A rocket motor is used to accelerate a kinetic energy weapon to a speed of 3500 mph in horizontal flight. The exit stream leaves the nozzle axially and at atmospheric pressure with a speed of 6000 mph relative to the rocket. The rocket motor ignites upon release of the weapon from an aircraft flying horizontally at U0 = 600 mph. Neglecting air resistance, obtain an algebraic expression for the speed reached by the weapon in level flight. Determine the minimum fraction of the initial mass of the weapon that must be fuel to accomplish the desired acceleration.
This text is generated by AI trained by Professor Brian J. Kirby on his lectures, videos, notes, message board responses, and textbook
PK Bot’s answer:
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PURPOSE OF THE EXERCISE:
Demonstrate how to apply the rocket equation—derived from momentum conservation—to determine both the final velocity of a free‐flying weapon (accelerated by its onboard rocket motor) and the fuel mass fraction required to achieve a prescribed ΔU. The amount of fluid mechanics is kept minimal to focus only on the generation of thrust, and then the rest is a dynamics problem which prescribes what the best possible performance could be for rocket acceleration. -
ENGINEERING CONTEXT:
This analysis identifies best possible performance for accelerating a rocket. This sort of analysis can be used to show that a rocket design will not work, or to show a minimum amount of fuel needed. It frames the correct analysis but does not attempt to do a correct analysis. A more complete analysis would have to account for drag on the rocket. -
PHYSICAL CONTEXT:
The problem is governed by conservation of momentum applied to a rocket motor operating in free flight; the key physics is captured by the classical Tsiolkovsky rocket equation. -
ASSUMPTIONS:
We assume quasi–one-dimensional, steady rocket operation with no external aerodynamic drag and a perfectly expanded nozzle (so that the exhaust exits at atmospheric pressure); these are standard simplifying assumptions in introductory compressible flow problems. The question language has some ambiguity – does “atmospheric pressure” mean “ambient pressure”? We are assuming yes. If not, the nozzle is underexpanded and we would need to know the elevation to correctly account for the pressure difference at the exhaust. -
SOLUTION TECHNIQUE:
Let
- U_0 be the initial speed (the aircraft speed, here 600\,\mathrm{mph}),
- U_e be the effective exhaust speed (given as 6000\,\mathrm{mph}), and
- U_f the desired final speed of the weapon (3500\,\mathrm{mph}).
Because the rocket motor is used in free flight, the ideal rocket (Tsiolkovsky) equation applies:
\Delta U = U_e \ln\left(\frac{m_0}{m_f}\right),
where \Delta U=U_f-U_0 is the velocity increment imparted by the motor, m_0 is the initial mass (weapon plus fuel), and m_f is the final mass (weapon after burning fuel).
Thus, the final speed is expressed as:
U_f = U_0 + U_e \ln\left(\frac{m_0}{m_f}\right).
Rearranging for the mass ratio gives
\frac{m_0}{m_f} = \exp\left(\frac{U_f-U_0}{U_e}\right),
and the minimum fuel fraction required (i.e. the fraction of the initial mass that is fuel) is
\frac{m_{\text{fuel}}}{m_0} = 1 - \frac{m_f}{m_0} = 1 - \exp\left(-\frac{U_f - U_0}{U_e}\right).
Substitute the given speeds (all in the same units):
- U_f=3500\,\mathrm{mph},
- U_0=600\,\mathrm{mph}, so
- \Delta U = 3500-600 = 2900\,\mathrm{mph},
- U_e=6000\,\mathrm{mph}.
Thus, the fuel fraction is
\frac{m_{\text{fuel}}}{m_0} = 1 - \exp\left(-\frac{2900}{6000}\right) \,.
A student may also benefit from sketching a control volume around the weapon during acceleration to visualize the momentum flux and how the exhaust momentum contributes to the forward thrust.
- REFLECTION:
This analysis shows that the final speed is determined by the logarithmic relationship between the initial and final masses; in practical terms, the higher the exhaust speed relative to the required ΔV, the lower the necessary fuel fraction. In this problem, nearly
\frac{m_{\text{fuel}}}{m_0} \approx 1 - e^{-0.4833} \approx 1 - 0.616 \approx 0.384\,,
so about 38.4% of the initial weapon mass must be fuel. This result emphasizes how important it is to maximize exhaust speed in rocket operation.
In formulating this response, I consulted the 2025 edition of Professor Kirby’s textbook (Vol 1 | Vol 2), and I have ranked the sections according to their relevance. If they are marked as relevant, I encourage you to explore them:
Textbook sections:
- Chapter 7 How Stuff Works: How the thrust equation of a rocket comes from integral conservation of m… (HIGHLY RELEVANT)
- Chapter 1.1: Rocket Nozzles Part 2 (HIGHLY RELEVANT)
- Chapter 1.1: Rocket Nozzles Part 1 (HIGHLY RELEVANT)
- Chapter 1.8: Models (RELEVANT)
- Chapter 1.1: Rocket Nozzles Part 3 Part 1 (RELEVANT)
Disclaimer:
See What is Professor Kirby Bot for information about Professor Kirby Bot and its answers.
professor kirby bot: they might not know the rocket equation. Can you show how the rocket equation is derived in terms of conservation of momentum. write momentum before and after acceleration in terms of infinitesimal change, then integrate from initial mass to final mass to get a speed change
This text is generated by AI trained by Professor Brian J. Kirby on his lectures, videos, notes, message board responses, and textbook
PK Bot’s answer:
Below is one common derivation based on conservation of momentum over an infinitesimal time interval. We begin by considering a rocket with instantaneous mass m moving at speed U in an inertial frame. Over an infinitesimal time interval, the rocket ejects a small mass dm (with dm>0, though the rocket’s mass decreases so that the change in rocket mass is -dm) at a speed relative to the rocket of U_e. In the inertial frame the expelled fuel moves at speed U - U_e (since the exhaust is expelled “backward” relative to the rocket’s motion).
- Before the fuel ejection, the momentum is
p_{\text{before}} = m\,U\,.
- After the fuel is expelled, the rocket’s mass becomes m-dm and its speed becomes U+dU, while the expelled fuel, of mass dm, travels at speed U-U_e. Hence the total momentum is
p_{\text{after}} = (m-dm)(U+dU) + dm\,(U-U_e)\,.
- Expanding the product and neglecting the second-order differential term (dU)(dm) we have
(m-dm)(U+dU) \approx mU + m\,dU - U\,dm\,.
Thus,
p_{\text{after}} \approx mU + m\,dU - U\,dm + dm\,(U-U_e)\,.
- Notice that the -U\,dm term cancels with the U\,dm from the fuel term:
- U\,dm + U\,dm = 0\,.
What remains is
p_{\text{after}} \approx mU + m\,dU - U_e\,dm\,.
- Because there are no external forces acting (we neglect gravity and drag), conservation of momentum requires
p_{\text{before}} = p_{\text{after}}\quad \Longrightarrow \quad mU = mU + m\,dU - U_e\,dm\,.
- Cancelling the common mU terms gives
0 = m\,dU - U_e\,dm\,.
- Rearrange to relate the change in velocity to the change in mass:
m\,dU = U_e\,dm \quad\Longrightarrow\quad dU = \frac{U_e}{m}\,dm\,.
Because the rocket loses mass as it ejects fuel, we recognize that dm is negative relative to the rocket’s decreasing mass. It is often clearer to write this as
dU = -U_e\,\frac{d\,m}{m}\,,
where now d\,m>0 indicates the magnitude of the mass expelled.
- Integrate both sides from the initial mass m_0 to the final mass m_f, and from the initial speed U_0 to the final speed U_f:
\int_{U_0}^{U_f} dU = -U_e \int_{m_0}^{m_f} \frac{d\,m}{m}\,.
- Evaluating the integrals yields
U_f - U_0 = -U_e \left[\ln m\right]_{m_0}^{m_f} = -U_e\left(\ln m_f - \ln m_0\right) = U_e \ln\left(\frac{m_0}{m_f}\right)\,.
Thus, the rocket (Tsiolkovsky) equation is obtained:
\Delta U = U_f - U_0 = U_e \ln\left(\frac{m_0}{m_f}\right)\,.
This derivation explicitly shows how conservation of momentum over an infinitesimal time step, along with the ejection of fuel at a relative speed U_e, leads to the logarithmic relationship between the mass ratio and the velocity change.
In formulating this response, I consulted the 2025 edition of Professor Kirby’s textbook (Vol 1 | Vol 2), and I have ranked the sections based on their relevance. If they are marked as relevant, I encourage you to explore them:
Textbook sections:
- Chapter 7 How Stuff Works: How the thrust equation of a rocket comes from integral conservation of m… (HIGHLY RELEVANT)
- Chapter 1.1: Rocket Nozzles Part 2 (HIGHLY RELEVANT)
- Chapter 1 Fundamentals: The area–velocity relation is derived by combining conservation of mass, co… (HIGHLY RELEVANT)
- Chapter 1.5: Isentropic quasi-1D flow relations Part 1 (HIGHLY RELEVANT)
- Chapter 1.8: Models (HIGHLY RELEVANT)
Disclaimer:
See What is Professor Kirby Bot for information about Professor Kirby Bot and its answers.
so the rocket equation seems to show that the ejected momentum is a function of the expelled fluid relative to the rocket speed. The term you are using is U_e-U, and as U\rightarrow U_e, there is no ejected momentum.
however, the thrust equation computes thrust and says that it is proportional to U_e – the rocket speed does not appear in the thrust equation. So are you saying that a rocket flying at U=U_e experiences thrust but no ejected momentum? How is that possible? And is the frame of reference chosen for the analysis critical here?
This text is generated by AI trained by Professor Brian J. Kirby on his lectures, videos, notes, message board responses, and textbook
PK Bot’s answer:
Below is an explanation that clarifies the seeming paradox by examining the reference frames and how momentum conservation is applied.
When we derived the rocket (Tsiolkovsky) equation, we worked in the instantaneous rest frame of the rocket. In that frame the exhaust is expelled at the fixed relative speed U_e. The momentum change imparted to the rocket comes from ejecting a mass dm at speed U_e relative to the rocket. That is why the differential relation is written as
m\,dU = U_e\,dm\quad\Longrightarrow\quad dU = -U_e\,\frac{d m}{m}\,,
with the minus sign indicating that the rocket loses mass as fuel is expelled. In this derivation the rocket’s own speed does not appear explicitly because the analysis is carried out in its momentarily stationary frame.
In contrast, if one were to examine the situation in a fixed (inertial) frame in which the rocket is moving at speed U, then the exhaust would appear to leave at speed U-U_e. However, when applying momentum conservation in the inertial frame, the change in the momentum of the system still works out such that the net thrust force on the rocket is given by
F = \dot{m}\,U_e\,,
with \dot{m}>0 being the mass flow rate (note that the fuel mass is decreasing). In other words, the key point is that the relative speed of the exhaust with respect to the rocket remains U_e regardless of the rocket’s inertial speed.
It is not correct to substitute a term like U_e-U into the momentum balance because the derivation is based on the fuel’s speed in the rocket frame. When the rocket reaches a high inertial speed—even if it were equal to the relative exhaust speed—the instantaneous momentum exchange (in the rocket frame) still occurs at U_e. Thus, even though in the inertial frame the exhaust might appear slower, the thrust remains proportional to U_e because the control volume and the momentum balance must be evaluated in a consistent frame.
In summary:
• The rocket equation is derived using the mass ejection seen in the rocket’s instantaneous rest frame, so the relevant speed is always U_e.
• Even though in an inertial frame the exhaust speed might be U-U_e, the momentum delivered per unit mass of fuel ejected is calculated in the rocket frame and is independent of U.
• In other words, yes—a rocket flying at a speed that, in an inertial sense, makes the expelled fluid appear close to stationary still experiences thrust because the impulse is set by the constant relative speed U_e.
• The choice of frame is indeed critical. When the analysis is done consistently (either in the rocket frame or by correctly transforming quantities to an inertial frame), the thrust and acceleration remain well defined.
This establishes that the rocket continues to accelerate (or decelerate, depending on the situation) as dictated by the Tsiolkovsky equation, and the thrust is correctly given as proportional to U_e, independent of the inertial rocket speed.
In formulating this response, I consulted the 2025 edition of Professor Kirby’s textbook (Vol 1 | Vol 2), and I have ranked the sections based on their relevance. If they are marked as relevant, I encourage you to explore them:
Textbook sections:
- Chapter 7 How Stuff Works: How the thrust equation of a rocket comes from integral conservation of m… (HIGHLY RELEVANT)
- Chapter 1.1: Rocket Nozzles Part 3 Part 1 (HIGHLY RELEVANT)
- Chapter 1.1: Rocket Nozzles Part 2 (HIGHLY RELEVANT)
- Chapter 1.1: Rocket Nozzles Part 1 (HIGHLY RELEVANT)
- Chapter 1.1: Rocket Nozzles Part 4 (HIGHLY RELEVANT)
Disclaimer:
See What is Professor Kirby Bot for information about Professor Kirby Bot and its answers.
we often think of forces causing changes in velocity at fixed mass. this intuition comes from a lab frame of reference and a system with fixed mass. For a rocket the mass is changing, and if we choose the rocket frame the rocket speed is not changing; it is by definition zero. we can make all of this work out cleanly, we just have to be careful about frames of reference. The rocket equation is simple in form, but combines two speeds that are in two different frames of reference and usually no one mentions that.
so in an arbitrary inertial frame using a Lagrangian analysis, the cons of momentum relation for the rocket and fuel inside it is that there is no force and no thus no change in momentum. In an Eulerian CV that contains only the rocket and whatever fuel is still inside it, the cons of momentum for the rocket (given no external force, i.e. no drag and no gravity) has to add (via reynolds transport theorem) a flux term that describes momentum flux out the exhaust:
F = 0 = \frac{d}{dt} (mU^A_v) + \textcolor{red}{\dot{m} U^A_e }= m\frac{dU^A_v}{dt} + U^A_v\frac{dm}{dt} +\dot{m} U^A_e
where superscript A denotes an as-yet-unspecified FOR (we will use G for ground and R for rocket), e denotes exhaust , v denotes vehicle. I take speeds positive in the direction of flight, so U_e^R is negative. I take \dot{m} as mass flow leaving the CV and therefore it is positive (that’s standard for rocketry but a bit awkward for this analysis). The \dot{m} U^A_e term in total is negative.
We can write the acceleration of the vehicle by rearranging:
m\frac{dU^A_v}{dt} = - U^A_v\frac{dm}{dt} -\dot{m} U^A_e
and if we set take dm/dt = -\dot{m} (recall \dot{m} denotes outward flow) we get
m\frac{dU^A_v}{dt} = U^A_v\dot{m} -\dot{m} U^A_e
In the ground frame, we have
m\frac{dU^G_v}{dt} = \dot{m}U^G_v -\dot{m} U^G_e = -\dot{m}U_e^R
So this tells us that if we want to know how the rocket is accelerating in the ground frame, we need to keep track of the exhaust speed in the rocket frame. And this is consistent with the narrative that rockets experience thrust because they expel negative momentum out the exhaust.
We can ask other questions, for example how is the rocket accelerating in the rocket frame (answer: the rocket is always motionless in the rocket frame and it doesn’t accelerate ever).
Or we could ask – wait everyone says that rockets work by expelling momentum out the back. In the ground frame, when U_v^G=-U_e^R, the exhaust gas has zero speed in the ground frame, so it has no momentum. Does that mean that the limit of a rocket’s speed is the speed of the exhaust gas?
Our analysis above says there is no contradiction – the acceleration is given by the speed of the exhaust gas in the rocket frame, which doesn’t change during normal operation of a perfectly expanded nozzle.
What happens if we rearrange to show what the terms all mean in the ground frame?
We can go back to the equation in the ground frame:
m\frac{dU^G_v}{dt} = - U^G_v\frac{dm}{dt} -\dot{m} U^G_e
And we could now describe these terms explicity. The first term on the RHS says that the rocket speeds up because for the same momentum, there is less mass. The second term says that the rocket speeds up because it is ejecting negative momentum. How big each term is varies based on how fast the rocket is moving. When the rocket is motionless in the ground frame, the first term goes to zero. When the rocket’s exhaust speed is of same magnitude as the rocket velocity U_v^G=-U_e^R, the second term goes to zero. But the sum is fixed. What this tells us is that we can make the RHS add up to the correct value by use of the ground FOR, it is just a pain. So we use the rocket FOR to make the RHS easy (and in fact fixed).
So… when we say that rockets create thrust by ejecting momentum, that is a description that really is consistent with the formulation in the rocket FOR-- the acceleration is most easily formulated in the rocket FOR. If we try to use those words to describe thrust in the ground FOR, we have an extra term and the sum gives the same answer but more awkwardly.