Where does the potential energy go?
July 13, 2015 1:54 PM Subscribe
So, imagine we have a 100 meter tall tower, a 100 kg weight, and a rocket. Consider the following two, very similar scenarios...
Scenario 1:
*The tower has nothing on it.
*The rocket is 100 meters in the air.
*The weight is on top of the rocket.
*The rocket's thrusters are firing, providing exactly enough thrust to keep the rocket and the 100 kg weight suspended in mid-air.
Scenario 2:
*The weight is resting on top of the tower.
*The rocket is 100 meters in the air, with nothing on it.
*The rocket's thrusters are firing with the same amount of thrust as before. Since there is no weight on top of the rocket, it moves upward.
My question: Both scenarios, as far as I can tell, start with exactly the same amount of mass and energy. But in scenario 2, the rocket gains potential energy as it rises upwards. Where is that energy in scenario 1?
Scenario 1:
*The tower has nothing on it.
*The rocket is 100 meters in the air.
*The weight is on top of the rocket.
*The rocket's thrusters are firing, providing exactly enough thrust to keep the rocket and the 100 kg weight suspended in mid-air.
Scenario 2:
*The weight is resting on top of the tower.
*The rocket is 100 meters in the air, with nothing on it.
*The rocket's thrusters are firing with the same amount of thrust as before. Since there is no weight on top of the rocket, it moves upward.
My question: Both scenarios, as far as I can tell, start with exactly the same amount of mass and energy. But in scenario 2, the rocket gains potential energy as it rises upwards. Where is that energy in scenario 1?
Isn't that the energy you used to carry the ball up the tower?
posted by mathiu at 2:11 PM on July 13, 2015 [3 favorites]
posted by mathiu at 2:11 PM on July 13, 2015 [3 favorites]
Is it increased strain on the structure of the tower in scenario 2?
posted by Pfardentrott at 2:14 PM on July 13, 2015
posted by Pfardentrott at 2:14 PM on July 13, 2015
In scenario 1, chemical energy from the rocket is used to counteract the downward acceleration of gravity acting on the weight. In scenario 2, you've removed the weight from the rocket, so that chemical energy is converted to potential energy in the form of increased height.
I think that part of your question focuses on the distinction between force (like the force of the tower supporting the weight against gravity) and work/energy.
posted by mercredi at 2:19 PM on July 13, 2015 [1 favorite]
I think that part of your question focuses on the distinction between force (like the force of the tower supporting the weight against gravity) and work/energy.
posted by mercredi at 2:19 PM on July 13, 2015 [1 favorite]
The tower is providing (ignoring its own mass) no upward force in S1 because it's empty. In S2, it's providing a constant upward force to support the weight which was removed from the rocket.
Basically, the tower is "thrusting" with "exactly enough thrust to keep the [tower] and the 100 kg weight suspended in mid-air", and S2 is exerting more total energy than S1 because there is more force being applied by the tower in S2 than in S1 while the rocket's force is unchanged.
posted by jpeacock at 2:22 PM on July 13, 2015 [1 favorite]
Basically, the tower is "thrusting" with "exactly enough thrust to keep the [tower] and the 100 kg weight suspended in mid-air", and S2 is exerting more total energy than S1 because there is more force being applied by the tower in S2 than in S1 while the rocket's force is unchanged.
posted by jpeacock at 2:22 PM on July 13, 2015 [1 favorite]
Best answer: If you replace the rocket with a machine gun firing downwards, in the second scenario you'll find the bullets are moving more slowly because they're being fire from a gun that's moving up. They may not even be moving at all.
So the energy that would have gone to lifting the rocket in the first scenario has instead gone to velocity of the spent fuel (and eventually into heat as the exhaust slows).
posted by justkevin at 2:26 PM on July 13, 2015 [3 favorites]
So the energy that would have gone to lifting the rocket in the first scenario has instead gone to velocity of the spent fuel (and eventually into heat as the exhaust slows).
posted by justkevin at 2:26 PM on July 13, 2015 [3 favorites]
I don't understand what's going on with the tower. Is it attached to the rocket or otherwise relevant to the issue of the rocket's thrust? Or, is it just so I can know where the 100 kg mass was placed when not on the rocket? For the purpose of this answer, I am assuming the tower is just standing to the side and only mentioned so I know where you put the 100 kg mass.
In any event, there is no missing potential energy in Scenario 1 because the rocket isn't moving against the gravity field. Gravitational potential energy is added or subtracted to a mass by moving it with or against the gravity field's vector. In Scenario 1, the rocket isn't moving so it doesn't gain or lose potential energy. The engines are generating an amount of thrust in order for the rocket to hover, which is counteracted by the gravitational force gravitational force in the opposite direction of the rocket's mass plus the 100 kg ball. (let's pretend the rocket has a mass of 100 kg, so the total thrust would be 19,600 newtons, equal to the force of gravity on the system). In Scenario 2, there is a net of 9,800 newtons of force pushing the rocket up, so it will accelerate upward at a rate related to the rocket's mass. Again, assuming the rocket has a mass of 100 kg, it would accelerate upward at 9.8 m/s^2.
Please let me know if I misunderstood your hypothetical. I feel like I am missing the relevance of the tower.
posted by Tanizaki at 2:31 PM on July 13, 2015
In any event, there is no missing potential energy in Scenario 1 because the rocket isn't moving against the gravity field. Gravitational potential energy is added or subtracted to a mass by moving it with or against the gravity field's vector. In Scenario 1, the rocket isn't moving so it doesn't gain or lose potential energy. The engines are generating an amount of thrust in order for the rocket to hover, which is counteracted by the gravitational force gravitational force in the opposite direction of the rocket's mass plus the 100 kg ball. (let's pretend the rocket has a mass of 100 kg, so the total thrust would be 19,600 newtons, equal to the force of gravity on the system). In Scenario 2, there is a net of 9,800 newtons of force pushing the rocket up, so it will accelerate upward at a rate related to the rocket's mass. Again, assuming the rocket has a mass of 100 kg, it would accelerate upward at 9.8 m/s^2.
Please let me know if I misunderstood your hypothetical. I feel like I am missing the relevance of the tower.
posted by Tanizaki at 2:31 PM on July 13, 2015
The way rockets work is by shooting hot gas downwards. Due to conservation of momentum, this causes an upwards force on the rocket. At the same time, some of the chemical potential energy goes into the kinetic energy of the gas, some goes into the kinetic energy of the rocket, and some goes into heat.
In S1, though, the rocket gains no kinetic energy. So what isn't in the momentum/heat of the expelled become heat in the rocket.
By analogy, consider–instead of a rocket–applying an upwards force with your hand, just enough to hold a weight steady. If you're holding a weight, the chemical potential in your arm will be dissipated into heat energy, no added gasses necessary. If you're not holding a weight, the chemical potential will be turned into kinetic + potential energy as your arm moves upwards.
The really interesting question is: since linear momentum has to be conserved, where does the momentum go?
posted by Maecenas at 2:39 PM on July 13, 2015
In S1, though, the rocket gains no kinetic energy. So what isn't in the momentum/heat of the expelled become heat in the rocket.
By analogy, consider–instead of a rocket–applying an upwards force with your hand, just enough to hold a weight steady. If you're holding a weight, the chemical potential in your arm will be dissipated into heat energy, no added gasses necessary. If you're not holding a weight, the chemical potential will be turned into kinetic + potential energy as your arm moves upwards.
The really interesting question is: since linear momentum has to be conserved, where does the momentum go?
posted by Maecenas at 2:39 PM on July 13, 2015
Both scenarios, as far as I can tell, start with exactly the same amount of mass and energy.
You're making a little bit of a fallacy here by equating the two scenarios - they're apples and oranges. A better way of looking at this would be:
Scenario 1, System A - tower
Scenario 1, System B - rocket + mass
Scenario 2, System A - tower + mass
Scenario 2, System B - rocket
In this case, the two "A" systems definitely do not have the same mass and (potential energy), and neither do the two "B" systems. You can't really add the tower and rocket together because it's kind of a nonsense answer.
Anyway, ignore the tower. Look at it even simpler - you have a rocket and a mass. In one system, the rocket is strapped to the mass and is firing against the gravity vector. In the second system, the rocket and the mass are separate and the mass is just hanging there by the force of who cares.
The total energy of those two systems are the same. You have the PE of the mass, the PE of the rocket, and the chemical energy of the rocket. None of these things change whether the mass is strapped to the rocket or not. Yes, when you detach the mass the rocket will start moving but now you're looking at a) two separate systems because the two pieces are not attached and do not influence each other and b) a transient problem, which is different than the first (static) situation you have posited.
Does it make sense that the definition of systems is flawed here? A good example would be filling a jerry can up with gasoline and leaving it in your garage while you go for a drive - yes, you start out with just as much energy as if you had dumped it in your car, but that doesn't make a fat load of good if the two are separate.
posted by backseatpilot at 2:55 PM on July 13, 2015 [3 favorites]
You're making a little bit of a fallacy here by equating the two scenarios - they're apples and oranges. A better way of looking at this would be:
Scenario 1, System A - tower
Scenario 1, System B - rocket + mass
Scenario 2, System A - tower + mass
Scenario 2, System B - rocket
In this case, the two "A" systems definitely do not have the same mass and (potential energy), and neither do the two "B" systems. You can't really add the tower and rocket together because it's kind of a nonsense answer.
Anyway, ignore the tower. Look at it even simpler - you have a rocket and a mass. In one system, the rocket is strapped to the mass and is firing against the gravity vector. In the second system, the rocket and the mass are separate and the mass is just hanging there by the force of who cares.
The total energy of those two systems are the same. You have the PE of the mass, the PE of the rocket, and the chemical energy of the rocket. None of these things change whether the mass is strapped to the rocket or not. Yes, when you detach the mass the rocket will start moving but now you're looking at a) two separate systems because the two pieces are not attached and do not influence each other and b) a transient problem, which is different than the first (static) situation you have posited.
Does it make sense that the definition of systems is flawed here? A good example would be filling a jerry can up with gasoline and leaving it in your garage while you go for a drive - yes, you start out with just as much energy as if you had dumped it in your car, but that doesn't make a fat load of good if the two are separate.
posted by backseatpilot at 2:55 PM on July 13, 2015 [3 favorites]
Actually, I'll go even one step farther and tell you that the static system of your scenario 1 is exactly the same as the transient system of scenario two at time t=0. They just won't be identical for very long.
posted by backseatpilot at 2:58 PM on July 13, 2015
posted by backseatpilot at 2:58 PM on July 13, 2015
Response by poster: Tanizaki: I posited the rocket as holding the weight in mid-air in order to make it clear that the rocket in scenario 1 is actually doing something (holding the weight up). But I wanted the two scenarios to be as similar as possible, so I posited a tower in scenario 2 to hold the weight at the same height. Then I put the tower in scenario 1 as well (again, to keep the two scenarios as similar as possible).
backseatpilot: Isn't it acceptable to speak of each scenario as a self-contained system? Suppose instead of calling them "scenarios 1 and 2", I call them "universe 1 and 2", and describe them as hypothetical universes which contain nothing at all except a tower, a rocket, a weight, and a featureless planet for everything to sit on top of. Surely such a hypothetical universe can be spoken of as a "system", containing an invariable quantity of mass/energy? And surely the two scenarios I described contain equal amounts of mass/energy, since the only difference between them is the horizontal position of the weight?
(Please be patient with me if I'm banging my head against a simple concept, or using these terms incorrectly. I am still learning physics.)
posted by CustooFintel at 3:17 PM on July 13, 2015
backseatpilot: Isn't it acceptable to speak of each scenario as a self-contained system? Suppose instead of calling them "scenarios 1 and 2", I call them "universe 1 and 2", and describe them as hypothetical universes which contain nothing at all except a tower, a rocket, a weight, and a featureless planet for everything to sit on top of. Surely such a hypothetical universe can be spoken of as a "system", containing an invariable quantity of mass/energy? And surely the two scenarios I described contain equal amounts of mass/energy, since the only difference between them is the horizontal position of the weight?
(Please be patient with me if I'm banging my head against a simple concept, or using these terms incorrectly. I am still learning physics.)
posted by CustooFintel at 3:17 PM on July 13, 2015
Best answer: Here's an even simpler pair of scenarios that avoids the weight: the rocket is inside the tower, which has a retractable ceiling.
Scenario 1: The ceiling is closed and the rocket is pushing against it but not going anywhere. (Instead of the weight keeping it down, the normal force from the ceiling is keeping it down.)
Scenario 2: The ceiling is open and the rocket rises, gaining potential energy.
posted by dfan at 3:29 PM on July 13, 2015
Scenario 1: The ceiling is closed and the rocket is pushing against it but not going anywhere. (Instead of the weight keeping it down, the normal force from the ceiling is keeping it down.)
Scenario 2: The ceiling is open and the rocket rises, gaining potential energy.
posted by dfan at 3:29 PM on July 13, 2015
CustooFintel - your question was fine. as far as i can tell you described things perfectly.
posted by andrewcooke at 3:33 PM on July 13, 2015
posted by andrewcooke at 3:33 PM on July 13, 2015
I'm going to take Maecenas's suggestion and look at linear momentum.
If the rocket isn't moving anywhere, but its exhaust gas is still coming out the back, how does the system conserve linear momentum? The answer, I think, is that the rocket exerts the same Fg on the earth as the earth does on it. Thus, it's "pulling" the earth slightly toward it with gravity, but then counterbalancing this force with its exhaust. Thus, the only thing that "actually happens" is that the exhaust velocity gets converted into heat, as it was always going to anyway. If the rocket is moving, we know that its Fg is smaller than the force of its exhaust, thus the earth moves a bit "downwards" and the rocket moves a lot upwards.
So I think the only difference is that in one scenario, the rocket and earth have changes in their net momentum, while in the other, they have no net momentum. What do we call motion that doesn't change net momentum? It's randomized -- so we call it heat.
Thus, in one scenario, the exhaust gases heat up the atmosphere slightly more than in the other scenario. I think this is the only difference.
posted by goingonit at 3:38 PM on July 13, 2015
If the rocket isn't moving anywhere, but its exhaust gas is still coming out the back, how does the system conserve linear momentum? The answer, I think, is that the rocket exerts the same Fg on the earth as the earth does on it. Thus, it's "pulling" the earth slightly toward it with gravity, but then counterbalancing this force with its exhaust. Thus, the only thing that "actually happens" is that the exhaust velocity gets converted into heat, as it was always going to anyway. If the rocket is moving, we know that its Fg is smaller than the force of its exhaust, thus the earth moves a bit "downwards" and the rocket moves a lot upwards.
So I think the only difference is that in one scenario, the rocket and earth have changes in their net momentum, while in the other, they have no net momentum. What do we call motion that doesn't change net momentum? It's randomized -- so we call it heat.
Thus, in one scenario, the exhaust gases heat up the atmosphere slightly more than in the other scenario. I think this is the only difference.
posted by goingonit at 3:38 PM on July 13, 2015
No, that's silly. Don't start conflating momentum and energy.
If you want to look at it as a whole universe, that's fine. My point about static vs. transient still holds - at the initial instant you start looking at the transient problem, it's exactly the same as the static problem. Once time starts moving you have two different problems.
posted by backseatpilot at 3:50 PM on July 13, 2015
If you want to look at it as a whole universe, that's fine. My point about static vs. transient still holds - at the initial instant you start looking at the transient problem, it's exactly the same as the static problem. Once time starts moving you have two different problems.
posted by backseatpilot at 3:50 PM on July 13, 2015
Best answer: You have primarily 3 kinds of energy at play here. Potential, Kinetic and Chemical.
In case 1 you are converting chemical into kinetic energy. This kinetic energy is entirely carried by the exhaust from the rocket. CE = KE(exhaust)
In case 2 you are converting chemical into kinetic and potential energy. The potential energy of the rocket increases as it goes higher. The remainder of the energy goes into kinetic energy of both the exhaust and of the rocket. CE = PE + KE(rocket) + KE(exhaust)
From this, using the top of the tower as your stationary reference, you can determine that the speed of the exhaust away from the top of the tower is larger in case 1 than in case 2. What you conserve in both cases is the amount of CE being converted, not the velocities.
posted by NoDef at 5:15 PM on July 13, 2015 [1 favorite]
In case 1 you are converting chemical into kinetic energy. This kinetic energy is entirely carried by the exhaust from the rocket. CE = KE(exhaust)
In case 2 you are converting chemical into kinetic and potential energy. The potential energy of the rocket increases as it goes higher. The remainder of the energy goes into kinetic energy of both the exhaust and of the rocket. CE = PE + KE(rocket) + KE(exhaust)
From this, using the top of the tower as your stationary reference, you can determine that the speed of the exhaust away from the top of the tower is larger in case 1 than in case 2. What you conserve in both cases is the amount of CE being converted, not the velocities.
posted by NoDef at 5:15 PM on July 13, 2015 [1 favorite]
Best answer: ok, so we've got two equivalent questions (one from dfan) which are basically equivalent. they both describe two different scenarios:
a - the rocket is free to accelerate upwards
b - the rocket is constrained so that it stays in the same place
and the problem is that it seems impossible for energy to be conserved in both. in particular, in (a) it seems reasonable that chemical potential energy in the fuel is converted into kinetic and gravitational potential energy in the rocket. that's cool. but then in (b) we are still "spending" the chemical potential energy - so what is increasing to "compensate"?
my answer, justkevin's, and NoDef's all argue that the problem comes from ignoring what is being squirted out of the rocket - that the exhaust has more kinetic energy in case (b). so that is where the chemical potential energy is going. i think justkevin's and NoDef's answers are probably clearer than mine, but am happy to explain mine in more detail if anyone has questions.
but this post is mainly about pointing out where other answers are wrong or confused. so here goes nothing...
@mathiu: we are starting to look at things after that point. if you prefer, look at dfan's (equivalent) problem, which has no weight.
@Pfardentrott: there is certainly some energy stored in the tower as it "gives" a little when the weight is placed on it. but that energy is fixed. it does not increase with time. in contrast, the discrepancy in our "problem" does increase with time - in case (a) the rocket gets higher and higher and faster and faster. in comparison the energy stored in the tower to hold up the wight is fixed (we know it is fixed because nothing is moving - change in energy is work done, which is force x distance moved, and if things are not moving, distance moved is zero). so it cannot explain our difference.
@jpeacock: it is true that the tower is providing an upwards force. but force is not the same as energy. it's quite unintuitive, because our muscles don't work that way, but the tower is not doing work to hod up the weight. and so it is not spending energy. as i said just above, work is only done when a force moves - in this case the force is stationary.
@Tanizaki: everything you say is correct, as far as i can tell, but you seem to have missed the point of the question (as i think you suspected). i tried to explain the question above - i hope it is clearer now.
@Maecenas: your answer also seems to be correct in most details, but you also seem to have missed the interesting part of the question, which relates to energy. there is an interesting puzzle here and your answer doesn't really provide an insight into why the energy balance is "odd", imho.
@backseatpilot: again, much of what you say is true, but you don't actually explain anything. if you can add energies, then there is a problem - you can't make it go away by just saying that things are "flawed".
@goignonit: i am a bit tired but i think what you say is true. but again it doesn't really answer the problem with energies.
posted by andrewcooke at 5:29 PM on July 13, 2015
a - the rocket is free to accelerate upwards
b - the rocket is constrained so that it stays in the same place
and the problem is that it seems impossible for energy to be conserved in both. in particular, in (a) it seems reasonable that chemical potential energy in the fuel is converted into kinetic and gravitational potential energy in the rocket. that's cool. but then in (b) we are still "spending" the chemical potential energy - so what is increasing to "compensate"?
my answer, justkevin's, and NoDef's all argue that the problem comes from ignoring what is being squirted out of the rocket - that the exhaust has more kinetic energy in case (b). so that is where the chemical potential energy is going. i think justkevin's and NoDef's answers are probably clearer than mine, but am happy to explain mine in more detail if anyone has questions.
but this post is mainly about pointing out where other answers are wrong or confused. so here goes nothing...
@mathiu: we are starting to look at things after that point. if you prefer, look at dfan's (equivalent) problem, which has no weight.
@Pfardentrott: there is certainly some energy stored in the tower as it "gives" a little when the weight is placed on it. but that energy is fixed. it does not increase with time. in contrast, the discrepancy in our "problem" does increase with time - in case (a) the rocket gets higher and higher and faster and faster. in comparison the energy stored in the tower to hold up the wight is fixed (we know it is fixed because nothing is moving - change in energy is work done, which is force x distance moved, and if things are not moving, distance moved is zero). so it cannot explain our difference.
@jpeacock: it is true that the tower is providing an upwards force. but force is not the same as energy. it's quite unintuitive, because our muscles don't work that way, but the tower is not doing work to hod up the weight. and so it is not spending energy. as i said just above, work is only done when a force moves - in this case the force is stationary.
@Tanizaki: everything you say is correct, as far as i can tell, but you seem to have missed the point of the question (as i think you suspected). i tried to explain the question above - i hope it is clearer now.
@Maecenas: your answer also seems to be correct in most details, but you also seem to have missed the interesting part of the question, which relates to energy. there is an interesting puzzle here and your answer doesn't really provide an insight into why the energy balance is "odd", imho.
@backseatpilot: again, much of what you say is true, but you don't actually explain anything. if you can add energies, then there is a problem - you can't make it go away by just saying that things are "flawed".
@goignonit: i am a bit tired but i think what you say is true. but again it doesn't really answer the problem with energies.
posted by andrewcooke at 5:29 PM on July 13, 2015
@andrewcooke I think the confusing thing about this explanation is how does the exhaust "know" to have a higher velocity in one case than the other. It's the same chemical reaction going on...
And presumably I could throttle the engine while shrinking the weight until the exhaust had exactly the same power as the no-weight version, but the rocket still didn't go anywhere. Or is this impossible in some way?
So I was trying to understand how more of the exhast energy converts into heat in case b than in case a, where as we know some of it converts into kinetic energy. In retrospect the momentum business was beside the point.
posted by goingonit at 6:14 PM on July 13, 2015
And presumably I could throttle the engine while shrinking the weight until the exhaust had exactly the same power as the no-weight version, but the rocket still didn't go anywhere. Or is this impossible in some way?
So I was trying to understand how more of the exhast energy converts into heat in case b than in case a, where as we know some of it converts into kinetic energy. In retrospect the momentum business was beside the point.
posted by goingonit at 6:14 PM on July 13, 2015
Oh but we are saying the same thing since more (eventual) heat is the same thing as higher velocity! I still don't quite grok it but I agree with you.
posted by goingonit at 6:24 PM on July 13, 2015
posted by goingonit at 6:24 PM on July 13, 2015
The very first answer, from andrewcooke, gets to the heart of it. You have to account for the kinetic energy of the exhaust gases from the propellant.
You can think of two extreme cases. In the first one, the rocket is fixed in position, as in your scenario 1. In that case, the rocket has no kinetic energy and the exhaust gases have all of the kinetic energy.
Then at the other extreme, think of the rocket moving at exactly the same speed as the exhaust. In that case, the exhaust gases come out of the rocket nozzle with no velocity and no kinetic energy and the rocket has all of the kinetic energy.
So for any case in between those extremes, there are varying amounts of kinetic energy partitioned to the exhaust gases or the rocket. The portion partitioned to the rocket is converted to potential energy as the rocket rises.
Total energy is the same in all cases. Some is potential energy of the rocket, some is kinetic energy of the rocket and the remainder is kinetic energy of the exhaust gases.
Keep in mind that kinetic energy is a scalar, meaning that direction is irrelevant. The sum of the kinetic energy is the same regardless of which portion is in the rocket going up or the exhaust gases going down.
posted by JackFlash at 9:02 PM on July 13, 2015
You can think of two extreme cases. In the first one, the rocket is fixed in position, as in your scenario 1. In that case, the rocket has no kinetic energy and the exhaust gases have all of the kinetic energy.
Then at the other extreme, think of the rocket moving at exactly the same speed as the exhaust. In that case, the exhaust gases come out of the rocket nozzle with no velocity and no kinetic energy and the rocket has all of the kinetic energy.
So for any case in between those extremes, there are varying amounts of kinetic energy partitioned to the exhaust gases or the rocket. The portion partitioned to the rocket is converted to potential energy as the rocket rises.
Total energy is the same in all cases. Some is potential energy of the rocket, some is kinetic energy of the rocket and the remainder is kinetic energy of the exhaust gases.
Keep in mind that kinetic energy is a scalar, meaning that direction is irrelevant. The sum of the kinetic energy is the same regardless of which portion is in the rocket going up or the exhaust gases going down.
posted by JackFlash at 9:02 PM on July 13, 2015
The velocity of the exhaust does not increase. The upward force on the rocket is equivalent to the rate of change of the momentum (p) of the exhaust (taken as a whole):
F_thrust = dp_exhaust/dt = (density of exhaust [kg/m^3]) * (area of exhaust pipe [m^2]) * (exhaust velocity [m/s])^2
The problem assumes the thrust of the rocket is constant within the timeframe we're considering, right? The upward force on the rocket is (assuming near Earth's surface) (assume m_fuel is much less than m_rocket)
F_gravity = (m_rocket + m_weight)*g = -F_thrust
The density of the exhaust and the area of the exhaust pipe aren't likely to change, and neither are the weight of the rocket or the weight of the 100kg weight. So the velocity of the exhaust must be the same.
Ok, so, where does the energy go, then? Well, in the non-weighted case, it becomes the rocket's kinetic energy, corresponding to the rocket moving upwards. In the weighted case, it still becomes the rocket's kinetic energy. Except this time, the rocket isn't moving around as a whole. But its individual molecules can have lots of kinetic energy, bouncing around all over the place, but if their velocities are randomly distributed, the net velocity of the rocket will be 0. But 0 net velocity + high kinetic energy is the same thing as heat!
The exhaust also has heat, so not all of the chemical energy will get turned into directed, or macroscopic, kinetic energy. This is related to the efficiency of the thruster--how much chemical energy does it take to get a given rocket to a certain velocity?
So, to sum up
-In the weighted case,
K_chemical = K_rocket (random) + K_thrust (directed) + K_thrust (random)
= E_rocket-heat + E_thrust-heat + K_thrust.
-In the unweighted case,
K_chemical = K_rocket (directed) + U_rocket-gravity + K_thrust (directed) + K_rocket (random) + K_thrust (random)
=K_rocket + U_rocket-gravity + K_thrust + E_rocket-heat + E_thrust-heat
posted by Maecenas at 9:11 PM on July 13, 2015
F_thrust = dp_exhaust/dt = (density of exhaust [kg/m^3]) * (area of exhaust pipe [m^2]) * (exhaust velocity [m/s])^2
The problem assumes the thrust of the rocket is constant within the timeframe we're considering, right? The upward force on the rocket is (assuming near Earth's surface) (assume m_fuel is much less than m_rocket)
F_gravity = (m_rocket + m_weight)*g = -F_thrust
The density of the exhaust and the area of the exhaust pipe aren't likely to change, and neither are the weight of the rocket or the weight of the 100kg weight. So the velocity of the exhaust must be the same.
Ok, so, where does the energy go, then? Well, in the non-weighted case, it becomes the rocket's kinetic energy, corresponding to the rocket moving upwards. In the weighted case, it still becomes the rocket's kinetic energy. Except this time, the rocket isn't moving around as a whole. But its individual molecules can have lots of kinetic energy, bouncing around all over the place, but if their velocities are randomly distributed, the net velocity of the rocket will be 0. But 0 net velocity + high kinetic energy is the same thing as heat!
The exhaust also has heat, so not all of the chemical energy will get turned into directed, or macroscopic, kinetic energy. This is related to the efficiency of the thruster--how much chemical energy does it take to get a given rocket to a certain velocity?
So, to sum up
-In the weighted case,
K_chemical = K_rocket (random) + K_thrust (directed) + K_thrust (random)
= E_rocket-heat + E_thrust-heat + K_thrust.
-In the unweighted case,
K_chemical = K_rocket (directed) + U_rocket-gravity + K_thrust (directed) + K_rocket (random) + K_thrust (random)
=K_rocket + U_rocket-gravity + K_thrust + E_rocket-heat + E_thrust-heat
posted by Maecenas at 9:11 PM on July 13, 2015
how does the exhaust "know" to have a higher velocity in one case than the other. It's the same chemical reaction going on...
it doesn't, but you have nailed what is so confusing about the question.
the exhaust gases come out with the same speed relative to the rocket. let's call that 100mph.
but when the rocket is moving upwards at 99mph that means that the exhaust is moving downwards at 1mph. and 1pmh is very little, so it has very little kinetic energy in "our" frame (on the earth, where we are doing the maths).
this is why there are machine guns in justkevin's answer - to try make this point clearer.
(also, in practice, in an atmosphere, the exhaust will eventually get diverted and mixed up and converted into diffuse heat as others are saying).
posted by andrewcooke at 3:02 AM on July 14, 2015
it doesn't, but you have nailed what is so confusing about the question.
the exhaust gases come out with the same speed relative to the rocket. let's call that 100mph.
but when the rocket is moving upwards at 99mph that means that the exhaust is moving downwards at 1mph. and 1pmh is very little, so it has very little kinetic energy in "our" frame (on the earth, where we are doing the maths).
this is why there are machine guns in justkevin's answer - to try make this point clearer.
(also, in practice, in an atmosphere, the exhaust will eventually get diverted and mixed up and converted into diffuse heat as others are saying).
posted by andrewcooke at 3:02 AM on July 14, 2015
« Older Help me automate a PDF-merge task! | Best Coloring Pencils For Adult Coloring Books? Newer »
This thread is closed to new comments.
it's either a very sneaky question or a deep one. i *think* it's mainly sneaky.
you are probably saying "there's just as much energy in the exhaust in scenario 2". but you are wrong (i hope). one way to see that you are wrong is to look at the exhaust in detail, at some point "later", when the rocket in 2 is whizzing upwards. in your frame (which is where you are calculating the energy) that exhaust gas is hardly moving. it's like shooting a cannonball out the back of a truck that is moving as fast as a cannonball moves.
i'll try think of a better explanation. i'll also ask my partner. this is going to cause arguments at dinner....
posted by andrewcooke at 2:08 PM on July 13, 2015