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The hopper proper, the water in the tank on top of the hopper, the water flowing between the tank and the drain hole, and the water that has left the hopper together form an isolated system. The total linear momentum of this isolated system is constant. EDIT. Something is seriously wrong with scienceforums.net. LaTeX is not working. I tried posting the math, but all that came out was [ LaTeX Error: One or more directories do not exist ] Is this related to the performance problems described in this post:http://www.scienceforums.net/forum/showthread.php?p=313047&highlight=performance#post313047 ? Bottom line: If the water that has left the vehicle has a non-zero momentum, the hopper must end up with a non-zero velocity to compensate. This is exactly what happens. CPL.Luke, Show the math that exhibits how the entire system starts at rest, the hopper ends at rest, but the exhaust water has a non-zero velocity.

Weigh three balls versus another three balls, then one ball against another ball. The choice of the balls for the first weighing is random. The choice for the second weighing depends on the outcome of the first: First weighing shows equal masses: Weigh the remaining two balls against each other. The odd ball is the heavier of the two. First weighing shows unequal masses: Weigh two of the three balls from the heavier group against each other. If the second weighing shows equal masses, the odd ball is the unselected ball in the heavier group. If the second weighing shows unequal masses, the odd ball is the heavier one.

Is this a homework assignment, or are you just looking for an answer out of curiousity? If it is the former, you need to show some work on the subject.

A very short growing season. Krummholz forests just below the tree line comprise short, twisted trees.

The water falls from the hopper with the same horizontal velocity as the hopper. Ignoring vertical motion, this is akin to a rocket for which [math]v_e=0[/math]. The intent of this contrived setup was to create the same equations of motion as those of the vehicle in space with a pair of opposing thrusters, one on each end of the vehicle. One very good reason for employing conservation laws in analyzing some system is that doing so enables one to avoid investigating the mechanisms that change behavior. I know that in the initial configuration (full tank, closed valve), the total system (vehicle+water) is at rest with respect to some inertial observer. In the final configuration (empty tank, water flowing under the tracks), the water flowing under the tracks has some momentum. To conserve total momentum the vehicle must have exactly the opposite momentum as that of the water. In this case, the mechanism that results in the change in vehicle momentum is opening and later closing the valve. The change in fluid momentum flow inside the vehicle changes the vehicle's momentum. Simulating real rockets. We want to ensure that the propagation techniques themselves are very accurate so that any discrepancies between the simulation and reality result from stochastic effects and modeling errors.

No, you have not. First, I'll examine what happens to the hopper car that leaks water to the ground. Nomenclature: [math]\dot r_{v_{flow}}[/math] Hopper car velocity while water is flowing [math]\dot r_{v_{final}}[/math] Hopper car velocity after water flow terminates [math]r_{v\rightarrow f}[/math] Location of the water tank wrt the center of the hopper car [math]m_v[/math] Dry mass of the hopper car [math]m_f[/math] Mass of the water in the tank prior to opening the valve [math]\dot m_f[/math] Constant rate at which water tank mass changes Scenario: A hopper car on a frictionless track starts at rest with respect to the ground with the valve to the water tank on top of the hopper car closed. The valve is opened at some time [math]t_0[/math], at which time water immediate begins to flow at a constant rate [math]\dot m_f[/math] from the tank toward the center of the hopper. The water drains from the hopper to a frictionless channel underneath ground. The velocity of the drain water with respect to the hopper car is zero. Water drains from the tank to depletion at some time [math]t_f = t_0 + m_f/\dot m_f[/math]. Analysis: Ignoring end effects (which I can make arbitrarily small by increasing the amount of water in the tank), the hopper car moves with a constant velocity while the water is flowing. Call this velocity [math]\dot r_{v_{flow}}[/math]. To conserve momentum, the hopper, the water in the tank, and the water under the track must have a velocity that opposes the flow inside the hopper: [math](m_v + m_f) \dot r_{v_{flow}} = - \dot m_f r_{v \rightarrow f}[/math] Note that [math]\dot m_f[/math] is negative. Thus the vehicle moves in the direction of the water tank while water is flowing. After tank depletion, the water under the track continues to flow with the velocity [math]\dot r_{v_{flow}}[/math]. To conserve momentum, the hopper must have a final velocity [math]m_v \dot r_{v_{final}} = -m_f \dot r_{v_{flow}}[/math] or [math]\dot r_{v_{final}} = \frac {\dot m_f\, m_f}{m_v (m_v+m_f)} r_{v \rightarrow f}[/math] In other words, the hopper does not stop when the flow stops. Instead, it changes direction. The exact same situation occurs with the dual-thruster rocket. The rocket moves one way while the thrusters are firing, then reverses direction when the flow stops. When applied to a realistic rocket, the result is that the velocity is always a tad smaller than the result acheived by integrating the momentum transfered to the exhaust gas. The effect on the velocity is small. The effect on position is not small -- at least to me. I have to certify that the propagation techniques used to simulate vehicle behavior are accurate to 1 part in 10-12 or so.

Mass and energy are conserved. Mass can be converted to energy, and vice versa. That does not mean they are the same thing. They are different things that can be converted to the other. If they are one and the same thing, then aren't the electron, positron, and the photons they create upon collision also one and the same thing?

Of course it is "invalid", in the sense that most simple models are "invalid". (More precisely, they are only approximately correct.) A lot of simplifying assumptions are made in deriving the rocket equation. Quoting from the Wikipedia article, Ignore all the details of timing. The rocket equation ignores the fluid flow momentum (and a host of other items, such as gravity, thrust variation, and thrust vectoring).

If you read that in my post, I apologize for the bad write-up. There is no third setup in my original post on this concept. There is no additional ramp. The fluid flow momentum from the tank to the ith thruster is [math]\dot m_{t_i}(\vec r_{s\rightarrow t_i} - \vec r_{s\rightarrow f})[/math] where [math]\vec r_{s\rightarrow t_i}[/math] and [math]\vec r_{s\rightarrow f}[/math] are vectors from some fixed point on the rocket to the ith thruster and the fuel tank. The flow momentum to the thruster further from the fuel tank is larger (in magnitude) than the flow momentum to the nearer thruster. It is this flow momentum that the rocket equation ignores. Nothing wrong with that per se; a lot of physics is knowing what effects to ignore and what effects need to be modeled. More later; my wife and kids are waiting on me.

CPL.Luke, In the first setup, the car will come to a rest as soon as the water flow is turned off. The main points of this setup were (1) to show that the vehicle does indeed move as a result of fuel flow and (2) to set things up for the variable mass situation. In the second setup, the car does not stop moving when the fuel flow terminates. The fuel flow inside the vehicle forces a residual velocity that balances the flow outside the vehicle. There is no third setup --- yet. The third setup follows. This setup does indeed involves rockets. Two of them on one vehicle. The vehicle is symmetric except for the location of the fuel tank. Two rockets are placed on the vehicle, one on each end, each thrusting outward. The ascii diagram below depicts this simple vehicle. The [] represents the off-center fuel tank location. >=====[]==< The two thrusters have identical specific impulses and fuel consumption rates. The net force on the vehicle is therefore zero. The asymmetric fuel flow dictates that the vehicle itself gain some momentum while the thrusters are firing. This momentum does not magically disappear when the fuel flow terminates.

They are not interchangeable. Two examples: Photons are massless but have energy proportional to their wavelength. You can catch a baseball thrown by 6 year old barehanded. Try doing that with the same ball thrown by an ace pitcher.

A bit off-topic, but why in the world should NASA be all over this one? NASA gets less than 1% of the federal budget. They are already stretched beyond the limit doing what the President and Congress asks them to do now. Last I checked, developing new power generation techniques is not under NASA's purview.

I assume you meant exhaust velocity when you wrote the above. This is indeed THE "rocket equation". Ragib said "escape velocity", which is quite a different concept from exhaust velocity. I couldn't agree more. While math and physics are beautiful in and of themselves, LaTeX makes papers written about math and physics beautiful to behold. Quoting (or misquoting) some equation written elsewhere doesn't take much ability at all. Deriving the rocket equation requires some simplifications. The ability to simplifying things (but avoiding oversimplification) is an important skill in applied physics.

It does not "simply cancel out". Here's a simple example. Drill holes in the center of the top of a covered hopper car (one of these). Assume the hopper is quite deep so that the center of mass of any fluid poured into the hopper will always be at the center of the hopper, lengthwise. Place a channel along the length of the top of the hopper car, sloped so that any fluid poured onto the track will run into the hole in the top of the hopper. Put a water tank over the channel and located some distance from the hole. Put the hopper car at rest on a frictionless railroad track. Open the spigot to the water tank so the water drains into the hopper. What happens to the hopper car? Answer: it moves. To an observer fixed on the hopper car, the horizontal location of the system (water+water tank+hopper car) center of mass changes as water flows from the off-center water tank to the hopper. However, no horizontal external forces act on the entire system, so to an observer fixed on the ground, the horizontal location of the system center of mass does not move. The hopper has to move to make the system center of mass stationary. Now repeat this experiment, but with a hole drilled in the bottom of the hopper as well. Now what happens to the hopper? It still moves, but with a different velocity than the original problem. Imagine that a frictionless channel is placed between the frictionless tracks so that any water that spills from the hopper maintains its momentum. The center of mass of the entire system (hopper car + water tank + water flowing on top of the hop + water under the hopper) is stationary. Think of this latter case as a zero-thrust rocket. The fuel flow on top of the hopper has to be included in the equations of motion. A falling rock, a rocket, an intricate set of pulleys and weights, any physical system for that matter, obeys [math]d\vec p/dt = m\,d\vec v/dt + dm/dt \,\vec v[/math]. Why? Because the equation is a tautology.

That is not the "rocket equation". That is simply the chain rule. Not quite, but closer. Google is your friend ... Google rocket equation BTW, the "rocket equation" is erroneous in the sense that it ignores the momentum of the fuel flow inside the rocket.

Where do these shoppers think the leather file cases and calfskin passport holders come from? Plants? The disection kit is honest. The complainants that came looking for leather goodies for Christmas are hypocrits.

Physicists model the elementary particles as point particles. However, they cannot be viewed as little points zooming around in space. The uncertainty principle comes into play. The point particles must be viewed as wave packets.

That, in a nutshell, is the problem with string theory.

That is exactly what an analogy is.

There is no gold standard nor silver standard for any Western nation's unit of money. The relative worth of the dollar, pound, mark, ... are determined on the international exchange market. Google Bretton Woods system.

Because gravity is axiomatic in GR, as is the case in Newtonian physics. It exists. Not a particularly satisfying answer, is it? Physicists are always looking for deeper meaning.

Yes. His name was (drum beat) Rep. Charles Rangel, D-N.Y. BTW, he voted against his own resolution and encouraged fellow Democrats to do the same when the Republicans called his bluff and put the resolution up for a vote. A news article on this is here: http://www.command-post.org/2004/2_archives/015787.html.

Farsight, you have an image of a $100 bill. Did you read the fine print? "This note is legal tender for all debts, public and private". That $100 bill is real money. Most money is in accounts of some sort. If it is in a US bank, the account is insured by the US government. Other countries have similar concepts. Just because you can't touch it, feel it, smell it doesn't mean it isn't real. People who play funny games with a bank's balance sheets tend to find how real that money was. I don't know of a single country that still uses the gold standard. Certainly no Western nation does. The dollar, the mark, the pound, etc, are worth what money traders are willing to pay for them. ========== edited to add: Farsight, you are far too enamored with analogies. Analogies are very helpful in making strange new concepts seem familiar. However, one can easily get caught in a false analogy if the analogy is extended too far. This has happened to you more than once.

I like it! Can I extend your hypothesis a bit? I conjecture that the little green gremlins not only eat space but also eat a little green light (they are green, after all). This will result in giving far distant objects a reddish tinge, a red shift, if you will.

Where, exactly, in that dictionary definition, does it say that a "particle" is a tiny billiard ball? In particular, under physics, I see the classical and quantum mechanical definitions of particle. Not bad for a dictionary definition. You might even see the term 'wave-particle duality' if you look a little deeper than a dictionary. BTW, weren't you the very person who disparaged a dictionary definition in another post? Hypocrit. Interesting logic. I see you found some handy Wikipedia entries: Misleading vividness Faulty generalization Red herring Ad hominem ======================================================== ParanoiA, I am answering your question in this same post. Do not take this to mean that I think you belong in the loony bin with Farsight. Just trying to obey site rules against double-posting ... The word particle does not imply mass. A particle is simply the smallest unit of some thing. An atom of gold is a particle in the sense that it cannot be cut into any smaller part that resembles gold. That atom of gold can of course be divided into electrons, protons, and neutrons. Protons and neutrons can be further divided into quarks. Electrons and quarks are elementary particles: they can neither be cut nor divided into some smaller entity. Light, like matter, also exhibits a wave-particle duality. Einstein's Nobel Prize was primarily for his work on the photoelectric effect, where this duality comes to forefront. Photons, the individual units of light are 'particles' in the sense that there is no such thing as half a photon. Photons are uncuttable and indivisible. They are particles in the sense of the word used by particle physicists.