What are the key concepts in rocket dynamics?

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This tutorial provides an elementary development of the key concepts
that describe the behavior of rocket propulsion systems.

The following concepts will be discussed in this tutorial

• Characteristics of a rocket engine
  This post characterizes a rocket engine.

• Equations of motion
  The next post uses these charactistics along with conservation of mass and momentum to develop the equations of motion of a rocket.

• The Tsiolokovsky rocket equation
  The third post in this series develops the Tsiolokovsky rocket equation. This equation explains why it took one of the most powerful machines ever built, the Saturn V rocket, to get a tiny vehicle to the Moon.

• Energy concepts
  The forth post in this series examines the energy involved in making a rocket accelerate. Rockets function by converting some form of potential energy (usually chemical) into kinetic energy.

Characteristics of a rocket engine
Rockets accelerate by ejecting mass at speed from the vehicle. Being an introductory level tutorial, only two parameters suffice to describe this ejected material. These parameters are

$$\dot m_e(t)$$ -- The exhaust mass flow rate and
$$\vec v_e(t)$$ -- The exhaust velocity relative to the rocket.

The exhaust mass flow rate is the rate at which the cloud of exhaust gas left by the vehicle gains mass. Note that this quantity is always positive when the rocket engine is firing and is of course zero when the rocket is quiescent. The suffix $e$ in $\dot m_e$ and $\vec v_e$ denotes the exhaust. There are two other items of interest -- the fuel and the rocket itself. The parameters that describe these other two items are

$$m_f(t)$$ -- The quantity of fuel remaining in the rocket,
$$m_v\;\,$$ -- The dry mass of the vehicle (including the empty fuel tanks),
$$m_r(t)$$ -- The mass of the rocket as a whole (dry mass + fuel mass), and
$$\vec v_r(t)$$ -- The velocity of the rocket expressed in some inertial reference frame.

The rate at which the fuel and vehicle mass changes is simply the additive inverse of the exhaust mass flow rate.

Specific impulse
Rocket designers and analysts often use specific impulse to characterize a rocket engine rather than exhaust speed. Specific impulse has units of time. There is a simple relationship between specific impulse and effective exhaust velocity:

$$I_{sp} = \frac{v_e}{g_0}$$

where $g_0$ is standard gravity, the acceleration due to Earth's gravity at sea level, or $9.80665 \mathrm{m} / \mathrm{s}^2$. This tutorial uses the exhaust velocity rather than specific impulse to make the connection to momentum more obvious. For example, the Shuttle main engines have a specific impulse of 453 seconds when operated in vacuum. This translates to an effective exhaust velocity of 4500 meters/second.

Further reading
This tutorial does not provide any of the underlying details that dictate how combustion of the fuel leads to a high velocity exhaust. A good place to start is the The Beginner's Guide to Rockets, http://exploration.grc.nasa.gov/education/rocket/index.html.

 

[D H]

Rocket equations of motion

This post develops the equations of motion for a simple rocket.

Simplifying assumptions
In a real rocket, the center of mass changes as the rocket burns fuel. Moreover, any thrust that is not directed through the center of mass will result in a torque as well as a force on the vehicle. In this tutorial the rocket is treated as a point mass. This simplifying assumption eliminates concerns regarding the motion of the center of mass within the vehicle and concerns of rotational behavior.

Gravity and other forces such as atmospheric drag contribute to a rocket's motion. All of these other forces are ignored in this tutorial. Imagine the rocket is somewhere deep in space, far from any massive object, so that there are no measurable external forces acting on the vehicle. As forces are additive, the extraneous forces can later be factored into the overall equations of motion to develop a more realistic description of a rocket's motion.

Conservation principles
For this derivation, I will use some math that mathematicians don't particular like but physicists use willy-nilly -- things $\Delta v$. Things can be done more formally using continuum physics (classical treatment of gases), but such a development is not appropriate for an introductory level tutorial.

Over a small time interval $\Delta t$, the rocket will eject a small mass of exhaust $\dot m_e(t)\,\Delta t$. At the start of the interval, the rocket has mass $m_r(t)$, velocity $\vec v_r(t)$, and momentum $m_r(t) \, \vec v_r(t)$. At the end of the interval, the rocket has mass, velocity, and linear momentum

$$m_r(t+\Delta t) = m_r(t)-\dot m_e(t)\,\Delta t$$
$$\vec v_r(t+\Delta t) = \vec v_r(t)+\Delta \vec v_r(t)$$
$$\vec p_r(t+\Delta t) = (m_r(t) -\dot m_e(t)\,\Delta t)\, (\vec v_r(t)+\Delta \vec v_r(t))$$

The bit of ejected fuel carries some momentum from the vehicle. The mass, inertial observer velocity, and momentum of the exhaust are

$$\Delta m_e(t) = \dot m_e(t)\,\Delta t$$
$$\vec v_{e_{\mathrm{inertial}}}(t) = \vec v_r(t)+\vec v_e(t)$$
$$\Delta \vec p_e(t+\Delta t) = \dot m_e(t)\,\Delta t\, (\vec v_r(t)+\vec v_e(t))$$

The momentum of the rocket and the ejected exhaust at the end of the time interval is thus
$$\vec p_{r+e}(t+\Delta t) = (m_r(t) -\dot m_e(t)\,\Delta t)\, (\vec v_r(t)+\Delta \vec v_r(t)) + \dot m_e(t)\,\Delta t\, (\vec v_r(t)+\vec v_e(t))$$

Dropping the second-order term $\Delta t \Delta \vec v_r(t)$ and simplifying,

$$\vec p_{r+e}(t+\Delta t) = m_r(t)\,\vec v_r(t)+ m_r(t) \, \Delta \vec v_r(t) + \dot m_e(t)\,\Delta t\, \vec v_e(t)$$

The rocket and the ejected fuel form a closed system by the isolated rocket assumption. Momentum is conserved in a closed system, so $\vec p_{r+e}(t+\Delta t)=\vec p_r(t)$:

$$m_r(t)\,\vec v_r(t)+ m_r(t) \, \Delta \vec v_r(t) + \dot m_e(t)\,\Delta t\, \vec v_e(t) = m_r(t)\,\vec v_r(t)$$

or

$$m_r(t) \, \Delta \vec v_r(t) + \dot m_e(t)\,\Delta t\, \vec v_e(t) = 0$$

Dividing by $\Delta t$ and taking the limit $\Delta t \to 0$,

$$\frac {d\vec v_r(t)}{dt} = - \, \frac {\dot m_e(t)} {m_r(t)} \, \vec v_e(t)$$

This is the equation for the acceleration of the rocket at time $t$.

 

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Tsiolkovsky rocket equation

This post expands on the simple equations of motion developed in the previous post. The result of this analysis is the Tsiolkovsky rocket equation, aka the ideal rocket equation, or just the rocket equation, first published in 1903 by Konstantin Tsiolkovsky.

The rocket equation
From the previous post, the acceleration of the simplified rocket is

$$\frac {d\vec v_r(t)}{dt} = - \, \frac {\dot m_e(t)} {m_r(t)} \, \vec v_e(t)$$

Integrating the left-hand side of the above equation, yields the change in velocity over some period of time:

$$\int_{t_0}^{t_1} \frac {d\vec v_r(t)}{dt} dt = \vec v_r(t_1) - \vec v_r(t_0)$$

Integrating the right hand side will yield an expression that gives this change in velocity:

$$\vec v_r(t_1) - \vec v_r(t_0) = - \int_{t_0}^{t_1} \frac {\dot m_e(t)} {m_r(t)} \, \vec v_e(t) dt$$

By conservation of mass, the time derivative of the rocket's mass is just the additive inverse of the exhaust mass flow rate: $\dot m_r(t) = -\dot m_e(t)$. The right-hand side is integrable if the relative exhaust velocity vector has both constant magnitude and constant direction.

$$\vec v_r(t_1) - \vec v_r(t_0) = \vec v_e \int_{t_0}^{t_1} \frac {\dot m_r(t)} {m_r(t)} \, dt = \vec v_e \ln\left(\frac{m_r(t_1)}{m_r(t_0)}\right)$$

Denoting the change in velocity as $\Delta \vec v_r$ and choosing a unit vector $\hat v$ aligned with this vector,

$$\Delta v_r \hat v = - v_e \ln\left(\frac{m_r(t_1)}{m_r(t_0)}\right) \hat v$$

The negative sign appears because the delta-v and exhaust velocity vectors are anti-aligned. This is a one-dimensional problem, so there is no need for vector notation so long as sign conventions are obeyed. Getting rid of the unit vector and exchanging the numerator and denominator to eliminate the negative sign yields the Tsiolokovsky rocket equation,

$$\Delta v_r = v_e \ln\left(\frac{m_r(t_0)}{m_r(t_1)}\right)$$

Analysis
Suppose the rocket initially loaded with some quantity of fuel $m_f$ fires its engines until all of the fuel is consumed. The mass of the vehicle at the end of the burn is just the dry mass, $m_v$. Applying the rocket equation to determine the change in the rocket's velocity,

$$\Delta v_r = v_e \ln\left(\frac{m_v+m_f}{m_v}\right)$$

Rockets are loaded with the amount of fuel needed to achieve some desired delta-v. Solving the above for the fuel mass required to achieve this delta-v,

$$m_f = m_v (e^{\frac{\Delta v_r}{v_e}}-1)$$

There is no limit to the attainable delta-v¹. The rocket can end up going a lot faster than the exhaust velocity. It just takes a large expenditure of fuel.

A related item of interest is the fuel mass ratio, the ratio of the fuel mass to the total vehicle mass $m_f/m_r= m_f/(m_f+m_v)$:

$$\frac{m_f}{m_v+m_f} = 1 - e^{-\,\frac{\Delta v_r}{v_e}}$$

Suppose we want to use a single-stage rocket with an effective exhaust velocity of 4.5 km/sec to launch a vehicle from the surface of the Earth to low Earth orbit. The delta-v cost from gound to LEO is 9.3 km/sec, making the mass ratio for such a rocket is 88.4%. Nearly nine tenths of the vehicle must be fuel. This is the brutal math that makes a single stage to orbit rocket a pipe dream. We can get very small things in low Earth orbit with a very large single stage rocket. Getting larger things into low Earth orbit requires creative work-arounds. Using multiple stage rockets mitigates some of the consequences of the rocket equation. Getting an object of any size to the Moon or beyond requires a very powerful rocket and a whole lot of fuel.

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Notes:
¹ The statement "there is no limit to the attainable delta-v" flies in the face of the theory of relativity. That's because the developments that led to the ideal rocket equation are based on Newton's laws, which implicitly ignore relativistic effects. The equations above remain valid for small velocities. That 9.3 km/sec orbital velocity is quite fast by terrestrial standards but remains quite small when compared to the speed of light. If you want to get a spacecraft to very high velocities, the http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html" must be used in lieu of the classical ideal rocket equation.

 

[D H]

The second post developed the equations of motion for a simple rocket from the perspective of conservation of momentum. This post begins by developing the equations of motion from the perspective of conservation of energy. The post proceeds with some simple analysese of energy-related concepts.

Conservation of Energy
Per the simplifying assumptions of the second post in this series, the rocket and exhaust gas cloud formed by the rocket collectively form a closed system. The total energy of the rocket+exhaust system is constant. The system's energy comprises potential energy in the form of unburned fuel and kinetic energy in the form of random (thermal) energy and mechanical energy. Taking the time derivative of the total energy leads to the conservation of energy equation

$$\dot E_{r+e} = \dot U + \dot Q_{r+e} + \dot T_{r+e} = 0$$

Both the rate at which the rocket's potential energy changes, $\dot U$, and the rates at which heat is transferred, $\dot Q_r+\dot Q_e$ depend solely upon the rate at which fuel is being burned in the rocket. These quantitites are independent of the rocket's velocity:

$$\nabla_{\vec v_r} (\dot U + \dot Q_{r+e}) = 0$$

where

$$\nabla_{\vec v_r} \equiv \frac {\partial}{\partial v_{r_i}}$$

Since $\dot E_{r+e} = 0$, the time derivative of the total mechanical kinetic energy must also be independent of the rocket's velocity, or

$$\nabla_{\vec v_r} \dot T_{r+e} = 0$$

Time Derivative of the System Kinetic Energy
Suppose the rocket's velocity as observed in some inertial reference frame is $\vec v_r(t)$. During a small interval of time $\Delta t$, the rocket ejects a small quantity $\dot m_e(t) \Delta t$ exhaust gas with a velocity $\vec v_e(t)$ relative to the rocket. The kinetic energy of this exhaust gas as assessed in the inertial frame is

$$\Delta T_e(t) = \frac 1 2 \dot m_e(t) \Delta t (\vec v_r (t)+ \vec v_e(t)) \cdot (\vec v_r(t) + \vec v_e(t)) = \dot m_e(t) \Delta t \left(\frac 1 2 v_r(t)^2 + \vec v_r(t) \cdot \vec v_e(t) + \frac 1 2 v_e(2)^2\right)$$

Dividing by $\Delta t$ and taking the limit $\Delta t \to 0$,

$$\dot T_e(t) = = \dot m_e(t) \left(\frac 1 2 v_r(t)^2 + \vec v_r(t) \cdot \vec v_e(t) + \frac 1 2 v_e(2)^2\right)$$

The above expression is the time derivative of the exhaust gas mechanical energy. An equivalent expression is needed for the rocket to form the derivative of the system mechanical energy. The rocket's mechanical kinetic energy is

$$T_r(t) = \frac 1 2 m_r(t) v_r(t)^2$$

Differentiating with respect to time,

$$\dot T_r(t) = \frac 1 2 \dot m_r(t) v_r(t)^2 + m_r \vec v_r(t) \cdot \frac{d\vec v_r(t)}{dt}$$

Summing the exhaust and rocket mechanical energy,

$$\dot T_{r+e}(t) = \frac 1 2 \dot m_r(t) v_r(t)^2 + m_r \vec v_r(t) \cdot \frac{d\vec v_r(t)}{dt} + \dot m_e(t) \left(\frac 1 2 v_r(t)^2 + \vec v_r(t) \cdot \vec v_e(t) + \frac 1 2 v_e(2)^2\right)$$

By conservation of mass, $\dot m_r(t) = -\dot m_e(t)$. Simplifying,

$$\dot T_{r+e}(t) = \vec v_r(t) \cdot \left(m_r \frac{d\vec v_r(t)}{dt} + \dot m_e(t)\vec v_e(t)\right) + \frac 1 2 \dot m_e(t) v_e(2)^2$$

Equations of Motion

Taking the gradient of the total mechanical energy with respect to velocity,

$$\nabla_{\vec v_r} \dot T_{r+e}(t) = m_r \frac {d \vec v_r(t)} {dt} + \dot m_e(t) \vec v_e(t)$$

This expression must be identically zero as a necessary (but not sufficient) condition for conservation of energy. Solving for the rocket's acceleration,

$$\frac {d \vec v_r(t)} {dt} = -\,\frac{\dot m_e(t)} {m_r(t)} \vec v_e(t)$$

which replicates the equations of motion achieved by using conservation of linear momentum.

Power
The time derivative of the total mechanical energy simplifies to

$$\dot T_{r+e}(t) = \frac 1 2 \dot m_e(t) v_e(t)^2$$

This represents the rate at which useful energy is extracted from the combustion of the fuel. The US Space Shuttle main engines consume 1475 kilograms of fuel per second at 104% power. The main engines have a specific impulse of 450 seconds in vacuum (4400 meters/second effective exhaust velocity) and 366 at sea level (3600 meters/second). Those exhaust velocities coupled with the immense fuel consumption rate translates to 14.4 gigawatts of delivered power in vacuum, 9.5 gigawatts at sea level.

 

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