Confused about kinetic energy

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  • #36
Drakkith
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@zanick Hmmm. I'm not sure I read your posts correctly. I'll wait to comment further on this until I get some sleep.
 
  • #37
jbriggs444
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that's right, at a constant HP (or fuel flow) acceleration goes down proportional to velocity... (which means so does thrust and force). So, for constant thrust, fuel consumption has to go up with speed.
Now Im confused!

You are attempting what amounts to an energy analysis. If you get more energy out, you must have put more energy (fuel) in. The problem with that analysis is that you need to account for all of the energy, not just the energy going into accelerating the rocket ship. You have to account for the energy in the exhaust stream. Re-read post #27 above by DaleSpam.
 
  • #38
gmax137
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... Re-read post #27 above by DaleSpam.

plus one to that! Dale's post is the clearest explanation of this I have ever seen.
 
  • #39
zanick
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You are attempting what amounts to an energy analysis. If you get more energy out, you must have put more energy (fuel) in. The problem with that analysis is that you need to account for all of the energy, not just the energy going into accelerating the rocket ship. You have to account for the energy in the exhaust stream. Re-read post #27 above by DaleSpam.
I was thinking about that before I went to sleep.... It was keeping me up. its the change in momentum of the fuel in the rocket and what is ejected out the rear.
The KE is going up with the square of speed, so doesn't the energy used have to go up as well? say the rocket in space is going 50,000m/s. at a constant speed, could it regain the same rate of acceleration by turning on the "jets" without using any more fuel than it did earlier? what I mean is, in space, how does it know how fast its going? Its like true air speed vs indicated air speed of a plane. its all relative, right? Doesn't the U2 use the same fuel rate at 140kts indicated (or near supersonic and near 650mph ground speed, but has a has of a lot more KE stored , or is that KE stored as potential energy because its so high in altitude?
in the case of the rocket, if it doesn't know the "relative " speed as it leaves earth, if it crashed into Mars (or similar weight and size asteroid), it certainly would realize all the KE as it crashed. :)
what am I missing here?
 
  • #40
zanick
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Here is an extremely simplified example:

Suppose that we have a spaceship of mass 1010 kg (including cargo) at rest in space. Now, to make things simple, instead of burning a steady stream of fuel, let's suppose that we throw a 10 kg moon rock out the back at a speed of -1000 m/s. Now, the momentum of the rock is -10,000 kg*m/s, so the momentum of the ship is +10,000 kg*m/s which corresponds to a speed of 10 m/s. If you work it out, the KE has changed from 0 to 5.05 MJ, with 5 MJ in the rock and 0.05 MJ in the ship.

Now, suppose we have the same scenario, but this time the ship is already moving at 100 m/s. This time the final speed of the rock is -900 m/s and the final speed of the ship is 110 m/s. If you work it out, this time the KE started at 5.05 MJ (5 MJ ship and 0.05 MJ rock), and changed to 10.10 MJ, with 4.05 MJ in the rock and 6.05 MJ in the ship.

So, the total change in energy is the same 5.05 MJ in both cases. In the first case 5 MJ went into the KE of the rock and only 0.05 MJ went into the ship, but in the second case 4 MJ went into the rock and 1.05 MJ went into the ship.

So the efficiency increased as less energy was being thrown away in the KE of the exhaust. The actual rocket equation is more complicated because it is continuous, but that is the basic principle.
That was one of the best explanation's ive seen too! so, its really about the mass changing in proportion to what was used as thrust, in the form of a part of that original mass. (so rocket has less mass and needs less thrust to create the same acceleration rate and in the end , the KE for the same rate of velocity change, is equal..... am I getting this right?)

what about a rocket in space at 0 velocity relative to earth. It accelerates out toward pluto and reaches 1000m/s where it runs out of fuel. We catchup to this space rocket in our own rocket... board it and fill it with the same level of fuel as it left with at the beginning of its journey....... It then ignites its fuel with the same thrust and accelerates at yet the same rate again. Wouldn't it be able to have the same rate of acceleration for its fixed thrust value it possessed from the start, even though its starting at a velocity of 1000m/s now, as it accelerates to 2000m/s? (same change in velocity it had from the start, but much more KE stored in the rocket as it reaches its new final velocity)
 
  • #41
jbriggs444
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That was one of the best explanation's ive seen too! so, its really about the mass changing in proportion to what was used as thrust, in the form of a part of that original mass. (so rocket has less mass and needs less thrust to create the same acceleration rate and in the end , the KE for the same rate of velocity change, is equal..... am I getting this right?)
That is not the effect that is being described.

Take the exact same rocket with exactly the same amount of fuel on board and consider it first from a reference frame in which it is moving at (for instance) 1000 m/sec and then from a reference frame in which it is moving at (for instance) 2000 m/sec.

The amount of kinetic energy gained by the rocket in the two frames will be different.
The amount of kinetic energy gained (or lost!!) by the exhaust in the two frames will be different.
The total amount of kinetic energy gained will be the same. That is the invariant quantity.

For completeness, consider a reference frame in which the rocket is moving backwards and thrusting forward. The "efficiency" of the motor will be negative in such a case -- it is consuming fuel and reducing the kinetic energy of the rocket. Nonetheless, the total kinetic energy produced will come out to the same invariant quantity.
 
  • #42
zanick
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That is not the effect that is being described.

Take the exact same rocket with exactly the same amount of fuel on board and consider it first from a reference frame in which it is moving at (for instance) 1000 m/sec and then from a reference frame in which it is moving at (for instance) 2000 m/sec.

The amount of kinetic energy gained by the rocket in the two frames will be different.
The amount of kinetic energy gained (or lost!!) by the exhaust in the two frames will be different.
The total amount of kinetic energy gained will be the same. That is the invariant quantity.

For completeness, consider a reference frame in which the rocket is moving backwards and thrusting forward. The "efficiency" of the motor will be negative in such a case -- it is consuming fuel and reducing the kinetic energy of the rocket. Nonetheless, the total kinetic energy produced will come out to the same invariant quantity.
so then its really about the mass departing the rocket, as was pointed out earlier in the discussion, right? Plus , all this is relative from some point in space, right?
 
  • #43
jbriggs444
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Multiple things were pointed out earlier in the discussion. A significant number pointing out the wrong effect.

Yes, this is relative. And that is the point. The increase in rocket kinetic energy resulting from a fixed "burn" is not invariant if you shift between coordinate systems that are in relative motion. The increase in total kinetic energy resulting from a fixed "burn" is invariant under such a shift.

I'm not happy with the wording "relative to some point in space". Try "relative to an inertial frame of referrence" instead.
 
  • #44
Drakkith
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I'd also remember that a rocket isn't a simple system that you find in introductory physics textbooks. You can't just take the basic work/energy/power equations and throw them in unless you know how to use them in this specific scenario. I think that's the source of pretty much all the confusion.
 
  • #45
Dale
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That was one of the best explanation's ive seen too! so, its really about the mass changing in proportion to what was used as thrust, in the form of a part of that original mass. (so rocket has less mass and needs less thrust to create the same acceleration rate and in the end , the KE for the same rate of velocity change, is equal..... am I getting this right?)
I am glad you enjoyed the explanation. The point was not about the mass changing, although it does.

The point is that if you want to look at conservation of energy then you have to consider the KE of the exhaust, not just the KE of the payload.
 
  • #46
Dale
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in the case of the rocket, if it doesn't know the "relative " speed as it leaves earth, if it crashed into Mars (or similar weight and size asteroid), it certainly would realize all the KE as it crashed. :)
what am I missing here?
This is the time-reverse of what I showed before. The asteroid and the ship both have KE and momentum. When they collide plastically there will be some change in the KE of the ship and some change in the KE of the asteroid. The sum will be the same in different frames, even though each part will be different.
 
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  • #47
sophiecentaur
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I'd also remember that a rocket isn't a simple system that you find in introductory physics textbooks. You can't just take the basic work/energy/power equations and throw them in unless you know how to use them in this specific scenario. I think that's the source of pretty much all the confusion.
I agree. Arm waving and 'chat' about a topic like this is bound to produce apparent paradoxes. Using the basic rules of conservation and doing the Math will give you the 'right answer', which will allow you to make good predictions.
 

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