Why kinetic energy is proportional to velocity squared

[learningphysics]

There are so many things wrong here I almost don't know where to start.



I'll start here at the original post. A constant force does not result in a constant acceleration because the vehicle is ejecting mass. Instead, a constant force results in an ever increasing acceleration as the vehicle becomes less massive.



Constant rate of fuel production does produce a constant force.

I apologize. Yes, what I wrote was wrong. I was thinking in very simple terms of an object with a fixed mass gaining kinetic energy at a constant rate...

Sorry Emanresu for any confusion I caused.

 

[learningphysics]

It is easier to see with a car than a rocket since people deal with them every day (and their mass doesn't change as they accelerate). Cars accelerate from 0-30 much, much faster than from 30-60. And even with roughly constant horsepower, you can see that the reason is that the force (the torque at the wheels) decreases as you go through theh gears.

Yeah, car would have worked, since it isn't driven by exhaust, but by the force of the tires on the road...

 

[animalcroc]

Er.. you need to re-read what has been written here. The reason why you can do that is the very reason why you can't assume that acceleration is a constant. There's nothing here that tells you that because you don't know that F is a constant to start with. The ONLY constant here is the energy transferred to the vehicle. As you can already see from the thread, this does not automatically imply a constant force. So you can't simply assume that acceleration is a constant.

You should also consider that using "calculus" is more general that what you have done here, which is more of a specific case. the use of calculus isn't the issue nor the cause of any of the disagreement here.

I suggest we WAIT for the OP to get back and clarify the exact scenario of the problem at hand.

Zz.

Of course calculus is more general but to explain a concept you start with the basics so it's more understandable. In fact, we don't even know the math level of the OP.
Intro physics book also use the constant acceleration approach. Many concepts are introduced without calculus because the math levels of readers are unknown and a better intuitive understanding may be developed without using calculus.
I only answered the first question of the OP (which was clear).

 

[Emanresu]

I apologize. Yes, what I wrote was wrong. I was thinking in very simple terms of an object with a fixed mass gaining kinetic energy at a constant rate...

Sorry Emanresu for any confusion I caused.

No problem. What you explained earlier made sense. But I am confused again as I didn't follow DH's argument. If now 'Constant rate of fuel production does produce a constant force' I am back to square one. Can you explain DH's argument for me ?

E.

 

[Emanresu]

Fine. Outfit our vehicle with a light sail. We'll power the light sail from a "fixed" source. Keeping this to one dimension once again for simplicity, and keeping velocities small to avoid relativistic effects, the momentum transferred to the vehicle over some time interval is \Delta p = 2 \dot n h/\lambda \Delta t, where \dot n is the photon flux. The factor of two results from the reflecting the incoming photon beam back toward the transmitter. The energy transferred to the vehicle is \Delta E = 2 \dot nh c/\lambda \Delta t, obviously proportional to the change in momentum. Since the mass is constant, the change in vehicle velocity is also proportional to energy transferred to the vehicle.

Suppose this were not the case: the change in velocity depends on the initial velocity. This means an observer in a inertial frame initially at rest with respect to the vehicle will see a different change in velocity than an observer in an inertial frame in which the vehicle has some non-zero initial velocity. As the relative velocity of the two observer frames is constant, both observers should see the same change in velocity.

You are a bit over my head here, but when you say 'Since the mass is constant, the change in vehicle velocity is also proportional to energy transferred to the vehicle', is there a constant force on the vehicle ?

E.

 

[D H]

Emanresu, we are having a bit of an issue with regard to your intent in your original post. As you asking about why energy is proportional to the square of velocity rather than velocity and using rockets to illustrate the issue, or are you asking about rockets and wondering about energy as a tangential issue, or both?

 

[Emanresu]

Emanresu, we are having a bit of an issue with regard to your intent in your original post. As you asking about why energy is proportional to the square of velocity rather than velocity and using rockets to illustrate the issue, or are you asking about rockets and wondering about energy as a tangential issue, or both?

Originally I thought that constant rate of energy supply would provide constant force so I was confused as to why energy out (KE) was proportional to v^2. Then as result of answers given here I was persuaded that a constant force would not result which explained why KE was not proportional to v. But you are saying something different which is similar to what I originally thought (constant energy supply WILL provide constant force) which in my mind does not tally with KE being proportional to v^2. I'm obviously not very good at explaining things, hence all the confusion, but if you could provide a simplified explanation of your light sail example, maybe that would help me.

E.

p.s. I only used a rocket because I thought it was the simplest concept (I was obvioulsy wrong).

 

[D H]

A rocket is far from the simplest concept. Let me pick at another post of yours:

Say you have a rocket ship at rest with a certain amount of chemically stored energy (fuel). To provide a constant force to constantly accelerate the rocket the chemically stored energy is used up at a constant rate. This energy gets mainly turned into kinetic energy and heat.

Energy cannot be created or destroyed. The fuel energy is being used at a constant rate and producing heat at a constant rate but producing kinetic energy at an ever increasing rate. It appears that energy is being created out of nowhere !

I am missing something here, what is it ?

You are missing the exhaust. Several things are going on here. First, a thrusting rocket loses mass. If the rocket does manage to generate constant thrust its acceleration will increases.

So, what about energy?

Examine what happens to the rocket and exhaust cloud over some small time interval \Delta t. (By small I mean sufficiently short in duration such that products of changes can be ignored). I am assuming the rocket is well removed from any massive bodies so I can ignore gravitational potential energy.

At the start of the interval, the rocket has kinetic energy E_{r-} = \frac 1 2 m_r v_r^2. Denote the initial kinetic energy of the exhaust cloud as some unknown and irrelevant quantity E_{e-}. Over the time interval, the rocket ejects \Delta m quantity of fuel as exhaust moving at a velocity v_e relative to the vehicle. The kinetic energies of the rocket and exhaust cloud at the end of the interval are

E_{r+} = \frac 1 2 (m_r-\Delta m)(v_r+\Delta v_r)^2<br /> = E_{r-} + m_r v_r\cdot\Delta v_r - \frac 1 2 \Delta m v_r^2 + O(\Delta^2)
E_{e+} = E_{e-} + \frac 1 2 \Delta m(v_r+v_e)^2<br /> = E_{e-} + \frac 1 2 \Delta m v_r^2 + \Delta m v_r\cdot v_e + \frac 1 2 \Delta m v_e^2

The total kinetic energy is thus
E_{+} = E_{-} + m_r v_r\cdot\Delta v_r - \frac 1 2 \Delta m v_r^2 + \frac 1 2 \Delta m v_r^2 + \Delta m v_r\cdot v_e + \frac 1 2 \Delta m v_e^2+ O(\Delta^2)

The change in kinetic energy is just E_{+} - E_{-}. Eliminating high order terms and simplifying,
\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

The rocket and exhaust cloud form a closed system subject to conservation of linear momentum. Applying conservation of momentum yields m_r \Delta v_r + \Delta m v_e=0. Applying this result to the change in kinetic energy yields
\Delta E = \frac 1 2 \Delta m v_e^2

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.

 

[learningphysics]

No problem. What you explained earlier made sense. But I am confused again as I didn't follow DH's argument. If now 'Constant rate of fuel production does produce a constant force' I am back to square one. Can you explain DH's argument for me ?

E.

A rocket... or an "exhaust-based' propulsion is much more complicated... since the mass of the fuel/exhaust ejected is significant (it has to be since the method of propulsion is shooting out exhaust... the greater the momentum of the exhaust that is shot out... ie the more exhaust is shot out... the greater the change in momentum of what's left of the rocket... due to conservation of momentum) And

mass of the exhaust can't be ignored... so the mass of the rocket is changing with time... and a significant portion of the energy in the fuel is lost with the exhaust.

As DH wrote, a constant exhaust velocity along with a constant rate of fuel consumption, leads a constant force on the rocket (whose mass is dropping)...

I don't think the rocket example is the best example to use to get across the idea that kinetic energy is proportional to v^2. A car (normal everyday car, not a rocket-type one) is a better example as russ mentioned... a car isn't driven forward by the exhaust but by the force of the tires against the road... the mass of the fuel and exhaust is very small compared to the mass of the car so the change in mass can be ignored...


The main idea is that an object with a fixed mass, gaining kinetic energy at a constant rate does not experience a constant force (rocket is not an example of such a situation since mass is changing)... what I posted before applies to this idea of a fixed mass and constant rate of kinetic energy (P) increase... I'll post it here too:

after a time t, the object has kinetic energy:

E = (1/2)mv^2
Pt = (1/2)mv^2
v = \sqrt{\frac{2Pt}{m}}

taking the derivative of Pt = (1/2)mv^2:

P = mv*dv/dt

P = mv*a

P = (ma)*v

P = F*v

F = P/v.

F = \frac{P}{\sqrt{\frac{2Pt}{m}}}

So F the force on the object changes with time...

I think the thread may have gone a little aside from your questions... I think that your main questions are rooted in why energy is defined as it is... why (1/2)mv^2, why \int Fdx etc... Hope my post #38 gives some idea of these ideas... the essential idea that leads to these I believe is that the fundamental forces are conservative...

 

[Emanresu]

I think the thread may have gone a little aside from your questions... I think that your main questions are rooted in why energy is defined as it is... why (1/2)mv^2, why \int Fdx etc... Hope my post #38 gives some idea of these ideas... the essential idea that leads to these I believe is that the fundamental forces are conservative...

Thanks again. I'm happy with the principle now, although I'm going to have a go at working through DH's rocket math just out of interest.

E.

 

[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 independant 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 \Delta v[/itex] that results from burning a mass of fuel \Delta m is<br /> <br /> \Delta v&lt;br /&gt; = v_e \ln \frac{m_0}{m_0-\Delta m}&lt;br /&gt; = 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 &quot;So when the rocket velocity increases to a certain point, it is actually traveling faster than the thermal velocity of the ejected fuel particles&quot;, 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.

 

[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

 

[rcgldr]

Even battery energy is rated in watt hours (or in amp hours and you have to multiply by the voltage). In this case energy = power x time.