Confused about kinetic energy

Drakkith

Staff Emeritus
2018 Award
Both facts are certainly important in a real rocket, but I think that the thing which causes the confusion of the OP is mostly forgetting the KE of the exhaust rather than the fuel-as-payload effect. You can contrive scenarios to neglect the fuel-as-payload effect, but the KE of the exhaust effect is always there if you look at different frames.
Hmmm. Just looking at it from a simple W=FD perspective, it seems pretty obvious that it takes more work to accelerate to higher velocities since you're applying a force over a larger distance than before. However, I have no idea how to include the fuel exhaust into this.

Dale

Mentor
Hmmm. Just looking at it from a simple W=FD perspective, it seems pretty obvious that it takes more work to accelerate to higher velocities since you're applying a force over a larger distance than before. However, I have no idea how to include the fuel exhaust into this.
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.

• gmax137, mic* and Drakkith

zanick

I think you have to remove all the variables from the set up of the question. call the rocket using a fuel that doesn't burn off. say it uses the photons from the sun. And with your imagination, say that the Force is equal to x..... as was said the force is constant, so the acceleration is constant, and the power goes up proportionally with the speed. the fuel use rate is exactly the same from 0-50m/s as it is from 50 to 100m/s. (in space) now, on land we have aero loses that go up with the square of velocity.
KE goes up with the square of speed..... so, for every second of change of speed, here is a HP-second associated with it. if the rate of change is the same, the hp-second (or HP x 746 =Joules) will be the same. as you go faster and faster, it all accumulates for higher and higher KE.

on earth, the aero dynamic forces go up with square, and power is force x speed, so the power required and used, goes up with the cube of speed. the KE only goes up with the square of speed. there is the difference.

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Joel A. Levitt

The equation for kinetic energy is:

k = 1/2*m*v2

This confuses me because this indicates it take more and more energy to accelerate...
Lets imagine a rocket in space (assume the mass of the rocket stays the same when accelerating) - the formula suggests it would take more energy (therefore fuel) to go from 50 to 100 km/s than it would from 0 to 50 km/s.

Is it true that it would take more fuel to accelerate from 50 to 100 km/s than it would from 0 to 50 km/s ?????
My guess is that you are a student in a first course in mechanics. Is my guess is correct? If my guess is correct, are you truly confused about kinetic energy or are you just having fun at the expense of the responders? If you are truly confused, I will try to explain work and energy.

zanick

My guess is that you are a student in a first course in mechanics. Is my guess is correct? If my guess is correct, are you truly confused about kinetic energy or are you just having fun at the expense of the responders? If you are truly confused, I will try to explain work and energy.
from the very first post, why didn't anyone just break out the math. KE change is the same from 0-50 as it is from 50 to 100. wouldn't that be the answer to the question right there?

Drakkith

Staff Emeritus
2018 Award
from the very first post, why didn't anyone just break out the math. KE change is the same from 0-50 as it is from 50 to 100. wouldn't that be the answer to the question right there?
I don't think that's correct. KE for a 1kg object at 50 m/s: 1250 joules. KE for a 1 kg object at 100 m/s: 5000 joules. Change from 0 to 50 is 1250 joules. Change from 50 to 100 is 3750 joules. That's 3 times as much change in KE.

zanick

I don't think that's correct. KE for a 1kg object at 50 m/s: 1250 joules. KE for a 1 kg object at 100 m/s: 5000 joules. Change from 0 to 50 is 1250 joules. Change from 50 to 100 is 3750 joules. That's 3 times as much change in KE.
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!

Drakkith

Staff Emeritus
2018 Award
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!
I don't think that's true either. Acceleration should remain constant, but the faster you go, the large the distance is that you apply the force over, meaning that work goes up.

zanick

actually, that is true.... acceleratrion is proportional to power, given by the Newtonian identity, acceleration = power/(mass x velocity)
so yes, as speed goes up, with constant power, acceleration will go down proportionately. for constant acceleration (or force) , power will have to go up with velocity.
Power = Force x Velocity

Drakkith

Staff Emeritus
2018 Award
actually, that is true.... acceleratrion is proportional to power, given by the Newtonian identity, acceleration = power/(mass x velocity)
so yes, as speed goes up, with constant power, acceleration will go down proportionately. for constant acceleration (or force) , power will have to go up with velocity.
Power = Force x Velocity
I don't think that applies to rockets, which is what we're talking about here. A rocket actually has to reduce power (and thrust) to maintain a steady acceleration.

Drakkith

Staff Emeritus
2018 Award
@zanick Hmmm. I'm not sure I read your posts correctly. I'll wait to comment further on this until I get some sleep.

jbriggs444

Homework Helper
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.

gmax137

... Re-read post #27 above by DaleSpam.
plus one to that! Dale's post is the clearest explanation of this I have ever seen.

zanick

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?

zanick

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)

jbriggs444

Homework Helper
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.

zanick

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?

jbriggs444

Homework Helper
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.

Drakkith

Staff Emeritus
2018 Award
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.

Dale

Mentor
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.

Dale

Mentor
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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sophiecentaur

Gold Member
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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