Why kinetic energy is proportional to velocity squared

[D H]

No one peer reviewed what I wrote. No guarantees on sign errors, etc. I wrote it from scratch. (BTW, modeling spacecraft dynamics is one of my jobs.)

Google "Tsiolkovsky rocket equation" for more info. The first several hits are fairly good, including the wikipedia article. Don't bother with the wolfram.com article; it's void of content.

 

[Emanresu]

Note that this result is independent of the rocket velocity. This quantity represents the useful energy obtained by burning fuel and ejecting exhaust from the vehicle. This quantity will alwys be less than the chemical energy in the fuel, as some of the chemical energy is lost in the form of an increased exhaust temperature.

Okay. I've worked through your math and am happy enough with it.

In order to get some insight I started plugging in numbers.

It appears that the faster the rocket is going the more energy is transferred to the exhaust gas

$$\frac 1 2 \Delta m v_r^2 + \Delta m v_r\cdot v_e + \frac 1 2 \Delta m v_e^2$$

because $$v_e$$ is constant and $$v_r$$ is getting bigger.

But that means less energy is being transferred to the rocket as it goes faster which means acceleration is decreasing.

Which means I've made a mistake somewhere (surprise !).

E.

 

[richard14]

Could someone please explain why kinetic energy is proportional to velocity squared and not just velocity.

A propulsion system that provides a constant force would typically use energy at a constant rate in order to accelerate at a constant rate and thus rate of change of velocity would be proportional to rate of use of energy. So why is rate of gain of energy not also proportional to rate of change of velocity ?

E.

in simple words emanresu the velocity is directly proportional to the kinetic energy the biger
the velocity the more kinetic energy obtain by the object let say a car.
was this helpful that's why V2

 

[ZapperZ]

in simple words emanresu the velocity is directly proportional to the kinetic energy the biger
the velocity the more kinetic energy obtain by the object let say a car.
was this helpful that's why V2

This is wrong. KE is proportional the v^2, not v.

Zz.

 

[PatPwnt]

Doesn't 2ad = v^2?

So 2a*d*m = m*v^2 -- a*m = N

2N*d = m*v^2

so N*d = 1/2m*v^2 -- N*d = one of definition of work/energy. Force through a
displacement.

Energy = Energy

a = acceleration d = displacement m = mass

It takes N Newtons through a distance d to stop a mass with velocity v. This is where 1/2mv^2 comes from.

Or... it takes N Newtons through a distance d to accelerate a mass to velocity v.

Can it be any simpler?

 

[ZapperZ]

Doesn't 2ad = v^2?

You should check how that expression was derived. You'll discover that you had implicitly assumed a constant acceleration situation, which may not be true in this situation.

Zz.

 

[rcgldr]

In the non-rocket situation, the frame of reference needs to be non-accelerating reference relative to the point of application of the force. In the case of a car, the point of reference would be the road, or a "fixed" point if you want to include the fact that the Earth is rotated backwards a bit by a car accelerating forwards.

In the case of the rocket, the point of application of the force is the rocket's engine, which is accelerating along with the rocket, accelerating a small part of the rockets mass in one direction (spent fuel) and accelerating the rocket and remaining fuel in the other direction. Note that kinetic energy of both the spent fuel and the rocket (and it's remaining fuel) are significant and part of the "system". As posted below, with a constant burn rate, and constant thrust, the sum of kinetic energy of spent fuel and rocket (and it's remaining fuel) increases linearly with time.

If the spent fuel is ignored, then this approach could be used: As the rocket accelerates, so does the remaining fuel, which increases it's "potential energy". So although the chemical energy remains constant, the total energy per unit of fuel increases with the square of velocity (relative to the starting velocity of the rocket). The net result is that the fuel will supply a fixed thrust per rate of fuel burned, regardless of the velocity of the rocket. The acceleration of spent fuel is always relative to a rocket's current velocity.

 

[D H]

It appears that the faster the rocket is going the more energy is transferred to the exhaust gas

$$\frac 1 2 \Delta m v_r^2 + \Delta m v_r\cdot v_e + \frac 1 2 \Delta m v_e^2$$

because $$v_e$$ is constant and $$v_r$$ is getting bigger.

But that means less energy is being transferred to the rocket as it goes faster which means acceleration is decreasing.

Which means I've made a mistake somewhere (surprise !).

The correct expression is
$$\Delta E = m_r v_r\cdot\Delta v_r + \Delta m v_r\cdot v_e + \frac 1 2 \Delta m v_e^2$$

By conservation of momentum, $m_r \Delta v_r + \Delta m \cdot v_e = 0$, so the change in energy is just

$$\Delta E = \frac 1 2 \Delta m v_e^2$$

=======================

As the rocket accelerates, so does the remaining fuel, which increases it's "potential energy". So although the chemical energy remains constant, the total energy per unit of fuel increases with the square of velocity (relative to the starting velocity of the rocket). The net result is that the fuel will supply a fixed thrust per rate of fuel burned, regardless of the velocity of the rocket.

As the fuel is used up, the mass of the rocket and remaining fuel decrease, and the rate of acceleration increases.

Thank you, Jeff.

The exact same logic also applies to automobiles, which, like rockets, carry the fuel that propels the vehicle. The reason it is harder to accelerate from 30 to 60 MPH than from 0 to 30 in consumer cars is not so much a work issue as it is an increase in aerodynamic drag and rolling friction and because of the relatively low power output of consumer-class automobiles. Racing cars are a different kind of beast. Their engines produce peak torque (and hence peak acceleration) at a much higher engine speed than do consumer-class vehicles.

 

[rcgldr]

To simplify the case of the rocket, imagine that the rocket is held in place (or that the rocket is the frame of reference). The only work done is to the spent fuel, which has a fixed terminal velocity (at a specific burn rate). Kinetic energy of the spent fuel increases linearly with time (velocity of spent fuel is constant, with only the amount of mass of fuel changing linearly with time). If the rocket is not held in place, and a non-accelerating frame of reference is chosen, then the increase in the sum of kinetic energy of spent fuel, and rocket (and it's remaining fuel) will also increase linearly with time, since the source of the work being done, burning of fuel, remains the same. I haven't checked, but the equations shown previously should show this to be true.

In the case of a car, the point of application of force is the pavement and driven tires of the car. This is a simpler case, power = force times speed, for example horsepower = force (lbs) times speed (mph) divided by 375 (conversion factor). Assuming no slippage of the tires, and that the power of the engine is constant, and that a continously variable transmission is used to keep the engine running at maximum power, then force will decrease linearly as speed increases, and kinetic energy will increase at a constant rate. The exhaust from the car is not a significant source of acceleration, and I'm ignoring it in this case.

Note that in both the case of the rocket and the car, kinetic energy of the systems were increasing at a constant rate. The rocket case isn't so special if you take into account the kinetic energy of the spent fuel.

 

[Emanresu]

The correct expression is
$$\Delta E = m_r v_r\cdot\Delta v_r + \Delta m v_r\cdot v_e + \frac 1 2 \Delta m v_e^2$$

By conservation of momentum, $m_r \Delta v_r + \Delta m \cdot v_e = 0$, so the change in energy is just

$$\Delta E = \frac 1 2 \Delta m v_e^2$$

No. I quoted what you gave for the EXHAUST GAS ONLY. Please re-read my question. I know my questions APPEAR stupid to you because you know the subject inside out, but I genuinely am just trying to understand this one bit at a time.

E.

 

[D H]

Emanuresu, you had the change in energy for the exhaust cloud correct. Sorry to have misread your question. The exhaust cloud gains mass, linear momentum, and kinetic energy because the rocket transfers each of these quantities to the exhaust. Because the rocket's velocity is increasing, the rate at which the exhaust cloud gains linear momentum and kinetic energy increases.What's going on here is simple: The exhaust cloud by itself does not form an isolated system. The conservation laws don't pertain to the exhaust gas by itself. They do apply to the rocket+exhaust gas system, but only if one assumes no other forces act on the rocket+exhaust gas. That's why I put the rocket in deep space, far removed from any massive bodies. When applying any of the conservation laws one has to take care in defining the system boundaries.

 

[Emanresu]

Emanuresu, you had the change in energy for the exhaust cloud correct. Sorry to have misread your question. The exhaust cloud gains mass, linear momentum, and kinetic energy because the rocket transfers each of these quantities to the exhaust. Because the rocket's velocity is increasing, the rate at which the exhaust cloud gains linear momentum and kinetic energy increases.


What's going on here is simple: The exhaust cloud by itself does not form an isolated system. The conservation laws don't pertain to the exhaust gas by itself. They do apply to the rocket+exhaust gas system, but only if one assumes no other forces act on the rocket+exhaust gas. That's why I put the rocket in deep space, far removed from any massive bodies. When applying any of the conservation laws one has to take care in defining the system boundaries.

So then the rate at which the rocket gains kinetic energy DECREASES ?
But the rocket's acceleration INCREASES ?
And the explanation for both of these is that the rocket's mass is DECREASING ?

E.

 

[learningphysics]

So then the rate at which the rocket gains kinetic energy DECREASES ?
But the rocket's acceleration INCREASES ?
And the explanation for both of these is that the rocket's mass is DECREASING ?

E.

Yes, this looks right to me. Both effects are due to mass decreasing.

 

[Emanresu]

To simplify the case of the rocket, imagine that the rocket is held in place (or that the rocket is the frame of reference). The only work done is to the spent fuel, which has a fixed terminal velocity (at a specific burn rate). Kinetic energy of the spent fuel increases linearly with time (velocity of spent fuel is constant, with only the amount of mass of fuel changing linearly with time). If the rocket is not held in place, and a non-accelerating frame of reference is chosen, then the increase in the sum of kinetic energy of spent fuel, and rocket (and it's remaining fuel) will also increase linearly with time, since the source of the work being done, burning of fuel, remains the same. I haven't checked, but the equations shown previously should show this to be true.

In the case of a car, the point of application of force is the pavement and driven tires of the car. This is a simpler case, power = force times speed, for example horsepower = force (lbs) times speed (mph) divided by 375 (conversion factor). Assuming no slippage of the tires, and that the power of the engine is constant, and that a continously variable transmission is used to keep the engine running at maximum power, then force will decrease linearly as speed increases, and kinetic energy will increase at a constant rate. The exhaust from the car is not a significant source of acceleration, and I'm ignoring it in this case.

Note that in both the case of the rocket and the car, kinetic energy of the systems were increasing at a constant rate. The rocket case isn't so special if you take into account the kinetic energy of the spent fuel.

Hi Jeff, I didn't give your posts due attention the first time I read them. They are very insightful (no math ). Thank you.

E.

 

[Mutos]

Hi all,I posted the same question as you just before I found this post, I missed it on my first searchs. Now after reading the posts, I feel I understand the point...

It's just a question of frame of reference and the keyword in this cas is measured energy and power. That is, if you measure the energy of the same object and the power output needed to make it accelerate, relative to different reference frames, you'll have different answers depending on the frame's own speed...

We are so accustomed to measure power output in an immobile reference frame that we intuitively think of it as a quantity independent from the referential. That is what makes us see things wrongly.

My question was formulated in the following terms :

Suppose the following situation : consider a spaceship under constant acceleration. Let's say it's a fictionary reactionless propulsion and not bother about fuel depletion, efficiency or other topics : ship mass stays constant, all energy is used for propulsion, there is no friction at all and we do not reach relativistic speeds.

Intuitively, you will say that this ship would use for its propulsion a constant power output. Thrust is the same at all time, forces and the physical mechanism that produce them are the same, so power output should be the same. Intuitively, a ship with a given power unit should be able to sustain constant acceleration infinitely, given we take out the fuel, efficiency and other problems.

But equations say that power depends on time as speed must be considered to compute the link between power and acceleration. They say P=m.a^2.t and so says the unit check : W=kg.m.s^-3, which is consistent with the formula.

Emanresu, it's basically the same question as you. I take out by advance all other considerations so that we don't err about rocket equations, mass depletion, friction, relativity and the likes. So I focus only on the acceleration to power question. And it's just about what we perceive intuitively VS what the equations say.

I don't doubt equations a moment and so what I perceive must be false but it's of no use trying to demonstrate it using maths as I already know they're exact... "Why is our perception false ?" is the actual question.

Now I think I understand and it brings back some memories, but even 15 years ago when I was studying engineering I never felt totally at ease with this problem of reference frames...

 

[colin9876]

Emanresu you first post was actually right!
Your question regarding KE proportional to v not v squared is actually a very clever one.

KE is actually first order (i.e. proportional to v) but has been approximated as a squared function - only works when speeds are small compared to the huge speeds our Earth is already moving at!
(this is becaue SQR(1+v*v) is approx 1+.5*v*v when v is small)

This leads to some paradoxes that can not be fully resolved with Newtonian physics, - so you have shown real genius in challenging something which was once considered a basic rule!

Consider Links paradox
300kg space platform hurtling thro space at average speed 2m/s/s
100kg man standing on platform

momentum conserved at (300x2 +100x2=800)
when man starts pacing round platform at 1m/s
so (a) when man pacing in same direction as platform moving relative to observer
platform 1.75m/s
man 2.75 m/s

(b) when man pacing in opp direction of platform movement
platform 2.75 m/s
man 1.75m/s

But K.E. in situations different ! can't be so?
(a) = .5*300*1.75*1.75+.5*100*2.75*2.75
(b)=.5*300*2.75*2.75+.5*100*1.75*1.75

 

[D H]

You aren't conserving momentum in case (b).

 

[gareth]

Maybe considering what drives the rocket might help, I'm guessing it's the force of the hot particles ejected from the rocket is what gives it velocity (acceleration initially),

So when the rocket velocity increases to a certain point, it is actually traveling faster than the thermal velocitiy of the ejected fuel particles,

the velocity has saturated at this point and the fuel only acts to counteract any resistive forces of the rocket and let it travel at constant v.

So initially the rocket's energy is increasing as 1/2 mv^2 at a constant fuel consumption, but the acceleration doesn't continue indefinately and saturates at a certain point.

If the rocket were perfect, and all the energy of the fuel goes into the kinetic energy of the rocket, I'm sure that the stored energy of the fuel used to get to velocity v would equal velocity v of the rocket (assuming there is no resitance to the flight)

As I re-read this post, I'm still a bit unsure, the rocket is still emitting energy after the velocity saturates but where is it going? Thermal energy left in space is my hunch...

Does this confuse matter further or give any insight?

 

[D H]

gareth,

A rocket's velocity doesn't saturate, classically. There is no upper limit on a rocket's velocity in non-relativistic physics. There is an upper limit in a practical sense. It takes an exponentially increasing amount of fuel to yield a linearly increasing change in velocity. This is the rocket equation. Suppose the rocket is to be fired in a constant direction and the rocket flies in an environment free of external forces. The change in velocity [tex]\Delta v[/itex] that results from burning a mass of fuel $$\Delta m$$ is<br /> <br /> $$\Delta v<br /> = v_e \ln \frac{m_0}{m_0-\Delta m}<br /> = v_e \ln \frac{m_f+\Delta m}{m_f}$$<br /> <br /> where $m_0$ is the mass of the rocket (including fuel) prior to firing the rocket and $m_f$ is the mass at the end of the burn.<br /> <br /> When you say "So when the rocket velocity increases to a certain point, it is actually traveling faster than the thermal velocity of the ejected fuel particles", you are implying that there exists a preferred reference frame. Suppose we have two inertial observers watching the rocket. The two observers are moving at the saturation velocity $v_{sat}$ with respect to one another. At some point in time, the rocket will have an instantaneous velocity equal to the saturation velocity with respect to one observer. The rocket will instantaneously be at rest with the other observer. If what you said about a saturation velocity were true, the rocket would be frozen in one frame but free to move in the other. There is no saturation velocity.[/tex]

 

[gareth]

I see, but if a rocket is burning fuel at a constant rate, why does the velocity not keep increasing? (negleting external forces and assuming the mass of the rocket stays constant)

And what is the physics behind the rocket equation? Why does it take more and more fuel to get a change in velocity?

Thanks

 

[D H]

I see, but if a rocket is burning fuel at a constant rate, why does the velocity not keep increasing?

The velocity of the rocket does keep increasing. This is why I objected to your "saturation velocity".

(negleting external forces and assuming the mass of the rocket stays constant)

You cannot assume the mass of the rocket stays constant. It is burning fuel.

And what is the physics behind the rocket equation? Why does it take more and more fuel to get a change in velocity?

I hope I have answered your questions in https://www.physicsforums.com/showthread.php?t=199087". In particular, see post #3. If you have additional questions feel free to ask.

 

[colin9876]

DH you object to point 2 'assuming the mass stays constant' - clearly this doesn't happen in practice but this is a hypothetical scenario to make a simpler approximation.
you could say 'assuming the mass of fuel burnt is small compared to the total mass etc.

Anyway I did make a maths error in my previous post - is it possible to edit posts in this forum - how do I do that please?

(But I still stand by the fact that in the whole scheme of things KE is proportional to V, its only when v is small compared to the high velocity we are moving at that it seems to be appoximated to half V squared)

 

[D H]

Anyway I did make a maths error in my previous post - is it possible to edit posts in this forum - how do I do that please?

You have to correct your errors quickly. The edit window is only open for a short period of time. (Too many abuses of editing.) You can post the correct maths in a new post.

(But I still stand by the fact that in the whole scheme of things KE is proportional to V, its only when v is small compared to the high velocity we are moving at that it seems to be appoximated to half V squared)

I gather you are talking about relativistic effects. You are correct that kinetic energy is not proportional to the square of velocity in special relativity. You are, however, completely wrong in saying that it is proportional to velocity. In special relativity, kinetic energy grows unbounded as the relative velocity approaches the speed of light.

 

[colin9876]

Actually I wasnt talking about relativity
I stand by the fact that KE proportional to v, its just that we are all traveling at speed u say, and sqrt(u*u +v*v) can be approximated by u+ .5v*v if v is small compared to u

But on the subject of relativity ...
Tell me this ... if (in theory) KE approaches infinity when speed approaches speed of light, why doesn't sunlight have infinite energy - sureley those photons are traveling at speed of light?

 

[D H]

Actually I wasnt talking about relativity
I stand by the fact that KE proportional to v, its just that we are all traveling at speed u say, and sqrt(u*u +v*v) can be approximated by u+ .5v*v if v is small compared to u

There is no square root term in the classical theory of kinetic energy. Kinetic energy in classical physics is
$$KE = \frac 1 2 m v^2$$
and for a collection of N particles,
$$KE = \sum_{i=1}^N \frac 1 2 m_i v_i^{\;2}$$

But on the subject of relativity ...
Tell me this ... if (in theory) KE approaches infinity when speed approaches speed of light, why doesn't sunlight have infinite energy - sureley those photons are traveling at speed of light?

Because photons are massless particles that have finite but non-zero momentum.

 

[colin9876]

I know in the classical theory KE proportional to .5mv*v but I am agreeing with the origonal post in this thread that KE really is proportional to v. Its only that we are moving at a speed u already that it seems like .5mv*v

By the way how do u do those proper symbols and summation signs? Can I get them from my keyboard?

By the way, so you say photons have mass=0, but they do have momentum (mc), why isn't 0*c zero then?
Do u believe that photons travel at speed c, or nearly approaching c?

 

[D H]

I know in the classical theory KE proportional to .5mv*v but I am agreeing with the origonal post in this thread that KE really is proportional to v. Its only that we are moving at a speed u already that it seems like .5mv*v

Energy has units of mass*length²/time². Things in the form of mass times velocity squared, or momentum times velocity, or force times length all have the correct units. Energy is testably proportional to the square of velocity.

If you are trying to say that energy is frame-dependent, that is correct. A speeding car in a frame at rest with the speeding car has zero kinetic energy. A passenger in a speeding car can open a soda and give it to the driver without the soda can tearing the driver's arm off. In the frame of the bridge that the driver is not paying attention and is about to hit because he is taking a soda from his girlfriend and not watching the road, different story. The car is going to transfer a lot of kinetic energy to the bridge.

However, in saying that "we are moving at a speed u", I think you are trying to say that there is some preferred reference frame in which all energy is (or should be) measured. That is demonstrably wrong. Galileo talked about this in the first theory of relativity, Galilean Relativity. Velocity (and energy) is relative.

By the way how do u do those proper symbols and summation signs? Can I get them from my keyboard?

Use LaTeX. See this thread: https://www.physicsforums.com/showthread.php?t=8997"

By the way, so you say photons have mass=0, but they do have momentum (mc), why isn't 0*c zero then?
Do u believe that photons travel at speed c, or nearly approaching c?

The momentum of a photon is not $mc$. You are applying classical physics in a realm where it simply is not valid. Photons have momentum $p=hf$, where h is Planck's constant and $f$ is the photon's frequency. The energy of a particle in some reference frame is given by
$$E = \sqrt{(pc)^2 + (m_0c^2)^2}$$
where $p$ is the momentum of the particle in the reference frame and $m_0$ is the particle's invariant (or rest) mass. For a photon with a zero invariant mass, this reduces to $E=pc$. For a massive particle at rest in the reference frame, this reduces to $E=m_0c^2$ (you may have seen this form of the equation).

Photons travel at c, which is why c is called the speed of light. They can only travel at the speed of light precisely because they have no mass. Massive particles, on the other hand, can only travel at speeds less than the speed of light.

 

[colin9876]

thanks for a very good answer.

Im on holiday at the moment so have all day to ponder about physics. I asked my girlfriend about kinetic energy yesterday and she just looked at me blankly and changed the subject so its great to have someone to help me try and get my head round this. lol!

Something about all this has never clicked with me - for example how can energy have time in the units? it doesn't make intuitive sense??

 

[D H]

Something about all this has never clicked with me - for example how can energy have time in the units? it doesn't make intuitive sense??

In the world at large, as opposed to the tiny world of quantum mechanics, there are only five basic quantities of nature: mass, length, time, electric charge, and temperature (and the last is dubious as a thing separate from the others). Everything else is derived from these basic quantities. A simple one of these derived quantities is area. Area is length squared. Volume is similarly length cubed. The velocity of an object is the distance (length) it covers per unit of time. Velocity is a derived quantity with units of length/time.

What about energy? From $KE = 1/2mv^2$, kinetic energy has units of mass times the square of the units of velocity. Note that the 1/2 drops out when doing http://en.wikipedia.org/wiki/Dimensional_analysis" because it is a dimensionless scale factor. Physicists write this as units(kinetic energy) = units(mass)*units(velocity)². Mass is basic, but velocity is not. Thus the basic units of kinetic energy are mass times the square of the basic units of velocity, or mass*length²/time².

One key concept of dimensional analysis is that units have to be consistent. While 1+1=2, it doesn't make any sense to compute the sum of one kilogram and one meter. Anything else that we call energy had better have the same basic units of mass*length²/time². Energy comes in many forms. For example, work (as defined by physicists) is energy via the work-energy relationship. We are doing work when we apply a force on something to make it move. The work done is the product of the force and the distance the object travels. So, units(work) = units(force)*units(distance). Force is mass times acceleration, and acceleration is change in velocity per unit time. Putting it all together, the units of work are mass*length²/time². Energy!

 

[gareth]

"Something about all this has never clicked with me - for example how can energy have time in the units? it doesn't make intuitive sense??"

Maybe it's because by definition, for something to have KE it must be moving. If something is moving, it has a velocity, if it has a velocity it must be traversing some distance L, since we have a speed limit in this universe (c), some time must elapse before we traverse a distance L.

So time must play a part in the description of energy, if there was no time involved everything would be standing perfectly still.

They are my thoughts