Thread split from [thread=424724]this thread[/thread]. Not true. The spray can is generating a constant force, not a constant power. As a result, the work done by the spray can depends on the distance it travels, and the power generated depends on the rate at which it is traveling. The farther the spray can travels, the more work is done.
Re: Zero-G reaction forces Careful there, cjl! You are mixing reference frames here. The spray can is generating a constant force in the inertial frame instantaneously co-moving with the person holding the spray can. It is not generating a constant force in all inertial frames.
Re: Zero-G reaction forces Yes it is. In all frames, the massflow from the spray can is the same, and the velocity difference between the spray can and the spray is the same. Therefore, in all frames, the force is the same.
Re: Zero-G reaction forces No, it's not. The force is d/dt(mv), and this a frame-dependent quantity when the mass is not constant (which it most certainly is not in the case of a rocket, or a spray can).
Re: Zero-G reaction forces Sure it is. The definition of force is F=dp/dt, not F=ma. In the case of a variable mass object the force acting on the object is a frame-dependent quantity.
Re: Zero-G reaction forces Nope. If the massflow from the can is constant, and the can's exhaust velocity is constant, then the force is constant. The acceleration will be variable if the can's mass is changing, but the force will not be variable. (Specifically, using your definition, dp/dt = dm/dt*v. If dm/dt is constant, and the exhaust velocity is constant, then dp/dt is constant. So, if dp/dt is constant, then by your own definition, F is constant).
Re: Zero-G reaction forces Wrong again. Now you are not only mixing frames but you are mixing quantities! Your v here is the velocity of the exhaust relative to the vehicle (or in this case, the velocity of the spray relative to DaveC). Do again without mixing things up. To avoid mixing things up, use [itex]m[/itex] is the (time-varying) mass of the vehicle, including the yet-unburnt fuel. [itex]v[/itex] is the velocity of the vehicle from the perspective of some inertial frame. [itex]p[/itex] is the momentum of the vehicle from this frame, [itex]p=mv[/itex]. [itex]u[/itex] is the velocity of the exhaust relative to the vehicle. [itex]\dot m[/itex] is the mass flow rate. F=dp/dt is the definition of force, and it is very much a frame dependent quantity here.
Re: Zero-G reaction forces Please state the transformation law for the force components from one Galilean frame of reference to another.
Re: Zero-G reaction forces Oh, I'm not the one mixing things up here. Vehicle mass is irrelevant, so m vanishes. Vehicle velocity is irrelevant, so v vanishes Vehicle momentum is irrelevant, since both terms it is dependent on are irrelevant. Exhaust velocity matters, so u is the first term out of your list that actually shows up. Exhaust flow rate matters, so [itex]\dot m[/itex] is the only other item on your list that actually shows up. This is easily demonstrable by the fact that from any frame, the vehicle's exhaust will have a constant speed relative to the vehicle (at the instant that it is emitted). Since the fuel was originally traveling with the vehicle, this relative velocity u is also equal to the [tex]\Delta[/tex]V of the fuel. The fuel mass flow rate [itex]\dot m[/itex] is then used to figure out the fuel's [itex]\dot p[/itex]. This is fairly easily done through the equation [itex]\dot p[/itex] = [itex]\dot m[/itex]*[tex]\Delta[/tex]V. Since [tex]\Delta[/tex]V is constant (as we defined exhaust velocity to be constant), and [itex]\dot m[/itex] is constant (as we defined massflow to be constant), the exhaust's [itex]\dot p[/itex] will always be the same, as viewed from any frame. Since conservation of momentum holds, the vehicle's [itex]\dot p[/itex] must be equal to -[itex]\dot p[/itex] for the fuel. Since the fuel's [itex]\dot p[/itex] is constant, the vehicle's [itex]\dot p[/itex] is also constant, and therefore F is constant. (Sorry for some of the strange formula appearances - I'm not that familiar with Latex)
Re: Zero-G reaction forces "Vehicle mass is irrelevant"? I know of no such provision in any reputable contexts of physical laws.
Re: Zero-G reaction forces That inertia or momentum is irrelevant is preposterous. Those aspects exist in all mass-systems. Never been proven otherwise.
Re: Zero-G reaction forces Short answer: Suppose inertial frame B is moving at velocity V wrt inertial frame A. The force (using F=dp/dt) on some object with non-constant mass m in frame B is [tex]F_B = F_A + \dot mV[/tex] In short, force is not a galilean invariant. Long answer: Dealing with variable mass systems are a bit tricky. The choices are Use F=dp/dt. This has the obvious advantage of being connected with the conservation laws. It has also the obvious disadvantage of not being a galilean invariant. Use F=ma. This is a galilean invariant, but it is no longer connected with the conservation laws. Using F=ma will lead to all kinds of problems. It is not valid for computing work, for example. Rewrite Newton's second law as [itex]F_{\text{ext}} = dp/dt - u\dot m[/itex]. For example, see Halliday and Resnick. (Did you sleep through freshman physics?) There is a slight problem here with work and momentum. This is essentially F=ma again. Throw up your hands in disgust and claim that Newton's laws only apply to systems with constant mass. For example, see Plastino & Muzzio, Celestial Mechanics and Dynamical Astronomy, 53:3 (1992) http://articles.adsabs.harvard.edu//full/1992CeMDA..53..227P/0000227.000.html. Oh yes you are. 'Nuff said. Well, almost 'nuff said. You are mixing frames. The exhaust velocity is only constant with respect to the vehicle, and the vehicle is accelerating. Arguing whether Newton's laws do or do not apply to objects with non-constant mass is, IMHO, being a bit too pedantic. Arguing whether Newton's laws do or do not apply in accelerating frames is a lot less pedantic. They don't -- at least not without the addition of fictitious forces.
Re: Zero-G reaction forces This is wrong. Your invalid conclusion is a consequence of trying to use the definition F = dp/dt. This equation has F_{net} on the left side and that it strictly holds for point particles. Mechanics teaches us that 2nd Newton's Law is not a definition of force. Forces are defined independently of it and depend on the nature of the bodies that interact and their relative position and velocity. For example, Newton's Law of universal Gravitation: [tex] \mathbf{F} = -G \, \frac{m_{1} \, m_{2}}{r^{2}_{1 2}} \, \hat{\mathbf{r}}_{1 2} [/tex]
Re: Zero-G reaction forces Good one! Tell me another. I have deadlines galore and humorous posts such as that do help.
Re: Zero-G reaction forces If you think this is humorous, then your deadlines are probably related to some accounting project.
Re: Zero-G reaction forces Rocket motors do not change in thrust just because you attach them to a different vehicle. The force they generate is purely a function of their massflow and their exhaust velocity. Hence, vehicle mass and velocity are irrelevant. (Specifically, F = [itex]\dot m[/itex]V_{e} for any rocket motor, in which [tex]\dot m[/tex] is the fuel mass flow rate, and V_{e} is the exhaust velocity)
Re: Zero-G reaction forces Of course if you're in the vehicle frame, you have to add fictitious (inertial) forces, but that's somewhat irrelevant to my point here. The rocket motor (since that's effectively what we're talking about here) is a constant thrust device in any frame. If you ignore drag, and operate in an inertial frame, this means that the net force applied to an object by a rocket motor is constant (assuming the mass of the rocket motor is negligible). In a non-inertial frame, there will be other forces, yes, but the rocket motor is still generating the same force as it always was, even in a non-inertial frame.
Re: Zero-G reaction forces Actually, my deadlines are very closely related to what has become the topic of this thread. This stuff is my job. In other words, "Why, yes, I am a rocket scientist." The reason your post was humorous is because it is exactly contrary to the modern interpretation of Newton's second law. This modern interpretation is that Newton's second law is not truly a scientific law. Instead, Newton's second law is definitional. In particular, it defines the concepts of momentum and force.