m/s (only in the opposite direction). I think we can apply the formula for special relativistic time dilation to their new situation: For small values of v/c, by using binomial expansion this approximates to: (ref. http://en.wikipedia....f_time_dilation ) In this case we would have: dilation = ~1-[(2605 m/s)2/(2)(2.998x108)2] = ~1-[(6.8x106)/(1.8x1017)] = ~1-(3.8x 10-11) This difference below 1 of about 3.8×10−11 represents the fraction by which Mary's and Sam's clocks differ. If we then multiply by the number of nanoseconds in a day we get: -(3.8x10-11)(60)(60)(24)(109) = ~-3283 ns per day, or about -3.3 microseconds per day. It's entirely possible that I may have this calculation all wrong, so corrections are welcome. Chris Edited to insert minus sign in final result
  3. I'm no expert on special or general relativity, but I'm beginning to see the nature of your dilemma. First, we can dispense with further discussion of why it is that the clock on a geostationary satellite runs faster than a clock on the ground. I take it that the previous replies have satisfactorily explained the general relativistic notion that a clock in a weaker gravitational field (the clock in the geostationary satellite) is not slowed down as much as a clock in a stronger gravitational field (the clock on Earth). Thus, when comparing the two clocks, both Mary and Sam will agree that Mary's clock is running faster than Sam's clock. The situation on the satellite and on Earth is asymmetric in regard to the gravitational field strength at their respective distances from the Earth's center. Now we get to the special relativistic notion that two observers - each in uniform inertial motion, but both moving relative to each other - can both detect that the other's clock is running slower than their own by communicating with signals back and forth between them. How is it possible for them both to say that the other's clock is running slow in this experiment? Once we find an answer to this question we can move on to the question of how it is that an observer on the Earth would agree that a clock in geostationary orbit (always directly overhead at the same distance) is running slower than the observer's Earth-bound clock, provided that we cancel out the aforementioned effect of general relativity. The fact that both the observer on the geostationary satellite and the Earth-bound observer agree on this difference in their respective clocks implies that there is not a symmetrical relationship between the two observers, as is required by the purely special relativistic example given above. Does this pretty much describe your questions? I would love to give you an answer, but I'm as baffled as you are by these questions. Chris
  4. Because Mary is in Geostationary orbit the following applies: (ref. http://en.wikipedia....eral_relativity ) Keep in mind that GPS satellites are placed so that the orbit Earth twice per day at an altitude of about 20,184 km above the surface of the Earth. A geostationary orbit, by definition, orbits the Earth exactly once per day (it stays over the same spot on Earth all the time) and thus has a higher orbital altitude of about 35,786 km above the surface of the Earth. Lets say for the sake of argument that Mary is on a GPS satellite when she transmits her message to Sam and it takes Sam 8.64 seconds to transmit his message back to Mary. This just happens to be 1/10,000 of a day. When they compare their times Mary will find that her clock shows 38 microseconds/10,000, or 3.8 picoseconds more on her clock than Sam. Mary's clock runs faster than Sam's because of the combined effect of both Special Relativity and General Relativity. If Mary is actually in geostationary orbit this effect will be even greater because she will be orbiting above the surface of the earth at greater distance and her orbital velocity will therefore be less than it is on the GPS satellite. Thus the Special Relativistic effect of slowing down Mary's clock relative to Sam's will be less. Also, she will be in a weaker gravitational field. Thus, the General Relativistic effect causing her clock to run faster than Sam's will be more pronounced. Chris
  5. Sorry, but I really don't know any more than you do about such things. There are others here who are quite knowledgeable, though. Hopefully this thread will catch their eye and they will be able to answer your questions. Chris
  6. Perhaps my understanding of entropy is too simplistic. For instance, suppose you take a tub of water with a partition down the middle and put hot water on one side and cold water on the other. If you put the "hot" end of a Stirling engine in the hot water side of the tub and the "cold" end of the engine in the cold water side of the tub you can run the Stirling engine and do nifty things like blow air on your face with a fan blade attached to the shaft of the Stirling engine. Once the temperature of the water on both sides of partition equalizes, though, the Stirling engine stops working. It can no longer extract any more energy from the tub. In other words, the entropy of the water in the tub has reached its maximum. This does not mean that there is no more energy in the water in the tub - there is no doubt lots of (thermal) energy still contained in the tub, but we can no longer extract any more energy from it as a closed system in order to do any work: (ref. http://en.wikipedia....tion_to_entropy ) In the far distant future, when thermonuclear reactions have ceased, the universe is filled with a uniform and gradually cooling radiation background and all gravitationally bound systems have either collapsed into their most compact form or dispersed due to the expansion of space, the world as we know it will have reached a stage of maximum entropy. The total amount of energy in the universe may still be very high (perhaps the same as it is at the present time), but there will be no energy available to do things like make new stars or power active galactic nuclei. This would be the so-called "heat death" of the universe. As far as I know there isn't any way to reverse this trend. It's a one-way street. The arrow on the street sign pointing in the direction that everything is going is the arrow of time. Chris
  7. The Wikipedia article you cite is (for me) the only understandable description I've found. If your looking for stable resonances I would direct your attention to the following passage in that article: (ref. http://en.wikipedia....es_of_resonance ) Preceding this passage there are several other stable orbital resonances mentioned. Chris
  8. Various military and civilian GPS navigation devices communicate with GPS satelites millions of times each day. The comparison of their clock times is as described in the Wikipedia article on "Error analysis for the Global Positioning System" that I cited earlier. Are you saying that you don't understand time dilation as predicted by relativity theory, or are you expressing your doubt that this effect is real? Chris
  9. Moving has to do with Special Relativity, which Wikipedia has a very good introductory article about here: http://en.wikipedia....cial_relativity If your looking for examples of how the strange consequences of Special Relativity have been tested, Wikipedia has that covered in another article here: http://en.wikipedia....cial_relativity Chris
  10. Regarding electron degeneracy pressure (white dwarfs) and the force needed to "shove an electron into the nuleus": (ref. http://en.wikipedia....generate_matter ) If you want more details than this, you'll probably have to find a physicist to explain it. Chris
  11. Before you get too wrapped up in conspiracy theories about how the scientific establisment is trying to fool the public, it would be a good idea to learn why the big bang model is generally accepted. There's a lot of experimental evidence and observations that support this model. Taking the word of scientists in a YouTube video as gospel is even more senseless than taking the word of "highly decorated scientists" as gospel. Study all the evidence and decide for yourself what to believe. A good place to start is Wikipedia: http://en.wikipedia.org/wiki/Big_Bang Chris
  12. I noticed that the quantity I arrived at in my post #3 calculation (2512 kg*m^2/s) is equivalent to 2512 J times 1 second: where N is the newton, m is the metre, kg is the kilogram, s is the second, Pa is the pascal, and W is the watt I'm not sure how to interpret this. If 1 Joule is equal to 1 Watt*second, what is 2512 J times one second equal to? Did I make a mistake in my calculation? I'm trying to figure out how much work it would take to bring the sphere I described to a halt using more familiar units, such as comparing it with the amount of energy consumed by a 100 Watt light bulb in one hour (0.1 kWh). I may be confusing angular momentum with rotational kinetic energy: (ref. http://en.wikipedia....Rotating_bodies ) If I use the same value I used previously for the moment if inertia (400 kg*m^2) and ω^2 instead of just ω, I would get: = {(400 kg*m^2)(6.28/s)^2}/2 = 7,887.7 kg*m^2/s^2 = ~7,888 W·s If I understand this formula correctly, the amount of energy it would take to slow the sphere I described down to a halt would power a 100-watt light bulb for about 1 minute and 19 seconds (if it could all be converted with 100% efficiency to electricity). Does this sound right? I'm still a bit confused about how angular momentum is applied to problems of this type. What sort of problems use angular momentum? Chris
  13. It's called Special Relativity. The Wikipedia "Introduction to special relativity" article is a good place to start if you're asking why clocks in relative motion to each other slow down (relative to each other). It has many links that can provide you with more in-depth information. It can be found here: http://en.wikipedia....cial_relativity Chris
  14. The short answer is that you don't understand relativity theory. The long answer - with all the gory calculations - can be found in the Wikipedia article on "Error analysis for the Global Positioning System", specifically the section on "calculation of time dilation" here: http://en.wikipedia....f_time_dilation Although this article relates specifically to GPS satelites, the calculations are also applicable to your scenario. Chris
  15. Since it's your theory it's incumbent upon you to check the predictions that it makes against experimental evidence. For instance, does the Earth travel through time faster than the Moon (because the moon is smaller than the Earth)? We've had astronauts on the Moon several times. It seems to me that the effect you describe would have been noticed. Perhaps this is something you should investigate. Chris
  16. The sophistication of the equipment and the quality of the results have improved considerably over the years: (ref. http://en.wikipedia....ly_20th_century ) The complete article cited goes into considerable detail on the subject of the speed of light, its measurement, and the history behind it all. It also contains many references for the technically-minded. Chris Edited to correct spelling errors
  17. Thanks for the link DrR. I'm afraid it's a bit over my head, though. I'll keep working on the basic concepts (and the math). Hopefully, one day a light will go off in hy head and I'll finally "get it". Chris
  18. As ajb points out, if you're thinking in terms of hypothetically adding or subtracting mass to a (spherical) rotating body, the equations for determining angular momentum will give you the the answer by assuming that the the mass of the body is uniformly distributed throughout and the radius of the body remains unchanged (it just becomes uniformly more dense). The formula for angular momentum for a rigid (that is, solid) body that's rotating around a fixed axis is: (ref. http://en.wikipedia....ntum#Definition ) So far, so good. Now the angular velocity (ω) is usually given in radians per second, although it may be measured in other units such as degrees per second, revolutions per second, revolutions per minute, degrees per hour, etc. This seems simple enough. Since there are 2*pi radians in a complete circle, one revolution per second would equal 2*pi radians per second (6.28 radians per second). Now we come to the moment of inertia part. This is a bit more complicated, but Wikipedia has simplified it for us: For a Ball (solid) of radius r and mass m: (ref. http://en.wikipedia....ents_of_inertia ) This is where things get sticky (for me, at least). We have the formula for angular momentum L, which is (). We know that ω (angular velocity) is in terms of radians per second. And we know that . For the sake of simplicity, lets say we have a 1000 kg solid ball that's 1 meter in radius rotating at 60 rpm (6.28 radians per second). The angular momentum would then be: L = {(2*1000 kg)*(1 meter)^2)/5}*(6.28/s) = (400 kg*m^2)(6.28/s) = 2512 kg*m^2/s The angular momentum of this object has a direction (the direction of its spin). I believe that makes it a vector quantity, although it may be a tensor (I really don't know what the difference between the two is). All of this is well and good (assuming I've done it correctly). Unfortunately, I'm stuck at this point. My guess is that just adding mass would actually slow down the rotation of the core in order for the result of the above equation (the angular momentum) to remain the same. If, however, adding mass causes the object to become more compressed (smaller radius), then it could conceivably cause the object to speed up. I don't think any of this will cause the core to "create" a magnetic field. If it produced a magnetic field before you added the additional mass, the magnetic field might become stronger or weaker - depending on whether the object ends up spinning faster or slower. As far as I know planetary magnetic fields are caused by conductive fluid motions in the interior of a planet - they're source is dynamically driven, as opposed to the static magnetic fields we normally associate with bar magnets. As always, I may have done this entirely wrong, so corrections are welcome. Chris Edited to correct spelling errors
  19. I'm a bit confused about the "gravity gravitates" concept. If the energy of gravity (I'm guessing that this would be the gravitational binding energy of the object) doesn't contribute to the stress-energy-momentum tensor of general relativity, in what way does it contribute to the nonlinearities in the field equations? I hope the explanation isn't too involved - my grasp of differential equations is, at best, rudimentary. Chris
  20. If you hope to have any sort of dialog in this forum you're going to have to be more specific than saying that you doubt "everything" about the big bang theory and that you don't think the theory "makes sense". The series of YouTube videos that you cite start out discussing quasars. Perhaps you can start there and explain what it is about the currently accepted theories about quasars that don't make sense to you. Alternately, you can pick some other astronomical observation that you feel has been misinterpreted by the scientific community at large. Chris
  21. I can't answer your question, but I have one that's related: In the gravitational collapse of a star, does the intensifying gravity of the collapsing core contribute to the stress-energy-momentum tensor? In the case of a stellar core collapse I'm thinking that the (negative) gravitational potential energy of the in-falling matter is being converted to (positive) gravitational binding energy. This is a bit off-topic, so if it requires more than a yes-or-no answer I'll put it in a separate thread. Chris
  22. Swansont is referring to the gravitational potential present at the location of the clock. Since you seem to doubt that time really does slow down due to the velocity that two different clocks may have relative to each other, or due to the position of one clock relative to the other in a gravitational field (that is, each is subject to a different gravitational potential), you may find this helpful: (ref. http://en.wikipedia....eral_relativity ) The complete Wikipedia article cited goes into more detail about how these effects are calculated and what steps are taken to compensate for them so that the clocks on GPS satellites stay "in sync" with the clocks in the receivers on the ground. These effects are real. If we didn't compensate for the combined effects of special relativity (time dilation due to relative motion) and general relativity (time dilation due to greater gravitational field strength) our Garmins and Tom-Toms would be useless - they would accumulate errors in the range of ten kilometers per day. How we "perceive" the flow of time has nothing to do with the proper operation of these devices. Chris
  23. (bold added by me) A person's time (his proper time in his frame of reference) is always the same: it passes at the rate of one second per second. If you say that time is ticking faster (or slower), it must be ticking faster (or slower) relative to someone else's frame of reference. Chris
  24. (ref. http://en.wikipedia...._of_discoveries ) Chris PS - sorry about the stretched-out lay-out. Copying the pdf format table seems to have messed up the coding in this post. I re-quoted the above passage - which seems to have been garbled by the coding conflicts in my original post. Chris
  25. As I said earlier, the use of the term "singularity" is often misinterpreted by the general public as the "impossible statement" to which you refer in which "...every known law of physics failed". The point is that the term "singularity" defines the state in which the known laws of physics fail. The term singularity, like dark energy and dark matter, is just a place-holder term for something about which we know nothing - something to which our known laws of physics cannot be applied without producing impossible and contradictory results. The big bang standard cosmological model describes the evolution of the universe from a very hot, dense state to the very cold, dispersed, homogeneous and isotropic state we see today. The very hot and dense state of the early universe can be logically inferred from the observed expansion of the universe defined by the Hubble constant (70 km/s per Mpc). This rate of expansion has been refined and verified many times over in the past 80 years. "Running the clock backward" produces the hot, dense state that constitutes the starting point for the standard cosmological model. The processes that the model describes are based on rigorous theories that produce testable results by which they've been validated. The "flatness" of the geometry of our observable universe, the abundances of the primordial elements, the observed large scale homogeneity and isotropy of the cosmos and the existence and characteristics of the cosmic microwave background all are predicted by and validate the various components of the standard cosmological model. Non-scientific people make assumptions about what came before the period of cosmic evolution that the standard cosmological model describes. They mistakenly jump to the conclusion that these assumptions must be part of the standard model. They are not. Chris
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