Wow. I'm learning more about angular momentum than I expected. It's interesting that I can find no examples of how mass loss effects (or dosen't effect) the angular momentum of a rotating body. So, if I am spinning in a chair holding a weight, and I drop the weight (no torques applied) my rotation rate will remain unchanged. So, if we do our experiment on the moon (no atmospher), and the rocket launches normal to the surface, the answer to the poll is exactly zero change. Yes?
Several things DH states I can agree with; for example, that without "strings," and gravity weakening so that current spin of Earth was able to throw off parts of outer shell of Earth, then the iron core remaining would maintain its current spin rate (assuming no "parting torques" during the separation of the outer layers) But on the main point I think Pete (and I) are correct. I.e. There is a very tiny decrease in the spin rate of Earth, cause by placing the satellite into orbit. Below I prove DH is wrong (I think) by simply considering the "before launch" and "after in orbit" states, without any need to go into details of the exhaust gases speeds etc. So let me offer a simple proof that there is a change in spin rate: Let the total angular momentum of the pre-lauch system, Ao, be decomposed into two parts, Ae, is everything but the rocket and satellite angular momentum, Ars. I.e. Ao = Ae + Ars, but for the initial conditions, subscript "o," the satellite is in the rocket on the launch pad, waiting for launch. Let the initial tangential speed of the of the launch pad, (which happens to be flush with surface of the Earth, for people who like to quible), in a non rotating reference frame fixed in the center of a spherical Earth of radius "r" be called: "Vo" and this speed is Vf after the satellite is in geostationary orbit at radius "R" Clearly R > r, as the notation implies. The question being discussed with DH is which of the following is true: (1) Vf = Vo or (2) Vf < Vo. The thread has assumed that (2), not (1), is true and DH has questioned, to say the least, this assumption. DH claims (1) is true. All agree angular momentum must be conserved, I.e. that Af = Ao so lets calculate (actually just express) both: Ao = Ae + Ars = f.r.Me.Vo + (r+h).Mr.Vo{1+ h/r} Where to avoid use of "I" and "pi" I have imagined that the entire mass of the Earth, Me, is traveling at speed Vo perpendicular to a radial line from center of the Earth, but is only f.r distant from the center of the Earth. I.e. f < 1, and its value makes the first term on the right hand side of equation above exactly Ae. Periods in equations represent multiplication. (Note Me does not include, Mr, the mass of the rocket plus satellite on launch pad.) For computational convenience, the satellite is assumed to be at the center of mass of the rocket and both are h above the surface of the spherical Earth, whose radius is r. The strange term inside { } is to account for fact that the center of mass of the rocket is not only farther than r from center of Earth, - i.e. at (r+h), but also has slightly greater tangential speed than Vo. The final angular moment of the system is: Af = f.r.Me.Vf + r.(Mr - 1000Kg).Vf + R.(1000Kg).Vg.....* Where Vg is the speed of the 1000Kg geostationary satellite in orbit at radius R from center of the Earth and as before, since the satellite is geostationary: Vg = Vf {R/r} so with this subsitution in last equation above and subtracting two equations: Af - Ao = 0 = f.r.Me.(Vf -Vo) + r.Mr.Vf - r.(1000Kg).Vf + R.(1000Kg).Vf.{R/r} -r.Mr.Vo.{1+ h/r} - h.Mr.Vo.{1+ h/r}. or 0 = (f.r.Me + r.Mr)(Vf -Vo) - Mr.Vo.h - h.Mr.Vo.{1+ h/r} - (1000Kg).Vf.{r - (R^2)/r} but the two terms of the first line of equation immediately above prefixed with negative signs are: - Mr.Vo.h - h.Mr.Vo.{1+ h/r} = - Mr.Vo.{2h + (h^2)/r)} and separating out the negative part of the final term and changing signs of all terms (one sign changed by moving first term to "zero side" of equation) we have: (f.r.Me + r.Mr)(Vf -Vo) = Mr.Vo.{2h + (h^2)/r)} + (1000Kg).Vf.{r - (R^2)/r} or "Eq. A" is: (Vf - Vo) = [Mr.Vo.{2h + (h^2)/r)} + (1000Kg).Vf.{r - (R^2)/r}]/(f.r.Me + r.Mr) For DH to be correct, both sides above of "Eq A" must be exactly zero. For Pete & me (and several others I think) to be correct, both sides must be negative. (Assuming I have not made mistake in algebra, which is surely possible.) The numerator in right side of "Eq.A" above has two terms. The first is clearly positive and because {r - (R^2)/r} = {r^2 - R^2}/r and R is much greater than r, the second term is clearly negative. Thus it is clear that DH can not be correct as if (Vf - Vo) were zero, then these two opposite sign terms in "Eq A" must ALWAYS satisfy: Mr.Vo.{2h + (h^2)/r)} = - (1000Kg).Vf.{r - (R^2)/r} or if Vo = Vf as DH contends, then Vo/Vf = 1 = - (1000Kg).{r - (R^2)/r} / Mr.{2h + (h^2)/r)} or, multiplying both sides by r: r = - (1000Kg).{r^2 - (R^2)} / Mr.{2h + (h^2)/r} or r = (1000Kg/Mr)[(r+R)(R-r) /{2h + (h^2)/r)}] = (1000Kg/Mr)[K], where K is some constant. The radius of the Earth certainly is NOT a linearly function of the mass of the satellite to mass of rocket ratio! Thus, either I have an algebraic mistake (quite possible, but not likely to change the conclusion) or DH is wrong. Showing that Pete and I are correct, by this approach may more difficult, but I will work on it IF I find time. --------------------------------------------- *In this equation, I am being kind to DH by assuming that all of the exhaust has condensed back onto the surface of the Earth, as if the rocket fuel were oxygen and aluminium dust and the resulting Al2O3 was resting on the ground in the final state. In reality the raising the fuel mass, converted to exhaust gases in the air, to some average altitude will by itself slow the Earth spin rate, just as a fast spinning ice dancer, with arms at her side, slows her rotation rate when she extends her arms out again.
Angular momentum is most certainly not conserved between the "before launch" and "after in orbit" states. The spacecraft throws itself into orbit by firing its rockets. Those rockets change the energy of the system; the conservation laws do not apply. Angular momentum is only conserved between the "instant before launch" and "instant after launch" states because the rockets haven't had any time to do any work.
If that were true, then you should be able to tell what "outside of system" torque has been acting. Recall that all of the exhaust gases and solid rocket parts are back on surface of Earth (by my "kind" aluminum dust+ oxygen fuel assumption discussed in footnote). If you need to, imagine that pre-launch Earth is far from sun, moon and all other objects. I.e. There is no outside agent to act. The angular momentum of a closed system such as this is conserved, with no torques from the outside the system acting NO MATTER HOW YOU RE-ARRANGE THE PARTS OF THE SYSTEM. No matter what forces one part of the system applies to another of the system to achieve the re-arrangement or how complex these forcres are, etc. If you do not believe the standard "no change in angular momentum unless system is acted upon by torque applied by outside agent" law, please state your different requirement or conditions under which angular moment is conserved.
I think BillyT and D H would be in agreement if the method of launch for the satellite had been defined as an equatorial earth-mounted cannon, pointed east, aimed toward the horizon at some angle. Certainly, in this case, there would be some slowing of the earth's period of rotation, however miniscule, (according to both D H and BillyT's hypotheses). If we allow the cannon to recoil back on wheels, intuitively it seems like the the earth's rotation would not be affected as much, but in theory it should still be affected by the same amount because of conservation of (angular) momentum laws. In a way, this seems analogous to fuel and the atmosphere in the original experiment. It is not easy (for me, anyway) to visualize all of the angular momentum being transferred back to the earth via the fuel particles settling out of the atmosphere.
Ok...it's been years since my classical mechanics class, but I believe DH is correct regarding launching a rocket from the surface of a rotating sphere. However, I believe you could easily fix the problem by modifying the question so that you are launching your satellites with some sort of cannon/railgun/whatever. Edit:woops, Neddy Bate beat me to it. Imagine a figure-skater spinning on her tows with her arms outstretched, holding a metal ball. Her angular velocity won't change if she simply releases the ball, even if the ball later accelerates under its own power; the ball no longer has anything to do with her angular velocity once it's released. On the other hand, if she actively tosses the ball away from her tangential to her direction of rotation, she would indeed slow down slightly.
It is good that you can not because it is not. The momentum of the satellite in geostationary orbit has increased from the value it had while sitting on the launch pad at the center of mass of the rocket. Ergo the angular momentum of all other parts of the system has decreased by an equal amount.* ---------------------------------------- *To be absolutely correct in this, one should catch all the falling rocket parts and Al2O3 dust on a platrom with altitiude h above the Earth, so that there has been exactly zero change in the in radial distantance of this mass (at height h both before and after satellite is in orbit at radius R.) Then the first paragraph's decrease of Ae is exactly compensated by the increase of the angular momentum in the satellite.
Perhaps that would make it easier for you to see /understand, but it makes absolutely no difference what forces are used to place the satellite into orbit,* so long as satellite is the only thing to permanently leave the earth and the mass distribution of all else remains the same. (No change in the moment of inertia of anything but the satellite. -See footnote of the post I just made replying to Ned.) ------------------------------------- * To be extreme (and have some fun): Assume rocket starts out below the sea and air pressure in a polaris sub's launch tube sends it to the surface. There it ignites, but seem to be sputting. Fortunately a cowboy in the "photograph helicopter" throws a rope arround the sputtering rocket. The force of sputtering rocket plus the lift of the helicoper get it to 10,000 feet, but it is wildly swinging and the sputtering exhaust burns thru the rope, so without the lift of the helicopter, it begins to fall back towards the sea, but fortunately at 1000 feet, enough fuel has been burnt so that even the sputteing fuel rocket can keep its self in level flight for 3 minutes. Then the sputtering stops and full rocket thrust begins. The guidance system get it headed upwards. At "flame out" it is finally going fast enough to coast to thru the last part of the atmosphere and coast into elliptic orbit. It's increditable luck continues as it is hit by another satellite and after this collision is in perfect circular geostationary orbit at R from the center of the Earth!!!! (The "other satellite" no doubt has its angular moment altered, as did the atmosphere, as did the sub, as did the sea, as did the helicoper, the rope, etc. but all these are in the term Ae, so these details as to how Ae is divided up do not matter.) If I may steal from DH: All these details do not make "one iota" of difference. What I have shown mathematically by considering only the initial and final states remain exactly unchanged! Ain't "conservation of momentum" wonderful (and powerful)! Please Register or Log in to view the hidden image!
Billy T, you are wrong, AGAIN. The rocket does not 'steal' angular momentum from the Earth to gain orbit. It is powered by its chemical engines. No angular momentum is lost from the Earth, no work is done by the Earth. No, I am not a physicist, but I knew this before D H gave his excellent analysis. That is the main reason I didn't vote in your gedankin, you had no 'zero' option.
:bugeye: So what if the (kinetic) energy of the system is changed? Angular momentum (and linear momentum, of course) must still be conserved.
Right. Unless whatever the ball uses as reaction mass is tied to the skater, or later comes to rest on the skater. Eg if the ball accelerates [EDIT] in a straight line [/EDIT] by ejecting ball bearings behind it which are caught by the skater, the skater will slow down, right?
Yes, if we ignore gravitational interaction Please Register or Log in to view the hidden image! Of course, I don't want to go there... because then the answer would depend on the relative angular position of the rockets, and to get really pedantic we might have to consider the interaction between the two rockets as well, which puts us in the realm of a three-body problem. Please Register or Log in to view the hidden image!
I think this captures the problem with the anology of the skater releasing a weight. Every action has to have an equal and oppisite reation. If there were no gravity or atmosphere maybe the anology would be correct - but it isn't is it. A better anology would be a girl made out of steel throwing magnets into orbit around her. There may as well be a string between the magnet and the girl - both pulling on eachother. You would have to take into consideration conservation of angular momentum.
Never said it did "Steal", only that the total system angular momentum is constant, so if one part increases (the satellite) the remained must decrease. You are welcome to your opinions, even if contradicted by mathematical analysis, as they are and have been.
Billy T, when I said the satellite does not 'steal' any angular momentum from the Earth, that is exactly what I meant. The satellite has exactly the same angular momentum sitting on the launch pad as it does in orbit, geostationary or otherwise. Angular momentum is conserved in both the satellite and in the Earth's rotation. It does not increase or decrease for EITHER body. The KINETIC energy of the satellite does increase, the conversion of gravitational energy (potential energy) to kinetic energy.
I don't think many people understand angular momentum or the conservation of it, and I don't think I want to argue or explain it. I'll just ask how is an orbiting body different then an ice skater that abducts her or his arms? An orbiting body in geosync. orbit is "tethered" by gravity as much as the skaters arms are tethered by bones and ligaments. http://en.wikipedia.org/wiki/Angular_momentum Oh, really. How would you like to be hit by boxer with 3 foot arms as opposed to a boxer with 1 foot arms (depending on type of punch thrown)? An object traveling around the planet has much more energy for the very same reason. Get 2 pieces of strings, one 6in long and one 3 feet long. Tie a small rock on the ends. Which do you think would have more energy if you swung them around your body as you were turning in a circle? Now after you've thought about this - do you really think a satellite has the same amount of energy on the surface of the earth as it does in space? If you are still not convinced, I have one more experiment for you: Step 1. Take the rocks with strings on them and make one 2" long and one 12" long. Step 2. Swing the one that is 2" long as hard as you can with your dominent hand and hit your self square on the top of the head with it. Measure the bump. Step 3. Repeat step 2 only use the one 12" long. Compare the size of the bumps and if bump number 1 is smaller then bump number 2 then you don't understand the conservation of angular momentum correctly. If bump number 1 is bigger then you may submit your evidence (the bumps on your head) to this forum by way of digital pictures, and if you're convincing enough, we may help you to rewrite the law of angular momentum!!
After flipping a coin, having destroyed my weejee board in a trash fire long ago, I have decided to guess that an object in orbit around the Earth has greater angular momentum than it has if it is sitting still on the surface of the Earth. After the thread is over, I might want to calculate it. Does anyone know who to ask for the definition and mathematical equation of angular momentum?
Perhaps you do not know the definition of angular momentum? For a point mass (as our satellite has been assumed to be) the magnitude of the angular momentum about a point "X" is the product of following three things: (1) Mass at the point, M. (2) Distance, D, along the vector line, L, from the point X to the mass point. (3) Speed, S, perpendicular to the "X" to "mass point" line. Thus the magnitude of the angular momentum about any point "X" is MDS. In our case of interest, "X" is the center of the Earth, but please note that the angular momentum is linear in the value of D, which would be different if the angular momentum were being claculated about some other point. For example, the angular momentum (neglecting quantum "spin") about the point mass itself is zero because then D = 0. Both "D" and "S" are much greater when the satellite is in orbit than when it was on the launch pad. Thus you comment is incorrect. Note, however, that angular momentum is a vector. How that vector is calculated is discused below: There are two different well-defined "multiplication operations" for vectors. One called the "cross product" and the other the "dot product." Put the word "vector" in front of these names and Google to read more. Speed, S, is not a vector, but velocity, V, is. (A very common convention, used here, is to make vectors bold.) The above names for these product come from their standard notation. I.e. for the two vectors given above: The dot product is written: L . V where the "dot" should be a little higher, but impossible to do here. L*V looks better, but does not make the origin of the name as clear. The result of the dot product is always a scaler. I will not explain it further, because it is of no use for this problem, but hope you will "google" to learn what it is, when it is useful, etc. I just note it is very useful, sometimes essential, in calculating the work done or changes in potential energy as something is moved in a force field etc. (Just to be complete, I should have "conservative" in front of "force field" as if force field is not "conservative," then the simple concepts of "potential" do not exist, but that is an entirely different lecture.) The cross product is written: L x V. The result of the cross product is always another vector, which is perpendicular to the plane formed by the two original vectors. (The side of that plane on which it extends requires too many words to describe here, but is well defined by the "right hand rule.") In this vector notation, the definition of angular momentum, A, of point mass about X is: A = M(L x V) The equation, given above, (A = MDS) for the magnitude only follows from this definition. (There is no other definition, only this one. - One should never give two different definitions for the same thing, unless you have a solid proof that the two definition are in fact identical, but only "look" different. How the A = MDS equation follows from the definition is illustrated near the end of this lecture.) If you think, even slightly, what the effect of L is, you will understand that the satellite's angular momentum is much larger when the satellite is far from the center of the Earth. Also it is much larger because the "perpendicular speed", S, which in this circular orbit case is the magnitude of the velocity vector, is much larger also. (On the equatorial lanch pad the satellite travels approximatel 25,000 miles around with the Earth in 24 hours, but much much farther in 24 hours when it is in the geostationary orbit.) Note speed is always the magnitude of the velocity, but only at four points in an elliptical orbit would the velocity be perpendicular to L. That is, at all other points in an elliptical orbit the following is true: |L x V| / |L| < S where the vertical bars indicate magnitude. (I think you can understand why this is true, if you try.) For the perfectly circular orbit, however, the following is always true: |L x V| / |L| = S. You should learn both what the definition of angular momentum is and how to use this powerful, compact, vector notation. I hope this helped you, but expect it will not; however, I like to teach, even though I am now retired, and think at least some will learn from this effort, even if you do not. To make another example of what you can do with conservtion of angular momentum, consider a comet at apogee and perigee using: A = M(L x V). Note this is true at ALL points on the eliptical orbit. Not only true, but A is constant in both magnitude and direction! In fact this is not restricted to objects in orbit, but follows from the definition for all objects at all times, in any type of motion or stationary! (It is the vector form of the definition of angular momentum - so must be true "by definition.") For "extended objects" you apply this to each "point" and integrate over the volume of the object to get the total angular momentum of the object about the point "X." Apogee and perigee are two of the 4 points in an elliptical orbit where: |L x V| / |L| = S so |A| = M|(L x V)| = A = M(|L|)S, but we have called |L|, "D" above, so for the comet at apogee and perigee (plus two other points): A = MDS as before, but note this is a scaler equation about magnitudes, not a vector one. Thus the ratio of the comet speed at apogee (point most distant from sun) to perigee (closest point on orbit to sun) is: (Va / Vp) = (Sa / Sp ) = |Lp| / |La| = Dp / Da, where the subscripts refer to the two respective points. Or in words: At apogee and perigee, the speed ratio is the inverse of the distance ratio. This comes directly from the definiton of angular momentum given above at the start of this lecture, and fact it is concerved. My "reducto-ad-adsurdum" proof that the satellite's Ab is not same as Af, where subscript "b" is "before" and "f" is "after", also comes from this definition and conservation of A. OPINIONS to the contray, do not count for much against math, logic (algebra) and definitions.
Hay! That looks about right. Thanks again, BillyT! Maybe 2inq' will get it. I have a confession to make. I already knew about angalur momuntem. I was trying to be sacrostic.
MY "lecture post" just prior to your request was not made in response to you intentionally as I was writing, editing and still modifying it while you posted, but it seems to be just what you were asking for. Please look at it again, especially the last parts of the "lecture" as more has been added. BTW, you are welcome.
