Our exchanges are passing each other in cyber space so I do not think we are changing Earth spin rate Please Register or Log in to view the hidden image! That is not exactly my question but surely has the same answer. You said you worked in this area (unless I am confusing you withsome one else) so you surely must know that a geostationary satellite is not launched into orbit to have a period equal to the pre-launch "length of one mean Earth day". They consdier the launch a great success if the period is within a minute of 24 hours. If it is, everyone who had anything to do with the satellite or launch is half drunk with celebration (except for a few "kill joys" who don't touch the stuff.) Satellite is now the operators / owners "baby." After the hang-over is gone, and they are back on job next day, they are asking "How did this schedule slip?" and "What are we going to do to catch up?" etc. They no longer care that the operators are firing tiny thrusters (which do change the total angular momentum of the satellite (and of course the Earth satellite system) as that exhaust mass escapes into deep space to get the period very close to "one mean Earth day." My question assumed that the launch crew was perfect and knew exactly what the effect of the launch of A would be on the length of the day so they made it that "after launch" period. What they did not know was another satellite was soon to go up and change the length of the day, causing A to be slghtly not geostationary anymore. Launch crew for B, knew exactly the length of day after A is in correct orbit, i.e. matches the rotation rate of Earth changed a little by its own launch (I am here assuming that none of the operators fine thrustors changes the Earths rotatation, because it was a "perfect launch" and did not give the operators anything to do.) Launch crew B also knows how their own launch will change Earth's rotation rate, so they, again with perfection, leaving the operators again with nothing to do, placed B into the new longer day's geostationary orbit - a SLIGHTLY higher one than A had before A's operator start firing small thrustors, (I believe) Question of the poll, is what is the value of this "SLIGHTLY"? I have already noted your differentiation is in error as it has no a^2 term, so will not reply to rest of your post.
You asked me to describe the mechanism that torques the earth to change its spin rate and I did so. Now I ask the same of you. Please describe the mechanism that torques the rest of the universe. In fact, lets make it easier, just tell how (mechanism) alpha centura's angular momentum is changed to be part of the "recoil torque' you think is applied there and else where in univers but not the Earth to change its spin by the mechanism I suggested. Support this OPINION with some numbers or reference. At least comare the effect to that of a big rain storm. I think smaller effect have been estimated by calculation as this one has been. Most times when a leap second adjust is made to the year (as at end of 2005) it is to add a leap second, but not always. Nearly random variation of the day length is orders of magnitude greater, so actually observation is of the average rate impossible. One can average the changes that have been required over the period in which atomic clocks have existed but for example, if he oceans are warming and expanding the current rate of slowing may be differrent than what is calculated. This is best investigated, not by telescopes or atomic clocks, but by the ancient records of in what cities the eclipse of the sun was total. Your value, if it were a constant average, would make eclipse totality occur in different cities than where the Chinese saw them several thousand years ago. Must go now will read rest and reply later.
Of course I know that. The orbit transfer is performed either using timed burns or using accelerometers to measure the accumulated delta-v. Timed burns are a terrible way to go; rocket thrust varies by 5-15 percent. Since accelerometers have an uncorrectable bias of 20 microG or so, even accumulated delta-v cutoff results in fairly large errors by the time the vehicle reaches GEO. The satellite is placed very precisely at some point in time. Who owns the satellite and exactly how it gets to its target orbit is a detail. Exactly. Who owns the satellite and exactly how it gets to its target orbit is a detail. At some point it is placed very precisely. That is all that matters -- except for the minor issue of where GEO is, and how launch and travel to GEO changes GEO altitude -- i.e., the point of this thread. My answer remains, none of the choices corresponds to "SLIGHTLY". With the cumulative effect of all of the spacecraft launched by mankind over the last 45 years, even a few micrometers change would be observable. Nanometers or picometers are SLIGHTLY, but you did not offer those as choices. I already did supply some numbers. I re-expressed your original list of changes to geocentric orbit radius to changes in length of day. The smallest entry (10e-4 meters) corresponds to a change in the length of a day of 300 nanoseconds; the largest (1 km) corresponds to a change of 3 seconds. The Earth's space agencies collectively launch hundreds of satellites per year. Multiplying your list by 1000 (100 launches per year * 30 years of launches*2/5 LEO/GEO = 1200 or 1000) yields a cumulative change of 300 milliseconds to 3000 seconds. Do you not think such a change would be observable? This is not the sound of an ant marching in a thunderstorm. Compared to the measurement accuracy of 13.5 decimal places for the length of a mean day, your list of choices corresponds to a large fleet of jets flying overhead at 100 feet. Ok, bright boy, what is the correct relationship? This is basic stuff! Check the units, Billy. My answer delta a = 2/3*a/T*delta T is correct. Units are length on both sides of the equation. An a^2 term would not yield correct units. This is an application of a general rule for derivatives of power models, and is very useful in modeling. Given some power relationship y(x) = a*x^p Differentiation with respect to "x" yields y'(x) = p*a*x^(p-1) Multiplying both sides by "x" yields y'(x)*x = p*a*x^(p-1)*x = p*a*x^p = p*y(x) or y'(x) = p*y/x The IERS does not mention your effect. They do however, have analyses such as this which shows the variations in length of day to sub-millisecond level from 1995 to 1999. I can see a long term drift in this plot, well explained by momentum transfer to the moon's orbital motion. I can see seasonal effects. I do not see any large jumps corresponding to satellite launches. There were a lot of satellite launches between 1995-1999. Where are the three millisecond jumps corresponding to a one meter change in geosynchronous altitude?
We are simply saying that it slows the earth, but to some extreemly exteemly small degree. You don't get somethig for nothing. Every action has an equal and opp. reaction. Why is this so hard to understand. It's a very simple concept! This is not rocke.....ummm ok it's rocket science but I sware that a 6th grader could understand the concept of every action has an equal and opp. reaction, and how people so smart could argue otherwise is way way beyond me. SHAME SHAME SHAME (finger furiously shaking) Having said this, I would like to point out the the effect would be so small, that I don't think any correction at all would be needed for the 2nd sat.
According to many in this thread, launching the satellite towards the East to take advantage of Earth's equatorial velocity slows the rotation of the Earth, they say to conserve angular momentum. OK, launch the same weight satellite into orbit, except launch towards the West. Yes, it takes more fuel to launch that way, but according to your thoughts, launch to the West will INCREASE the Earth's rotation rate, correct? The satellite will increase in angular momentum regardless of direction of launch. Now, if angular momentum is conserved in the satellite/Earth system, the Earth will still have to decrease its angular momentum regardless of the direction of launch, if the satellite gains in its amount of angular momentum. Same for a polar launch, the satellite will gain angular momentum acording to calculations after launch. Do you still maintain this 'increase' in angular momentum by the satellite comes from a decrease in Earth's angular momentum? Edit: removed 'geostationary' from post
Agreed. But you have to draw the boundaries of the system correctly. Drawing them incorrectly leads to incorrect results. The question can then be broken into three parts: Does launch itself change the rotation rate of the Earth? That is, is there any change in the rotation rate in the few seconds it takes for the vehicle to clear the pad? Does the motion of the free-flying vehicle from nanometers off the ground at Earth-rotation rate to LEO at 1 rev/90 minutes change the rotation rate of the Earth? Does the orbital transfer of the free-flying vehicle from LEO (1 rev/90 minutes) to GEO (1 rev/day) change the rotation rate of the Earth? I agree that the system boundary for the first question is indeed the Earth+satellite system. However, launch itself does not change the rotation rate of the Earth. I disagree with the system boundary being the Earth and satellite for the third part. The same strings (gravity) that couple the Earth and satellite also couple them with everything else in the universe. The Earth+satellite are not an isolated system. Arbitrarily drawing the boundary around the Earth and satellite for the transfer from LEO to GEO leads to incorrect results. The second part is no different than the third, except that the vehicle and the exhaust will transfer some momentum to the atmosphere (and thence to the Earth). Drawing the boundary around the Earth and satellite is incorrect once the vehicle is off the ground. In the end, this is a bunch of arguing over how many angels can dance on the head of a pin. Playing devil's advocate here, let's assume that the angular momentum of the Earth+satellite system is conserved. The angular momentum of the Earth is 8E37 kg-m^2 * 2*pi/86400 seconds, or 6E33 kg-m^2/s. The angular momentum of a 1000 kg satellite in GEO is 1.3E12 kg-m^2/s. Using delta T/T = delta L/L or delta T = delta L/L * T yields a change of 2E-17 seconds in the length of a day. Finally, using delta a = 300 m/s * delta T at GEO altitude yields a change of 6E-15 meters for the change in altitude of a geosynchronous orbit. The effect won't be noticeable until we have launched about half a million satellites into orbit.
Angular momentum is a (pseudo) vector quantity. So an "increased" angular momentum in the opposite direction is essentially simply a large negative angular momentum. -Dale
That is a really interesting plot. I assume that the seasonal effects you mention are the local minima near the half-year marks. Do you know what that is from? I think the satellite-launch effects are obviously undetectable, particularly with such large natural variations. I don't think that the thread is intended to be anything other than an exercise in estimation. Assuming no other effects than the satellite launch, assuming no rocket exhaust escapes, assuming a perfect solid spherical earth, etc. can you estimate the magnitude of the effects to a couple orders of magnitude? -Dale
Do you read your own posts? Moment of inertia times angular velocity is the same as linear momentum times radius. See CANGAS's post for the proof if you don't believe NASA. Anyway, it's still easy to plug in the numbers no matter what forumla you choose. Why don't you do it?
we agree on all this. I wanted people to only consider the thrusting of the rocket with the assumption that all the exhaust gases were retained on Earth I think we agree that where on Earth is not very important, but to be specific I suggested aluminum + oxygen making Al2O3 dust exhaust that fell to Earth. when worrying that some might get bothered by fact that the CoM of fuel before launch was slight higher than the Al2O3 would be after satellite is in orbit, I suggested the Al2O3 could be caught on a platform that circled the Earth at the original height of the CoM of the fuel in rocket pre launch. I did not want to have people worring about the change in the total system angular momentum that is associated with exo-atmospheric burns of tiny thruster need to correct exactly the orbit, not only for changes in length of the day but solar pressure, residual drag, magnetic interactions with Earth's field etc. Most people understood the question posed, correctly I think. yes i did so offer. your estimates are in the first group, one I have several times called bottom less. I think you voted for this group with the majority and may be correct. I will only list 2inqusitive as voting for the zero group as he clearly thinks it is exactly zero change. That is the displayed count of the bottomless group is correctly including your vote. I think our disagreement has been reduced to the observability of the effect. You claim it is observable as timing is very precise. I claim it is not observable as there is much greater variation in larger know effects (signal to noise problem makes effect of even many, certainly one, satellite launch on Earth's spin rate unobservable.) You have given a number for the rate of spin decrease and from this conclude, with your analysis, that because such a small rate of decrease is known the effect of satellite launch would be observed, if I understand your position correctly. I can only note that current decade average rate of spin decrease, is not the same as the average observed over several thousand years. (This by noting where on Earth the total eclipse, whose time of occurrence even 3000 years ago is very accurately known, was actually observed - written down by Chinese scholars while my ancestors and most of those active here in English were ill-literate fools thinking the gods were angry or something like that if they happened to experience a solar eclipse.) I have already congratulated you on your use of the a^3 ~ T^2 law. I probably would have done the same, but doubt I would have made it even simpler, A LOCALY LINEAR equation, as you did with a first order Taylor series expansion about the 24-hour day point. That was clever. I.e. For people here who have no idea as to what you did, let me put it this way: Effectively, DH plotted the "a vs. T" curve for a small region very close to T = 24 hours and then used the slope of that curve to estimate the effect of a change in A. He found that a one-meter change would correspond to roughly 0.003 seconds on the length of the day. As he knows that the long-term average rate of change is much smaller, he concludes that this great (0.003s) a change is impossible, and he may well be right. I note, however, that the three of us in the 1cm to 1M band are betting that the change is between 0.000,3 and 0.003 seconds. DH's argument has persuaded me, if his numbers are OK and they seemed to be, that our only hope to be correct is near the low end of this band. For example if 0.000,3 seconds is the change, then in 365 days we have about 0.11 second longer year for each 1000KG put into geostationary orbit that year. 2005 did require the addition of one "leap second" at end of year but if I could vote again, armed with HD's information, and assuming it correct, I would at least move down to the 0.1 to 1cm band if not join him and others in the "bottom less " majority. Anyone want to argue for a band other than the "bottomless band"? I think the E-4 to E-2M band surely has a chance still. - I only rejected it as stated earlier because it is 100 times smaller than the next greater one. It has been fun, and for me at least educational.
I think to many on this post this means that DH agrees that there is an effect. To me the question was: Is there an effect. The answer is yes. But I agree with DH in that the effect is so very small that it could go unnoticed for a very long time. Certainly the effect is not large enough to have to correct for after launching one satellite.
This kind of linearization is widely used. Non-linear equations are a bear to work with. It is so much easier to work with linear equations. So, when faced with a non-linear system, a common solution is to linearize (and then of course pay attention to the errors that result). You mentioned in an earlier post that spacecraft are overjoyed if the spacecraft is delivered even close to the target. The targetting algorithms used to bring the spacecraft from "close" to spot-on are linearized equations called the "Clohessy-Wilshire Equations" (or just CW).
Pseudo-vectors don't transform like "true" vectors. In particular, a true vector transforms to its additive inverse on reflection (negation of the coordinate axes). A pseudo-vector stays the same on reflection.
No, they wouldn't know about it because the size of the effect is orders of magnitude smaller than other unpredictable effects. That's right, because the effect is insignificant beside the other causes. Oh! My vote was a guess! A wildly wrong guess. By my calculation (which I'm not sure I trust, but it's better than a guess), the correct answer is perhaps a tenth of a millimeter. Sorry about that. Please Register or Log in to view the hidden image! It means we know the mean length of a day now to within 0.1 milliseconds, and we know the mean length of a day then to within 0.1 milliseconds. This means that we know the average daily change over the past century to 13 decimal places... but that's not much use when the actual daily change varies in the order of millseconds, is it? Including exhaust, right? I'm still disagreeing (but you knew that Please Register or Log in to view the hidden image! ). Are you maintaining that the gain or loss in the Earth+satellite system angular momentum comes from other astronomical bodies? How, exactly? If a satellite was launched from a planet in an otherwise empty universe, would the angular momentum of the planet+satellite be conserved?
Just when I thought I brought DH into the real world of cause and effect and a laws hundreds of years old Pete points out that he did say that he said: DH is right that the satellite is part of a much larger system as is every thing else in the universe including the earth. Take venus for example or murcury Venus is in some sort of resonence with the earth so that it shows the same face to earth whenever we are nearest to it. Murcury shows the same face to the sun all the time. But DH misses the point that in our earth satellite system the earth is already slowing so any additional slowing would not be counteracted by the sun or the moon or anything else likely. I cast my vote with Pete.
I agree but note that both DH and I concur that the launch only get the satellite close - I.e. entire launch crew celibrate if they turn a satellite over to controllers with a period of 24 hours plus or minus one minute. Point being that all geostationary satellites have, and use often, small thrustors to "station keep"* against much greater problems than the lauch of another geostationary satellite changing the Earth's spin rate. To do this they will have probably made some exo- atmospheric final burns, that as Dale has explicitly noted should not be considered in the guessing of the answer (not that they would make any difference, except in principle they violate conservation of angular momentum of our system if any of their exhaust goes away from Earth. I mentioned some reasons before why it is necessary to have some small thruster to keep on your assigned geostationary spot so here will just tell that once, when some one else was sick, it fell my lot to go the NASA, in Greenbelt MD I think it was, to place small magnets on a satellite to try to zero out the residual magnetic field satellite had. That is an art, not a science, and I needed help by NASa's local experienced "magnetic balancing artist." I initially thought: OK they just measured the net field and orientations for me, so I will put and "equal and opposite" strength magnet there to cancel the measured field out. That idea does roughly work at first on killing the gross initial field, but the new magnet induces fields in most materials, even those not normally considered magnetic. This satellite's main job was to carefully measure the magnet field of the Earth, solar wind effects on it, etc. so the field of the satellite was taking to space had to be almost killed. After about 15 much smaller magnets than my first try, had been added to the satellite with help of the "magnetic artist" the satellites own field at the end of the extended boom where the magnetometer would be was “nulled out.” I had to go to that NASA facility as they have a test room with large coils that took away the Earth's field. ------------------------------------------------- A station in the geostation 360 degrees is quite valuable and the UN has given rights to various nations. I believe a small spot is "owned" by some UN member - a tiny island country in South Pacific, not on the tourist routes, and renting its spot to telcom companies is its major source of foreign exchange!
This raises the question, What is inertial mass, and why is it the same as gravitational mass? The most satisfactory answer that I have found is that inertial mass of some object (and hence its inertia) is the sum of the gravitational interaction between the object and all of the other objects in the universe. The combined gravitational acceleration toward every other object in the universe is what makes an object resist change to its state. Turning the table, how exactly does a thrusting rocket change the Earth's rotation rate? Please don't answer with conservation of angular momentum, as I am of the opinion that you have drawn the boundaries of the system incorrectly. Suppose the rocket is in LEO and performs a Hohmann transfer to GEO. Does that action change the Earth's rotation rate? How? Now suppose the rocket is orbiting Jupiter. Does firing the rocket's engines change the Earth's rotation rate in this scenario? What is the difference? Given my answer to the first question, I don't think so. I suspect the laws of conservation of linear and angular momentum arise because the universe is so large and so uniformly distributed.
