studiot

I'm sorry you are so quick to challenge, rather than ask for further detail as to why I said what I said. No it is not exactly the same as the work of the others, although there are commonalities. If you were not so hasty you would see that you even misquoted my statement. I did not say Mdv/dt any more than I used m or F = dp/dt anywhere. I hoped you would follow my extended commentary instead of reaction with a Putin style knee-jerk. The whole point is to distinguish between M and m as well as V and v. If you did not understand it ASK.

I will answer your question, even though you haven't answered mine. Note this is your thread in Speculatrions and the rules you recently accepted when you joined require your answer. The constancy of G was established by Cavendish and confirmed by many subsequent more sensitive experiments. Also indirect confirmation comes from the experimental fact that a large part of theory in Astrophysics and Astronomy yield results consistent with constancy rather than variability. How many observations and experiments have you conducted ?

How am I supposed to take your question seriously ? First your claim is that g/G is the distance to the Sun. Then you question if G is constant. If you think G is not constant how cna it be related to a sensibly constant distance ?

OK then to continue. I am presenting this a Physics exercise, not a formal maths one, although maths is involved. Physicists are allowed to bend the maths rules a bit. Let us consider a rocket travelling along. The trick to analysing this is to use the fact that momentum is constant for the whole rocket system (rocket plus exhaust), although it is variable for each part separately. Thus system momentum = p = a constant. But we analyse this by considering the parts of the system separately. At some time t the rocket has mass M and velocity V. At some time (t +dt) these have changed so the rocket now has mass (m + dm) and velocity (V + dv). Note dm is negative . Furthermore -dm is the mass of the exhaust, which has a velocity v relative to the rocket or velocity (v-V) to a ground observer Splitting the system mass in this way saves introducing m for the exhaust mass and simplifies the maths. Here is a diagram showing these essentials. So the system momentum at time t = p MV (first figure).................1 Note this excludes the need for considering the exhaust momentum at this stage But the system momentum at time (t + dt) is also = p = (M + dM)(V + dV) - dM(V-v) .......................2 Multiplying out the terms and taking care of the signs then either subtracting equation 1 from equation 2 or simply using the fact that the right hand sides are both equal to p we can equate them directly leads to the expression MdV + vdM = 0 .........3 Two notes here. This is a momentum equation, not a force equation. This equation is the springboard to answer questions about the desired properties of the rocket system. So for instance we can remember our schoolboy calculus and note it can be rewritten as (after dividing through by M) dV = -v d(lnM) where lnM is the natural log of M But you have asked for the force on the rocket also known as the thrust. To get this we proceed as follows We divide equation3 through by dt to obtain [math]M\frac{{dV}}{{dt}} + v\frac{{dM}}{{dt}} = 0[/math] If we let [math]\alpha = - \frac{{dM}}{{dt}}[/math] Then [math]M\frac{{dV}}{{dt}} = \alpha v[/math] This is the required force on the rocket.

What works ? I have shown you that it does not work on other planets for example Jupiter. I thought exchemist's die analogy was rather good. Didn't you? For your information g is specific to local conditions, even on Earth. It is derived from G by applying these local conditions. This is why G is called the Universal Gravitational Constant - it applies everywhere. G is used to calculate something called gravitational potential, and also the gravitational force exerted between any two bodies with mass, and that calculation involves the values of both masses. g is used to calculate the force on a singlebody with mass in the vicinity of Earth, but g already incorporates a factor for the mass of the Earth and varies fromplace to place on the Earth's surface or height above it.

That's good I can use them then. But so as I don't spring this idea on you let me explain why I am using r not x for distance. Consider Newtons second Law Force = mass times acceleration. Easy peasy yes ? But there is a twist. This refers to a single body (or system) that remains intact and does not shed (or gain) material. In fact it refers to the centre of mass (COM) of that body. And r is a common symbol for the position of that COM in your coordinate system. So when we quote Newton's 2nd law we really mean that The product of the mass of a body and the acceleration of its COM equals the force imposed on it. Swansont refers to this as the net force which is important for a rocket system, because the net force is exactly zero, in the absence of drag, gravity etc. This means that the acceleration of the COM of the rocket system must be zero since the mass is not zero. This is both good and bad. First the bad news: we cannot use f = ma directly since it only tells us the 0 = 0. Now the good news: we can use it to form a suitable momentum equation to work with. ( I will post this next time) I have slyly slipped in the phrase 'rocket system' there. That is because we normally speak of rigid body dynamics. The point about a rigid body for the purposes of Newton's 2nd law is that however the COM accelerates, so will any other point of the body. But we don't have a rigid body. We have a system of bodies that deforms with time. So your title has some truth in it, but not quite for the reasons you propose. You are however right to be suspicious of some of the explanations offered online.

I don't necessarily disagree with you. In fact I think the translation barrier has not helped added to the fact we have got several members offering different approaches and/or explanations. This must be very confusing. I also note that different texts derive different (not necessarily incompatible) conclusions and results from the same material and starting point. This fact is why I asked you what you want to calculate. Since you have told me several times that you do not want to calculate anything I now think perhaps you are actually asking to understand certain terms in one version of 'the rocket equation'. Another sidetrack is the introduction of 'generalised coordinates' and 'generalised momenta', both of which are coordinates and momenta in name only. This really is sidetracking a steamroller to crack a nut and quite over the top. Studying this updating of simpler mechanics is really important to Physicists but takes at least a year of hard study. You have a track record on not answering the one or two questions I aks in a post. They are all designed to help us (especially me) help you so please answer this one. Are you familiar with the dot and double dot notation for derivatives with respect to time ? [math]\mathop r\limits^{ \bullet \bullet } [/math] and [math]\mathop r\limits^ \bullet [/math]

[math]\mathop r\limits^ \bullet [/math]

New figures from the Office for National Statistics.

Firstly let me say that I don't think this belongs in the Relativity section. The extract is very short and I am not familiar withe the source but I think the subject concerns only the ordinary time we measure and live with on the surface of the Earth. The clue is in the statement "it’s inconvenient for noon to occur in the middle of the night in some parts of the world,". I would agree that the phraseology is rather poor but I think that 'coming apart from' describes the fact that 'noon' is related to the Sun being overhead and that this will only happen at 12:00 on the Grenwich Meridian. Everywhere else it will happen at some other time, unless the residents there use a locally adjusted time other than GMT. It is a pity also that his example of the use of standard GMT elsewhere in the world is military. Surveyors, navigators and astronomers have for centuries used it as the basis for star tables for peaceful purposes as well (and still do). Does this help ?

So would you say this was crackpottery ?

There is indeed additional theory to apply relativity to rockets. But I would suggest that you deferred this until you have sorted out non relativistic theory. I say non relativistic because relativity is now counted as part of classical physics. When doing this sort of thing it is useful to make a list of symbols and what they represent, to avoid using the same symbol for different things in different places. So first of all what do you mean by F ? You have one overall system and two subsystems viz the exhaust, the rocket itself and the combination of rocket and exhaust. That gives the possibility of 9 forces acting to choose from. Some of these may be zero or equal to other forces. Each of these 3 (sub)systems will have their own mass, acceleration, velocity and momenta as a result. Now consider the case where the rocket itself has zero acceleration. What do you think this means ?

Perhaps @MigL knows the answer to this but I would be very suprised if any useful rocket was designed for constant mass exhaust. Different accelerations and velocities will also be required at different stages of the flight. This implies that the force applied to the rocket will vary over the flight time. These are parameters that can be worked out for a desired flight history.

It is the experience of all my family and friends that curry, casserole and cottage pie all taste better the second time around. So my question is Is there any way to skip the first day ?

I feel this thread is wandering further and further off topic which is (not was) the formal logical soundness of a point particle. The Cosmologist has not answered my question to him about quantum numbers. In my oldfashioned day we had two quantum numbers concerned with 'angular momentum. There were 4 in all labelled, n, l m and s respectively. The first 3 had a classical interpretation, although they mark stationary point solutions to the Schrodinger equation. These describe the energy, position size and shape of the obitals. The final one, s, has only a quantum interpretation although it is common to refer to it as the spin or quantum spin of the electron about its own axis, the actual behaviour is not exactly that of a mechanically spinning anything. The second one, the azimuthal quantum number, l, can be associated with classical angular momentum of the electron as a point particle rotating about the nucleus. So where are we in relation to point particles ? Incidentally as regards the so called infinite limit that also seems to be causing trouble, We have no problem with density being a point function limit of mass over volume, nor of pressure being the force over zero area in the limit. I respectfully suggest crossing swords with a real organic chemist about chirality would be difficult. Chirality embraces a host of phenomena of which handedness is only one, but I'm sure exchemist would run rings round any of us on this subject.

Surely you dont need a course in basic calculus ? Definition momentum = mass x velocity p = mv Therefore if both m and v are variable [math]\frac{{dp}}{{dt}} = m\frac{{\partial v}}{{\partial t}} + v\frac{{\partial m}}{{\partial t}}[/math] Where both m and v are functions of t. To work this out you would need to have equations for both m and v as functions of t, unless you could devise an equation connecting m and v. Do you have such equations ? Some treatments specify that the mass flows out at a constant rate, that is [math]\frac{{\partial m}}{{\partial t}} = C[/math] Alternatively you can calculate the changing mass flow required to produce constant acceleration.

You said To which I replied Yes if the lifetime is 1 second* the muon expeience 1 second of 'life'. But the rest of what I said means that the lab observer sees (measures, experiences) the moving muon taking much longer than this to die, whilst the muon in his hand take exactly 1 second to die. *(Actually the time is about 2 microseconds, not 1 second)

Just saying it doesn't make it so. I don't understand why you have this problem as an electrical engineer you must be familiar with the equation Instantaneous power = instantaneous voltage x instantaneous current. The analogy is exact Instantaneous force = instanteous mass x instantaneous voltage. So spit out your objection fully please.

That was a particularly high handed dismissal of my request for you to describe what you see as 'suspicious' You made the claim, not I, so it is up to you to justify it.

By the way. There is nothing in classical EM theory to connect charge directly to energy. If there were it could be like heat and we would observe convertion as well as exhaustion of the source. In that sense charge is more like gravity. Both are inexhaustible supplies of their particular effect.

I don't think so. That would be true if and only if 'space' was embedded in a higher dimensional manifold. Current theory is that this is not the case and that the physical manifold we call space is not embedded in anything. That is possible because of Gauss famous little theorem - The theorema egregium - which distinguishes between intrinsic and extrinsic curvature.

Why should that be suspicious ? All the equation tells us is that force is the product of two independent quantities. Independent means that each or both can be varied without reference to the other. Setting one or both constant just makes for easier maths.

We have certainly not exhausted or explored all the rooms in 'dimension house' yet. But we have found some phenomena that run counter to 'common sense' or our ordinary experience. Nor have we arrived at a consistent definition of dimension, suitable for all purposes. We have also has a couple of threads of discussion here at SF about the subject.

It must be because the thing you and I call space does not accelerate, or even move.

I understand what you are getting at but Trying to play the smartass with everybody has led you to post some ridiculous English. Nothing 'happens to points of view'. They do not change. However the muon experiment is really good in that it allows comparison of effects in two different frames and introduces of the second fundamental point of relativity. Of course it does. That is exactly your point and it is correct as far as it goes. But it is not complete because there are two different frames involved. The ground or laboratory observer measures different times in his laboratory for the same effect to occur when the muons with him in the laboratory, and when they are approaching it at speed.