losfomot Posted May 6, 2005 Share Posted May 6, 2005 1- Is an object in geostationary orbit, pointing in a steady direction toward the Earth (not rotating) considered to be at rest WRT the Earth? 2- If X and Y are at rest WRT eachother, and Z and Y are at rest WRT eachother, then Z and X must be at rest WRT eachother. True or False? Thanks. Link to comment Share on other sites More sharing options...
swansont Posted May 6, 2005 Share Posted May 6, 2005 1- Is an object in geostationary orbit' date=' pointing in a steady direction toward the Earth (not rotating) considered to be at rest WRT the Earth?[/quote'] Not really easy questions. A geostationary orbit is not an inertial frame. You can define a frame in which you say the earth is not rotating, and pretend it's inertial, but then you will have weird effects, like moving a clock along a line of latitude causes it to acquire phase (i.e. a time offset). People do this, of course, because of the overall convenience. (it's not unlike having pseudoforces like the Coriolis force, when we pretend the earth is an inertial frame in classical physics) One other reason that the ECI (earth-centered inertial) frame is useful is that the rotation effects of time dilation cancel the gravitational redshift changes from the oblateness (the equatorial bulge), so all ideal clocks on the geoid (basically the average sea-level) run at the same rate, regardless of latitude. So the rotation rate doesn't matter with that definition. What that all means is that you can't really say that either the earth or the satellite is not rotating, and with that conflict, you have to be careful what you mean by "at rest," since that implies an inertial frame of reference. If 1 was true, then 2 would not be true. Link to comment Share on other sites More sharing options...
geistkiesel Posted May 7, 2005 Share Posted May 7, 2005 Not really easy questions. A geostationary orbit is not an inertial frame. You can define a frame in which you say the earth is not rotating' date=' and pretend it's inertial, but then you will have weird effects, like moving a clock along a line of latitude causes it to acquire phase (i.e. a time offset). People do this, of course, because of the overall convenience. (it's not unlike having pseudoforces like the Coriolis force, when we pretend the earth is an inertial frame in classical physics) One other reason that the ECI (earth-centered inertial) frame is useful is that the rotation effects of time dilation cancel the gravitational redshift changes from the oblateness (the equatorial bulge), so all ideal clocks on the geoid (basically the average sea-level) run at the same rate, regardless of latitude. So the rotation rate doesn't matter with that definition. What that all means is that you can't really say that either the earth or the satellite is not rotating, and with that conflict, you have to be careful what you mean by "at rest," since that implies an inertial frame of reference. If 1 was true, then 2 would not be true.[/quote'] The SATS in orbit are force free. It would be most difficult to measure any say 20 meter stretch of orbit trajectrory and measure "curvature" . For all practical purposes the SATS are in inertial frames, especialy when in geostationary orbits. What ever the statement means that "you can't really say that either the earth or the satellite is not rotating", the SAT and the ground station are at rest wrt each other are they not? This is not that complex a geometry problem is it. Are there attributes of "motion" we do not see here? The biggest problem in GPS, AFAIK is the correction of position due to the non-zero time of flight of the SAT position data down linked to the ground station. This requires a massive calculation and comparison scheme to get the proper positions in a state of useful equilibrium. Well you did say that the rotational affects of time diolation are cancelled by gravitational red shiftaffects. Here is another opinion. I like this question and noone has answered so I will blurt out a guess and hope someone will correct or add incite: My thought is no because if any speed matched/compensated then all speeds would. Having said that there must be some angle (perpendicular component) such that it would work out where the blueshift is exactly compensated for by the total speed. What is that angle? I think it would be somewhat profound if it was 45 degrees so that is my guess. Geistkiesel {Note:I tried to use VbCode to reference MacSwell above but couldn't get it to work. Any suggestions?} Link to comment Share on other sites More sharing options...
J.C.MacSwell Posted May 7, 2005 Share Posted May 7, 2005 The SATS in orbit are force free. It would be most difficult to measure any say 20 meter stretch of orbit trajectrory and measure "curvature" . For all practical purposes the SATS are in inertial frames' date=' especialy when in geostationary orbits. What ever the statement means that "you can't really say that either the earth or the satellite is not rotating", [b']the SAT and the ground station are at rest wrt each other are they not?[/b] This is not that complex a geometry problem is it. Are there attributes of "motion" we do not see here? The biggest problem in GPS, AFAIK is the correction of position due to the non-zero time of flight of the SAT position data down linked to the ground station. This requires a massive calculation and comparison scheme to get the proper positions in a state of useful equilibrium. Well you did say that the rotational affects of time diolation are cancelled by gravitational red shiftaffects. Here is another opinion. Geistkiesel {Note:I tried to use VbCode to reference MacSwell above but couldn't get it to work. Any suggestions?} They are but it is definitely not an inertial frame although it could be a close approximation locally. How much higher is the satellite than the ground from the center of the Earth? Link to comment Share on other sites More sharing options...
J.C.MacSwell Posted May 7, 2005 Share Posted May 7, 2005 1- Is an object in geostationary orbit' date=' pointing in a steady direction toward the Earth (not rotating) considered to be at rest WRT the Earth? 2- If X and Y are at rest WRT eachother, and Z and Y are at rest WRT eachother, then Z and X must be at rest WRT eachother. True or False? Thanks.[/quote'] If X and Y are at rest WRT each other in an inertial frame that they are at rest in, and Z and Y are at rest WRT each other in that inertial frame, then Z and X are at rest WRT each other in that frame. Link to comment Share on other sites More sharing options...
swansont Posted May 7, 2005 Share Posted May 7, 2005 The SATS in orbit are force free. It would be most difficult to measure any say 20 meter stretch of orbit trajectrory and measure "curvature" . For all practical purposes the SATS are in inertial frames, especialy when in geostationary orbits. What ever the statement means that "you can't really say that either the earth or the satellite is not rotating", the SAT and the ground station are at rest wrt each other are they not? This is not that complex a geometry problem is it. Are there attributes of "motion" we do not see here? Not complex from a geometry standpoint, but then, we are discussing it in the context of relativity and have the ability to do some fairly precise measurements. So for all "practical" purposes, in this context, they are not in inertial frames. The earth is in fact rotating, and the satellite is in fact revolving around the earth. The biggest problem in GPS' date=' AFAIK is the correction of position due to the non-zero time of flight of the SAT position data down linked to the ground station. This requires a massive calculation and comparison scheme to get the proper positions in a state of useful equilibrium.[/quote'] How is it a big problem? The time delay of the signal is what measures the distance, so that's hardly a problem, and the satellite ephemeris data is downloaded quite easily. It works. Link to comment Share on other sites More sharing options...
geistkiesel Posted May 7, 2005 Share Posted May 7, 2005 They are but it is definitely not an inertial frame although it could be a close approximation locally. How much higher is the satellite than the ground from the center of the Earth? J.C.MacsewellI, I believe the last number I saw was 26 KM. If the SATS are not inertial frames what attributes of nature remove it from the inertial classification? Certainly no measureable forces are sensed by the SAT hardware once in orbit, but the SAT did experience considerable acceleration forces getting up there which could and probably does affect timing circuits. The SAT is changing direction at a rate of 360/24x3600 = 1.4 x 10^-4 degrees/sec which is not such a forceful number that cannot be easily corrected with the algorithims. I have seen the references to the non-inertial frame status of SATS, so how can there be any discussion of the application of SRT to GPS calculation My research tells me that SRT is the fundamental scheme in GPS calculations. Geistkiesel Link to comment Share on other sites More sharing options...
geistkiesel Posted May 7, 2005 Share Posted May 7, 2005 If X and Y are at rest WRT each other in an inertial frame that they are at rest in, and Z and Y are at rest WRT each other in that inertial frame, then Z and X are at rest WRT each other in that[/b'] frame. Is there any obvious way that x and y being at rest wrt each other can be in separate frames, meaning one of the two is moving wt the other? Ditto for Z and Y. Maybe I have been up too long. I cannot see that which motivated your reply.. If x and y are motionless wrt each other then x and Y can be considered welded together can they not? If z and y are also at rest wrt each other then the welding seam includes Z abd X also. If x and y are on a moving frame then Z would also have to be on that frame in order to be at rest wrt y wouldn't it? I am not arguing your point here. Please, if I missed something let me know. Geistkiesel Link to comment Share on other sites More sharing options...
swansont Posted May 7, 2005 Share Posted May 7, 2005 J.C.MacsewellI, I believe the last number I saw was 26 KM. If the SATS are not inertial frames what attributes of nature remove it from the inertial classification? Certainly no measureable forces are sensed by the SAT hardware once in orbit, but the SAT did experience considerable acceleration forces getting up there which could and probably does affect timing circuits. GPS orbits are around 20,000 km (26,000 km from earth center). Geostationary orbits are around 36,000 km (42,000 km from earth center) They are not inertial because they are in a rotating coordinate system. Space-qualified hardware in such satellites is tested extensively to ensure that the forces present during launch do not affect their behavior afterwards. It's not the timing circuits. The SAT is changing direction at a rate of 360/24x3600 = 1.4 x 10^-4 degrees/sec which is not such a forceful number that cannot be easily corrected with the algorithims. I have seen the references to the non-inertial frame status of SATS' date=' so how can there be any discussion of the application of SRT to GPS calculation My research tells me that SRT is the fundamental scheme in GPS calculations.Geistkiesel [/quote'] Your research is wrong. Link to comment Share on other sites More sharing options...
Johnny5 Posted May 7, 2005 Share Posted May 7, 2005 GPS orbits are around 20' date='000 km. Geostationary orbits are around 42,000 km [/quote'] 36000-42000 actually. The average i think is 39,000. Link to comment Share on other sites More sharing options...
swansont Posted May 7, 2005 Share Posted May 7, 2005 36000-42000 actually. The average i think is 39' date='000.[/quote'] Sorry, the number I gave for the geostationary was for r, not the altitude. Altitude is indeed about 36,000 km. But it can't have a range of values - there's only one solution, for a circular orbit above the equator. Link to comment Share on other sites More sharing options...
Johnny5 Posted May 7, 2005 Share Posted May 7, 2005 Sorry' date=' the number I gave for the geostationary was for r, not the altitude. Altitude is indeed about 36,000 km. But it can't have a range of values - there's only one solution, for a circular orbit above the equator.[/quote'] Yes I know this, and was going to carry out the calculation. Do you know it offhand? I seem to have forgotten it, though I remember basically how to get it. I only did it once, and that was long ago. Link to comment Share on other sites More sharing options...
J.C.MacSwell Posted May 7, 2005 Share Posted May 7, 2005 J.C.MacsewellI' date=' I believe the last number I saw was 26 KM. If the SATS are not inertial frames what attributes of nature remove it from the inertial classification? Geistkiesel [/indent'] This frame requires pseudo force "corrections" depending on the accuracy required. The speed of light is not constant in this frame although it may be close enough locally for most purposes. Pluto is much faster in this frame than Mars and yet Pluto still manages to stay in orbit. Link to comment Share on other sites More sharing options...
swansont Posted May 7, 2005 Share Posted May 7, 2005 Yes I know this' date=' and was going to carry out the calculation. Do you know it offhand? I seem to have forgotten it, though I remember basically how to get it. I only did it once, and that was long ago.[/quote'] Centripetal force = gravitational force, and the period is a sidereal day. Link to comment Share on other sites More sharing options...
Johnny5 Posted May 7, 2005 Share Posted May 7, 2005 Centripetal force = gravitational force, and the period is a sidereal day. Thank you Dr Swanson, I will re-perform the calculation once again. Probably monday Dr. Kind regards Set centripetal force equal to gravitational force, and period is one sidereal day. That'l keep till monday. Link to comment Share on other sites More sharing options...
J.C.MacSwell Posted May 8, 2005 Share Posted May 8, 2005 Is there any obvious way that x and y being at rest wrt each other can be in separate frames' date=' meaning one of the two is moving wt the other? Ditto for Z and Y. Maybe I have been up too long[b']. I cannot see that which motivated your reply.. [/b] If x and y are motionless wrt each other then x and Y can be considered welded together can they not? If z and y are also at rest wrt each other then the welding seam includes Z abd X also. If x and y are on a moving frame then Z would also have to be on that frame in order to be at rest wrt y wouldn't it? I am not arguing your point here. Please, if I missed something let me know. Geistkiesel My motivation was to give an example that I was confident was correct. I did not mean to imply an "if and only if" only that I suspected it could be wrong otherwise. If you and I are at rest or "welded together" in one inertial frame (Let's say you at the start of the Indiannapolis 500 and me on my butt in front of my computer) we are not necessarily at rest in all inertial frames. Given enough spatial separation in one that we are at rest in, you could choose a continuum of inertial frames that correspond to you starting, racing and finishing the Indiannapolis 500 all with me sitting in front of my computer hitting the letter "x" (all the same moment for me, and each frame containing one different moment for you). We of course would not be at rest wrt each other in all those frames. Link to comment Share on other sites More sharing options...
geistkiesel Posted May 9, 2005 Share Posted May 9, 2005 Sorry' date=' the number I gave for the geostationary was for r, not the altitude. Altitude is indeed about 36,000 km. But it can't have a range of values - there's only one solution, for a circular orbit above the equator.[/quote'] Swansont, Not to be counting angels doing the texas two step on the head of a pin, but aren/t mountains and equatroial bulges, for example, exceptions to the otherwise circular trajectory rule? Geistkiesel Link to comment Share on other sites More sharing options...
swansont Posted May 9, 2005 Share Posted May 9, 2005 Swansont' date=' Not to be counting angels doing the texas two step on the head of a pin, but aren/t mountains and equatroial bulges, for example, exceptions to the otherwise circular trajectory rule? Geistkiesel [/indent'] Small perturbations. I was speaking of the first-order solution. Link to comment Share on other sites More sharing options...
losfomot Posted May 9, 2005 Author Share Posted May 9, 2005 OK, How about this... If you were to set up a mirror on a stat. sat. and then shot a laser at it (from Earth) to reflect back down at you, would the laser be shot back at a single spot, or would it move around? Link to comment Share on other sites More sharing options...
swansont Posted May 9, 2005 Share Posted May 9, 2005 OK' date=' How about this... If you were to set up a mirror on a stat. sat. and then shot a laser at it (from Earth) to reflect back down at you, would the laser be shot back at a single spot, or would it move around?[/quote'] I think it would move, though I'm not sure of what the scale of position vs. time would be. AFAIK even geostationary satellites need some station-keeping if they need to be over a particular spot. You can't get a perfectly circular orbit. Link to comment Share on other sites More sharing options...
geistkiesel Posted May 10, 2005 Share Posted May 10, 2005 My motivation was to give an example that I was confident was correct. I did not mean to imply an "if and only if" only that I suspected it could be wrong otherwise. If you and I are at rest or "welded together" in one inertial frame (Let's say you at the start of the Indiannapolis 500 and me on my butt in front of my computer) we are not necessarily at rest in all inertial frames. Given enough spatial separation in one that we are at rest in' date=' you could choose a continuum of inertial frames that correspond to you starting, racing and finishing the Indiannapolis 500 all with me sitting in front of my computer hitting the letter "x" (all the same moment for me, and each frame containing one different moment for you). We of course would not be at rest wrt each other in all those frames.[/quote'] This is old business, I am just cleaning up loose ends. Would it be fair to say that the instantaneous velocity difference between you and I effectrively alows us to assume we are at rest wrt each other wrt to one of the infinite frames available in the universe that you described? Link to comment Share on other sites More sharing options...
J.C.MacSwell Posted May 10, 2005 Share Posted May 10, 2005 This is old business' date=' I am just cleaning up loose ends. Would it be fair to say that the instantaneous velocity difference between you and I effectrively alows us to assume we are at rest wrt each other wrt to one of the infinite frames available in the universe that you described?[/quote'] Yes, and nicely stated. Link to comment Share on other sites More sharing options...
geistkiesel Posted May 11, 2005 Share Posted May 11, 2005 Yes, and nicely stated. Having said all that, J.C., and agreeing as we did, I run into my drunken uncle who got kicked out of SRT school and he mutters something about,"but you and J.C.MacSwell were in realtive motion wrt each other, and the fact is he was stationary wrt the earth frame that you were moving over 200 mph wrt. So what physical reality is there in the use of the infinite number of inertial frames that you are aware if only because an SRT expert told you about these frames? Can you ever see, one of them, touch one of them or have any kind of observation of one of them? Then he took the last few gulps from his bottle of Glenfiddich and keeled over stoned drunk. How do I answer him when he wakes up? Link to comment Share on other sites More sharing options...
J.C.MacSwell Posted May 11, 2005 Share Posted May 11, 2005 Having said all that' date=' J.C., and agreeing as we did, I run into my drunken uncle who got kicked out of SRT school and he mutters something about,"but you and J.C.MacSwell [b']were[/b] in realtive motion wrt each other, and the fact is he was stationary wrt the earth frame that you were moving over 200 mph wrt. So what physical reality is there in the use of the infinite number of inertial frames that you are aware if only because an SRT expert told you about these frames? Can you ever see, one of them, touch one of them or have any kind of observation of one of them? Then he took the last few gulps from his bottle of Glenfiddich and keeled over stoned drunk. How do I answer him when he wakes up? You accept the consistency of the math, but not the experimental results? Or not the physical interpretation? Link to comment Share on other sites More sharing options...
geistkiesel Posted May 14, 2005 Share Posted May 14, 2005 OK' date=' How about this... If you were to set up a mirror on a stat. sat. and then shot a laser at it (from Earth) to reflect back down at you, would the laser be shot back at a single spot, or would it move around?[/quote'] The instantaneous angle of the laser beam to the perpendicular to the plane of the mirror at the instant the beam struck the mirror would define the spot where the beam would refkect to. If the laser was a continuous beam that tracked the SAT (and mirror) the reflected beam would then trace out a line defined by the laser/perpendicular direction. It would be a neat trick to have the beam reflact back to the laser source. Link to comment Share on other sites More sharing options...
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