You can’t forget about the change in mass; that’s the whole point! The moving rocket gains more kinetic energy than the stationary rocket by decreasing the kinetic energy of its fuel; but the fuel’s kinetic energy had to come from somewhere: i.e., from burning fuel (or otherwise doing work on the rocket) earlier in the rocket’s history.
To make the calculation easier and avoid all those nasty differential equations, let’s just suppose that our rocket burns its fuel in three rapid explosions; each explosion splits the rocket into two parts of equal mass, moving with relative velocity of 100 m/s. Thus the four-stage rocket, initially at rest, splits into the first-stage fuel mass (moving at -50 m/s) and the remaining three stages (moving at +50 m/s), and then again into the second-stage fuel (moving at 0 m/s) and the remaining two stages (moving at +100 m/s), and finally into the third-stage fuel (moving at +50 m/s) and the remaining payload (moving at +150 m/s). But the total fuel mass required here is not just 3 times the third-stage fuel mass, but 1+2+4=7 times. (The payload mass is 1; the third stage also has mass 1; the second stage has mass 2; and the first stage has mass 4.)
You’re incorrectly equating fuel with energy. Fuel is not energy, it’s just mass. But, when the fuel is accelerated out of the back of the rocket, it creates a force on the rocket equal to the mass of the fuel times its acceleration. F = ma, right?
Imagine standing on a skateboard and throwing a baseball. The ball goes one way and you go the other. If you throw the ball harder (give it more acceleration), you’ll go faster, having more energy. But it’s the same ball, with the same mass. You created more energy by accelerating the mass more, creating a larger force.
Oh yeah. The next time OPEC creates an embargo on oil, and your friend talks about an energy crisis, you can correct him by saying "It’s not an energy crisis, it’s a fuel crisis. " And then your friend will walk away thinking “what a nerd…”
The second case has 9 times the energy because it has exerted the same force over 9 times the distance. Accelerating from 0 to 150 m/sec takes three times as long as accelerating from 100 to 150 m/sec. In accordance with the equations v = at and s = ½at[sup]2[/sup] it takes 3 times as many seconds to accelerate by 150 m/sec as it does to accelerate by 50 m/sec. But 3 times the time means 9 times the distance. We have exerted the force ma at constant a and constant m, inaccordance with your wish to hold mass constant, through 9 times the distance. Since energy = force*distance we obviously have 9 times the energy.
Well, in expending three times the fuel it has changed momentum by a factor or three. But when you change the momentum of a body of constant mass by a factor of three you have changed velocity by a factor of three and energy by a factor of nine. The reason is tied up in the fact that at the higer velocity you have traveled further and, as noted before, energy is proportional to the distance through which a force is exerted as well as the amount of the force.
The process gains efficiency as the velocity increases. At the beginning you are burning fuel and expending energy at a furious rate but not much of it appears as energy in the rocket because the thing hasn’t gone very far. The faster it goes the further it goes in a given amount of time and for a given amount of fuel. So much more of the fuel energy is converted to rocket energy.
Please explain this to me. I don’t understand how a given amount of fuel can impart more energy to a rocket at rest than to a rocket in motion unless you factor in the reduction in mass from burned fuel.
What you think energy should be intuitively (namely, mv) IS defined, but as momentum, not energy. Since force is directly proportional to acceleration, it takes a certain amount of force over a certain amount of time (i.e. a certain impulse) to create a specific change in momentum, but it requires a force applied over a particular distance to change the kinetic energy of the object. If the object is slowing down or speeding up (which it definitely is, if there’s a force on it) then time and distance will not be directly proportional, so the change in momentum must be different from the change in energy, implying that momentum and energy are not equivalent concepts.
Well, let’s assume that the fuel develops full thrust at the instant of ignition. In the first .01 seconds a certain amount of fuel energy has been expended but the rocket has moved very little so little of that energy has been converted to rocket energy. Assume the usual 50 m/sec[sup]2[/sup] acceleration and a 100 kg rocket.
In increments of 0.01 seconds the velocity, motion and rocket energy history look like this:
The thrust of the rocket motor is constant so the force is constant and so is the acceleration assuming constant mass. The amount of fuel burned in each 0.01 second is the same so the input energy is the same in each time interval. Because distance increases by the square of the time so does the kinetic energy of the rocket. In the first 0.05 sec. the rocket energy increased by 312.5 kg-m. In the second 0.05 seconds it increased by 937.5 kg-m. for the same amount of fuel burned.
If we allow the mass of the rocket to decrease by the fuel burned the kinetic energy will decrease linearly because of that. However if the thrust is constant the acceleration will increase linearly at the same rate and so will the distance traveled. The two effects cancel.
And if you read my post #26 again you’ll see that I said that less of the fuel’s energy is converted to rocket KE at low velocities than is the case at high velocities.
Imagine a stationary pickup-truck weighing 1000kg carrying a 10kg cannon ball (total weight 1010kg). If I have a cannon mounted on the bed pointed backward and fire the cannon ball backward at 100m/s:
[ul]
[li]Conservation of Momentum says the truck should accelerate to a velocity of 1 m/s.[/li][li]This blast imparted 500 joules of kinetic energy to the truck.[/li][/ul]
Now, imagine the same experiment in a truck moving 20 m/s, as observed by someone alongside the road:
[ul]
[li]The ball as it sat in the cannon was going 20m/s forward, then after the blast went 80 m/s backward, a net momentum change of 1000kg-m/s, same as the first case. Conservation of momentum says the truck will still accelerate 1m/s.[/li][li]To the observer, the truck’s kinetic energy increased by 20500 joules.[/li][/ul]
The problem then is why are only 500 joules added in the inertial frame–which is equivalent to the energy needed to get the stationary truck moving at 1m/s–but 20500 joules are required to bump the speed by 1m/s when the truck is already moving 20m/s? Many here feel that it should be the same amount of energy, since it is the same physical activity imparting momentum to the truck.
The answer is because we are mixing frames of reference; these two scenarios are NOT equivalent, and there is no physical requirement that observers in separately moving frames measure the same kinetic energy, only that they both observe conservation of momentum.
To an observer on the truckbed, any blast of a cannonball will impart the same 500 joules of kinetic energy to the vehicle (since he/she always remains stationary with respect to the truck). To him/her, each cannonball produces the same amount of acceleration, noticing no difference whther he’s stationary or traveling 100 m/s (I’m ignoring the mass-reduction complication for the moment to keep things simpler; each cannonball would of course reduce the mass of the truck, giving later shots an extra oomph).
The person along the road does see a difference in the kinetic energy increase with each cannonball because the truck is moving faster with each shot. This is because the force of the blast is acting on the truck over a greater distance in his frame (because the truck is moving instead of stationary).
If the road observer claims the cannon shots are getting more powerful as truck velocity increases, we can conclude this is an illusion because of his frame of reference. Similar illusory observations are common when reference frames are mixed. in A person sitting on a spinning merry-go-round may claim there is a centrifugal force pushing him/her outward, but an observer outside the merry-go-round would say “No, the real force is the merry-go-round pulling you inward–a centripital force.” The outside observer would be correct; centrifugal force is an illusion (We can prove this by seeing what happens if the rider lets go of the merry-go-round. Centrifugal force appears to be radial, but upon letting go the rider flys off in a tangential direction 90 degrees away from the direction of the force, contrary to Newton’s 2nd law).
The important thing about energy is that difference in energy is what counts.
Observers on the truck measure only 500 joules because the cannonball imparted to them a change in energy just like it imparted a change in energy to the truck so the difference between their new energy and the cannonball energy is 500 joules.
The velocity that counts in computing kinetic energy is the velocity difference between the two objects in question. For example in the cannonball problem, those on the truck would measure the ball’s energy at 50000 J (10100[sup]2[/sup]/2). The person on the side of the road would measure it at 32000 J (1080[sup]2[/sup]/2). A ball that struck someone following on a truck also going 20 m/sec would deliver 50000 J while the ball striking someone on the side of the road would only deliver 32000 J.
Reaction engines (jets and rockets) all work that way. When the vehicle has zero speed, they produce zero power…or better to say they impart zero energy to the vehicle. Since work is force * distance, and the thing isn’t moving, all that force (thrust) does no work…Like pushing with all your might against a brick wall…sure you get tired, but you are not supplying any power to the wall. In the case of the jet or rocket, all the power goes into the exhaust stream. This is one reason jet propulsion is not used on low speed aircraft. To get economical amounts of power/vs fuel consumption you must be going fairly fast, that term being relitive to the exhaust velocity. If you are slow compared to the exhaust velocity, you get little power. When you are traveling at the exhaust velocity, (the exhaust stays stationary in relation to your inertial reference) then you have reached the limit of how much power you can get from the engine. To make jets more economical on slower aircraft, you need to change the exhaust stream to have more mass flow at lower velocity. That sort of design is a fan-jet or a turboprop. These options don’t exist for a rocket operating in a vacuum. (no source of mass to increase mass flow.
So the answer to your confusion is that the engine operates at increasing effiency as speed increases. You aren’t getting more power from your fuel, your just wasting less of it on the exhaust stream.
This isn’t mysterious or esoteric an any sense, guys. And it applies to all means of propelling a massive object into motion.
Irrespective of the means of propulsion a 1 kg mass accelerated at 1 m/sec[sup]2[/sup] for 1 sec will have traveled ½ meter in that first second. The force involved is 1 N so the energy of the mass will be ½ J. At the start the energy was 0 so it has gained ½ J in 1 sec.
That same mass accelerated for 10 sec will have traveled 50 m and the force will still be 1 N. The energy will be 50 J. When accelerated for 11 seconds it will have traveled 60.5 ft and the energy is 60.5 J. In that 1 sec it has gained 10.5 J.
In a drag race the drivers rev the engine, pop the clutch and for the first second most of the engines ouput quite literally goes up in smoke from the tires. As the dragster proceeds down the course more and more of the engine output in 1 sec. is converted to vehicle energy, and thus speed, in that 1 sec…
