Rockets and 1/2 mv^2

[Mr. Peterman]

Okay, I've heard this argument that says that the fuel energy of a rocket is mv rather than 1/2 mv^2. I was wondering if this is what is happening. Maybe the first amount of fuel explodes and pushes the rocket. Another 1/2 mv^2 would only get the rocket to the squart root of 2 times the first velocity. But it would be true that the rest of the fuel would get an increase in kinetic energy from the first amount of fuel, and so the next liter of fuel coming out would have more kinetic energy per liter than the first amount of fuel. Then this would still result in 1 liter corresponding to 100 mph and 2 liters getting the rocket going to 200 mph. Is this how it would work out?

[Janus]

Okay, I've heard this argument that says that the fuel energy of a rocket is mv rather than 1/2 mv^2. I was wondering if this is what is happening. Maybe the first amount of fuel explodes and pushes the rocket. Another 1/2 mv^2 would only get the rocket to the squart root of 2 times the first velocity. But it would be true that the rest of the fuel would get an increase in kinetic energy from the first amount of fuel, and so the next liter of fuel coming out would have more kinetic energy per liter than the first amount of fuel. Then this would still result in 1 liter corresponding to 100 mph and 2 liters getting the rocket going to 200 mph. Is this how it would work out?

 

I'm not quite sure what you mean by "fuel energy". But if you want to know the velocity that can be reached for a rocket for a given amount of fuel, you use the rocket equation:

 

V_f = V_e ln(MR)

 

Where Ve is the exhaust velocity and MR is the Mass Ratio, or the ratio of the mass of the fully fueled rocket to the mass of the rocket after the fuel has been expended.

[CraigD]

Okay, I've heard this argument that says that the fuel energy of a rocket is mv rather than 1/2 mv^2. I was wondering if this is what is happening.

As with any other mechanical system with speeds low enough for relativistic effects to be negligible, the laws of conservation of momentum and energy apply to a rocket, but as it’s a complex system of the rocket, the reaction mass still contained in it, and the reaction mass it’s expelled, each bit of it traveling at a slightly different velocity than the other bits, it’s difficult to visualize intuitively. However, if you simplify the system – for example, by having a rocket that expels its reaction mass as a small number of “bullets”, it’s possible to account for the energy and momentum involved using basic, though somewhat lengthy, arithmetic.

 

Accounting for the energy of a rocket and its reaction mass (in a chemical rocket, almost exactly the same thing as its fuel) is complicated. It’s been discussed a good bit over the years in these forums, in posts such as A short answer, and a lot of numbers and long threads such as A problem with KE = ½mv². Help.

 

Typical rockets in friction-free settings are well described by equations derived from the basic laws of motion such as the simple the idea rocket equation:

[math]\Delta v = v_\text{e} \ln\left( \frac {m_0} {m_1} \right)[/math]

where [math]\Delta v[/math] is the change in the rocket’s velocity,

[math]v_\text{e}[/math] is the velocity at which it expels reaction mass (exhaust velocity),

[math]m_0[/math] is the mass of the rocket + reaction mass, and

[math]m_1[/math] is the mass of the empty mass of the rocket.

 

A key point to note here is that the power of the rocket – the rate at which it expels reaction mass – doesn’t affect its net [math]\Delta v[/math]. In a gravity and friction free setting, although it takes it longer, a rocket with a lower-power motor reaches the same final velocity as one with a higher-power motor, if both motors have the same exhaust velocity, and the rockets’ empty and reaction masses and initial velocities are the same.

[enorbet2]

Okay, I've heard this argument that says that the fuel energy of a rocket is mv rather than 1/2 mv^2. I was wondering if this is what is happening. Maybe the first amount of fuel explodes and pushes the rocket. Another 1/2 mv^2 would only get the rocket to the squart root of 2 times the first velocity. But it would be true that the rest of the fuel would get an increase in kinetic energy from the first amount of fuel, and so the next liter of fuel coming out would have more kinetic energy per liter than the first amount of fuel. Then this would still result in 1 liter corresponding to 100 mph and 2 liters getting the rocket going to 200 mph. Is this how it would work out?

 

Hello Mr. Peterman

It might be helpful in answering your question if it was clearer what you mean by "fuel energy" since the amount of useful energy contained in a given mass of fuel depends greatly on how one utilizes it (efficiency). Not to be silly but actually to illustrate, if one were in a weightless, essentially frictionless environment (say launching from orbit) and was standing on the hull of a craft in a spacesuit and threw fuel in one direction that would indeed have an effect on you and the craft. Whether or not it would move would depend on your throwing arm and the inertia of the craft. Obviously that would be extremely inefficient, not to mention, tiring, since the burned fuel would be food in the thrower's body.

 

On the other end of the scale if the fuel mass were converted into thermal energy directed in a single direction in a manner releasing nuclear energy, with the most obvious winner being through the introduction of antimatter, the greatest efficiency and therefore thrust would result, assuming a working matter-antimatter containment and nozzle.

 

Furthermore I am confused regarding how you perceive the kinetic energy of the fuel increasing in a manner that increases thermal efficiency beyond having a preheated combustion chamber and there are physical limits to how much this gain can be.

 

So for the purposes of this discussion until you say differently I will assume you are referring to "fuel energy" as that of a chemical reaction, partly since the ion engine is the only successfully utilized loosely nuclear rocket as other successes have been confined to test stands.

 

Since CraigD has done an excellent job of explaining the proper formula to use for a calculable acceleration assuming lab perfect efficiency of chemical to thermal conversion, I would like to direct you to the realities of efficiency.

 

There are numerous mechanical systems efficiency to be considered in converting a high pressure - low velocity chemical reaction into a low pressure - high velocity reaction engine with usable thrust. Pump systems, fuel injection nozzles, combustion chamber geometry and composition, and particularly nozzle geometry are major influences on conversion efficiency. In solid fuel rockets the burn surface area (shaped charges are common) is important.

 

However beneath it all and utterly regardless of the fuel type, amazingly, it was in the mid 1820's that the definitive work was completed that describes how any and all heat engines function. Incidentally it even covers the gains possible from preheating as mentioned above. Meet the Carnot Cycle Carnot cycle - Wikipedia, the free encyclopedia

 

I am also unsure of what you imply by

1 liter corresponding to 100 mph and 2 liters getting the rocket going to 200 mph

 

since even in an automobile it can be seen that there are losses to figure, that it requires some fuel just to keep the reaction going, that some fuel will not burn usefully, etc. In any case it is possible the effect seen with multi-stage rockets may be what you mean in that a first stage starts from zero speed, while subsequent stages start from speed, and unlike a relay race, also have less mass to overcome and of course no energy benefits are passed with the baton - not additive. It is not true that a given mass of fuel will deliver or is even capable of delivering a substantial degree of increased energy simply because it is a late arrival to the combustion chamber. The only additive kinetic energy is exhaust velocity and that velocity is only additive until the vehicle reaches equilibrium, ie: the velocity of the exhaust. Burning more fuel will have no propulsive effect once that is achieved. This is why solar sails presently have the greatest potential for high speed limit, despite exceedingly low thrust.

 

If any of this fuels your interest or desire to narrow nomenclature or expand the question of your question I'm confident this could grow a healthy thread.

 

Thanks, this is fun

[Qfwfq]

Perhaps the meaning is tied to the fact that, for a given [imath]\Delta v[/imath] due to fuel, [imath]\Delta E[/imath] is given by [imath]\Delta\left(\frac12mv^2\right)=mv\Delta v[/imath]

[arkain101]

Also,

 

It's better to think of the velocity of the exhaust relative to the ship, and not earth or some other location when you are trying to work out or think about the "thrust" and "ship velocity" (relative to earth etc).

[enorbet2]

There seem to be some complications of which I was unaware. I am and was familiar with the Oberth Effect Oberth effect - Wikipedia, the free encyclopedia that governs the nature of propulsion near gravitational fields including so-called "gravity assist" or "slingshotting" but hadn't seen how the math looked.

 

The complication came form the link within to Propulsive efficiency - Wikipedia, the free encyclopedia where it states that

 

Rocket engines have a slightly different propulsive efficiency ηp than airbreathing jet engines as the lack of intake air changes the form of the equation somewhat. This also means that rockets are able to exceed their exhaust velocity..

 

Since this is wiki it is either mistaken or, more likely, that I just can't wrap my head around that yet despite the nice graph. Any assistance appreciated here if it is not considered OT.

[modest]

The complication came form the link within to Propulsive efficiency - Wikipedia, the free encyclopedia where it states that

 

Rocket engines have a slightly different propulsive efficiency ηp than airbreathing jet engines as the lack of intake air changes the form of the equation somewhat. This also means that rockets are able to exceed their exhaust velocity..

 

Since this is wiki it is either mistaken or, more likely, that I just can't wrap my head around that yet despite the nice graph. Any assistance appreciated here if it is not considered OT.

 

If you think of a jet going 100kph through the atmosphere and expelling its propellant at 100kph then no energy has been transferred to the vehicle. The air is motionless to the ground before and after going through the engine. Assuming the air itself comprises all of the propellant the jet gains no kinetic energy by catching it and ejecting it both at the same velocity.

 

If a rocket is going 100kph through the atmosphere and expels its propellant at 100kph then it has delivered it at maximum efficiency. The propellant started at 100 kph relative to the ground and was motionless to the ground after leaving the engine. The change in kinetic energy is transferred to the vehicle.

 

So the difference is that rockets carry their own propellant. This can be shown using the equations for thrust of an air-breathing jet and a rocket:

 

For a rocket:

[math]\mbox{Rocket Thrust} = F_R = mV+(P_e-P_0)A_e[/math]

The first term ([math]mV[/math]) is the mass of the propellant times the exhaust velocity which is a positive number regardless of the velocity of the vehicle. The second term ([math](P_e-P_0)A_e[/math]) is the pressure thrust term and is zero at peak efficiency.

 

The air-breathing jet on the other hand is:

[math]\mbox{Jet Thrust} = F_J = mV - m_0V_0 + (P_e-P_0)A_e[/math]

The first term ([math]mV[/math]) is the mass of the propellant times the exhaust velocity relative to the vehicle. Subtracted from that is the second term ([math]m_0V_0[/math]) which is the mass of the incoming air times the velocity of the incoming air (relative to the vehicle). Assuming that the two masses are the same this will subtract to zero when the vehicle is going the speed of the exhaust velocity. The last term can again be assumed zero in which case the jet achieves no thrust when it is traveling the speed of its exhaust velocity.

 

These equations are given here for jets and here for rockets.

 

~modest

 

P.S. This probably is a little OT, but an excellent question.

[CraigD]

There seem to be some complications of which I was unaware. I am and was familiar with the Oberth Effect Oberth effect - Wikipedia, the free encyclopedia that governs the nature of propulsion near gravitational fields including so-called "gravity assist" or "slingshotting" but hadn't seen how the math looked.

One of the best non-technical explanations of the Oberth effect I’ve encountered – better than Landis’s, which is linked from the wikipedia article – appeared in Clarke’s 1982 novel 2010. Paraphrasing, it goes like this:

The best place to “lighten your load” – expend reaction mass – is at toward the bottom of a gravity well, because the kinetic energy converted to gravitational potential energy when you climb out of it is proportional to your total mass.

 

Another way to imagine this (not mentioned in Clarke’s novel) is to consider the kinetic energy, relative to an observer on the planet you’re dropping toward then away from, of the reaction mass (exhaust) you’re expelling. As you’re expelling it a direction opposite of that you’re traveling, its velocity, and thus its kinetic energy, will be less when you’re traveling at a greater speed, at the low-point of your pass at the planet, than when you’re speed is lower, before or after that point. As energy is conserved, this means more of your system’s kinetic energy is in the form of the velocity of your ship, less in the velocity of your exhaust.

 

It’s important to note that the Oberth effect is not the same thing as a “gravity assist/slingshot swing-by”. A swing-by doesn’t require any use of a rocket motor, or a change in magnitude of the spaceship’s velocity (speed) relative to the slingshotting planet. Rather, only the direction of the ship’s velocity is changed. Since the slingshotting planet has a velocity relative to the ships intended target (another planet, etc), however, it’s direction and speed relative to that target is changed.

The complication came form the link within to Propulsive efficiency - Wikipedia, the free encyclopedia where it states that

Rocket engines have a slightly different propulsive efficiency ηp than airbreathing jet engines as the lack of intake air changes the form of the equation somewhat. This also means that rockets are able to exceed their exhaust velocity..

Since this is wiki it is either mistaken or, more likely, that I just can't wrap my head around that yet despite the nice graph.

The wiki’s correct.

 

An helpful way to wrap your head around this one is to consider a simplified “rocket” consisting of you, holding a softball while standing on a stationary skateboard on the smooth floor of a truck traveling down a smooth, straight highway. You and the skateboard mass 70 kg, the softball 0.2 kg. The truck is traveling 20.95 m/s. You throw the softball toward the back of the truck at 21 m/s (your “exhaust velocity”), which per conservation of momentum causes you and the skateboard to begin rolling toward the front of the truck at 0.06 m/s ([math]21 \mbox{m/s} \cdot \frac{0.2 \mbox{kg}}{70 \mbox{kg}}[/math]). From the perspective of someone watching from the side of the road, your speed increased from 20.95 m/s to 21.01 m/s. You’ve just “exceeded your exhaust velocity”.

[enorbet2]

Thanks to Modest and CraigD for helping to make the concept of exceeding one's exhaust velocity more intuitive. I had made a mistake of plugging simple, small, whole numbers into the formula, for example in Delta v = v_text{e} ln frac {m_0} {m_1} assuming initial (fully loaded) mass, m_0 = 2, and fuel expended mass m_1 = 1 to "see" the formula in action. It is at very high ratios however when this effect of exceeding exhaust velocity is more noticeable. (See: Brussard Ramjet)

 

I was also a little hung up with my concern with the actual engineering in that at very low ratios as with some relatively miniscule thrust engines can have much of their energy lost as heat in merely flexing the hull or hold down clamps in the case of an engine test stand where the mass is essentially that of the Earth.

 

Damn. I love this place! :)

[Qfwfq]

I am and was familiar with the Oberth Effect Oberth effect - Wikipedia, the free encyclopedia that governs the nature of propulsion near gravitational fields including so-called "gravity assist" or "slingshotting" but hadn't seen how the math looked.

The slingshot effect is a distinct thing, which doesn't require the rocket's thrust. Of course, you might ask where the energy comes from, without the fuel.

 

If you think of a jet going 100kph through the atmosphere and expelling its propellant at 100kph then no energy has been transferred to the vehicle. The air is motionless to the ground before and after going through the engine. Assuming the air itself comprises all of the propellant the jet gains no kinetic energy by catching it and ejecting it both at the same velocity.

Well, not quite, The mass of the fuel is added, so [imath]\Delta p[/imath] isn't zero. Of course efficiency decreases dramatically as you near exhaust velocity (according to the ratio of fuel mass to air mass) so cruising speeds are wisely less than it, especially to commercial purposes. In principle, however, you can exceed it a little, until the fuel's [imath]\Delta p[/imath] is offset by that of air in the forward direction (plus aerodynamic resistance).

[enorbet2]

Just a note since I have already enjoyed how this thread has evolved with differing perspectives adding to the thought-visuals in my head and my understanding of Oberth's excellent concept. I'm sure even anyone unfamiliar with the Laws of Motion can get a handle now. Please note emboldened text and then respond and elucidate if that is an improper association.

 

There seem to be some complications of which I was unaware. I am and was familiar with the Oberth Effect Oberth effect - Wikipedia, the free encyclopedia that governs the nature of propulsion near gravitational fields including so-called "gravity assist" or "slingshotting" but hadn't seen how the math looked.

[CraigD]

The slingshot effect is a distinct thing, which doesn't require the rocket's thrust.

Very true. The majority of gravity slingshotting goes on continuously and undetected, as the greater and lesser planets change the orbits of countless small, natural bodies – in some cases enough to eject them completely from the Solar System.

Of course, you might ask where the energy comes from, without the fuel. :)

Bonus points to anyone who doesn’t already know the answer to this question who can figure it out without resorting to wiki/google/etc-ing it!

 

It’s one of those answers that seems glaringly obvious once you’ve seen it – but has profound consequences on the long-term evolution of the Solar System, and opens the door to a host of still unanswered astrophysical questions having little to do with rockets.

[Ben]

 

Bonus points to anyone who doesn’t already know the answer to this question who can figure it out without resorting to wiki/google/etc-ing it!

How's this for a shot at it - hand on heart, 'tis mine own.

 

First place an object in an orbit around our planet at sufficient distance that the planet's gravitational field is "felt" by our object. Then there will be an optimal angular velocity that keeps our object in some stable orbit around our planet.

 

Then I am guessing, but for any other angular velocity, in the same orbit, the following scenarios, and only these, must apply; our object spirals back to our planet (insufficient angular velocity) or our object spirals away from our planet (excess angular velocity)

 

It the latter case, all that is required is that our orbiting object be initially boosted to the required angular velocity, and thereafter no further boost i.e. no further fuel energy is required to escape the planet's gravitational field.

 

Of course the "direction", if this term makes sense, of our object as it escapes is not predicted by my scenario, but I don't doubt that this has been calculated by NASA (or even you clever guys)

[modest]

If you think of a jet going 100kph through the atmosphere and expelling its propellant at 100kph then no energy has been transferred to the vehicle. The air is motionless to the ground before and after going through the engine. Assuming the air itself comprises all of the propellant the jet gains no kinetic energy by catching it and ejecting it both at the same velocity.

Well, not quite, The mass of the fuel is added, so [imath]\Delta p[/imath] isn't zero.

 

Correct. If you follow the first link of my post it says:

Most operate at low fuel-air ratios, and some of the high-pressure air is bled off to fan the auxiliaries. Thus we can assume that the flow rates m

₂

and m

₀

are approximately equal

This allows a simplification of the thrust equation, but is only an approximation. The text continues:

This form of the thrust equation reveals an interesting characteristic of all air-breathing propulsion systems. As their flight speed approaches the exhaust velocity, the thrust goes to zero. Even long before reaching this point, the thrust drops below the drag force (which is increasing rapidly with flight speed). Because of this, no air-breathing propulsion system can ever fly faster than its exit jet.

Of course efficiency decreases dramatically as you near exhaust velocity (according to the ratio of fuel mass to air mass) so cruising speeds are wisely less than it, especially to commercial purposes.

I don't think M₀ = M₂ is that bad of an approximation where it should come into play when considering the efficiency of a jet engine. Efficiency is at it's greatest when the speed of the plane is nearest the exhaust velocity as possible. This is revealed in the propulsive efficiency equation given by Enorbet's wiki page:

[math]\eta_p = \frac{2}{1 + \frac{c}{v}}[/math]

 

Max efficiency is at v=c where c is the exhaust speed and v is the speed of the airplane. This makes sense because moving a large amount of air very little is more efficient than moving a small amount of air a lot. What that equation does not reveal is that the maximum thrust power is greatest where c = 2v, where the exhaust velocity is twice the intake velocity. The book I've been quoting from comments:

This shows that the propulsive efficiency for air-breathers continually increases with flight speed, reaching a maximum where v=1 (or when V

₀

= V

₂

). This is quite reasonable since under this condition the absolute velocity of the exit jet is zero and there is no exit loss [see equation (12.49)].

 

At this point you can begin to see some of the problems involved in optimizing air-breathing jet propulsion systems. We showed previously that maximum thrust power is attained when V

₂

= 2V

₀

. Now we see that maximum propulsive efficiency is attained when V

₂

= V

₀

, but unfortunately, for the latter case the thrust is zero.

But, getting back to your point: Jet engines have a very high exhaust velocity by design. Since efficiency increases as the plane nears that exhaust velocity they run more efficiently as their speed increases.

Remember that the relations in this section apply only to air-breathing propulsion systems. Equation (12.59) further confirms the natural operating speed range of the various turbojet engines. Recall that a pure jet provides a large velocity change to a relatively small mass of air. Thus, as stated earlier, to have a high propulsive efficiency (ν → 1) it must fly at high speeds.

 

~modest

[Janus]

How's this for a shot at it - hand on heart, 'tis mine own.

 

First place an object in an orbit around our planet at sufficient distance that the planet's gravitational field is "felt" by our object. Then there will be an optimal angular velocity that keeps our object in some stable orbit around our planet.

 

There is no one "optimal" velocity for a stable orbit for a given distance from the planet. There is a range of velocities that will work. The velocity must be less than escape velocity for the distance and large enough that the periapis of the orbit doesn't graze the planet.

 

Then I am guessing, but for any other angular velocity, in the same orbit, the following scenarios, and only these, must apply; our object spirals back to our planet (insufficient angular velocity) or our object spirals away from our planet (excess angular velocity)

 

[/i]If you increase the velocity of an orbiting object it just climbs into an orbit with a higher mean distance from the planet(up the the limit mentioned). If you decrease it, it will drop to a lower orbit. The only way to make it spiral away or in is for there to be a constant force acting on the object.

 

It the latter case, all that is required is that our orbiting object be initially boosted to the required angular velocity, and thereafter no further boost i.e. no further fuel energy is required to escape the planet's gravitational field.

 

See above, Unless the boost is in itself enough to get the object to escape velocity, it will only climb to a higher orbit. If it is enough, it will not spiral out, but will follow a parabola away from the planet.

 

A gravity assist behaves nothing like this, and more akin to an elastic collision.

[Jay-qu]

If it is enough, it will not spiral out, but will follow a parabola away from the planet.

 

A parabola or a hyperbola? I thought non bound objects follow a hyperbola - is this a case were the trajectory becomes a parabola?