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.
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 in to 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 \(\Delta m_e(t) = \dot m_e(t)\,\Delta t\) \(\vec v_{e,{\mathrm{inertial}}} = \vec v_r(t)+\vec v_e(t)\) \(\Delta \vec p_e(t+\Delta t) = \dot m_e(t)\,\Delta t\, (\vec v_r(t)+\vec v_e(t))\) 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 \(t\).
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 [itex]\hat v[/itex] 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-v[sup]1[/sup]. 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. ==================================== Notes: [sup]1[/sup] 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 relativistic rocket equation must be used in lieu of the classical ideal rocket equation.
Energy concepts 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, \(\dot T_{r+e}(t) = \vec v_r(t) \cdot \left(m_r \frac{d\vec v_r(t)}{dt} + \dot m_e(t)\vec v_e(t)\right) + \frac 1 2 \dot m_e(t) v_e(t)^2 \) 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.
horizontal flight characteristics I follow your treatise with no problem. Could you explain why modern rocketeers direct the launch vehicle to a horizontal direction of flight as soon as is possible after launch?:shrug:
This is a good question, but perhaps it is to get some lift as well as thrust? Or they just want to get on a good trajectory to get into orbit?
A rocket does not turn to a horizontal direction of flight as soon as possible after launch. Instead, a rocket starts to turn to the horizontal (pitch down) shortly after launch. However, it doesn't become fully horizontal until it reaches orbital altitude 8-10 minutes after launch. This slow pitch rate accomplishes two ends. First and foremost, it ensures the vehicle has achieved orbital velocity. A circular orbit has zero vertical velocity. The velocity at the end of the ascent phase should be close to horizontal. If the rocket instead went straight up it would come straight down. This is what we jokingly call an orthogonal orbit. By pitching the vehicle down, the thrust vector aligns more and more with the horizontal. If gaining horizontal velocity is the foremost objective of a rocket launch, why don't we turn the rocket horizontally immediately? The second reason for the slow pitch maneuver is that this acts to maximize orbital altitude. A circular orbit inside the Earth's atmosphere isn't of much more use than an orthogonal orbit. The vehicle needs to be out of the bulk of the atmosphere by the time its engines cut off.
D H, for horizontal flight of a rocket, where should the COG be in order to produce a predictable arc-like path? I have experimented with fins that cause the rocket to spin, like that of a bullet. I have also experimented with many different nose-to-base weight ratios, but with inconsistent results. This could have been due to sudden wind vectors and other discrepancies, but this is causing my results to be inconclusive and to seem misleading. Mathematically, the weight ratios should have produced a stable rocket. Thank you.
question: if the rocket ascends to 100km above Earth perpendicularly to the ground (no horizontal movement at all) all the thrust was used only to get straight up to 100km in straight line...after the thrust turns off, will that rocket orbit the Earth? Also, what equation to use to calculate the minimum altitude required for x number of orbits around Earth?
Another question: if the rocket is one stage, with a mass to be deployed at an altitude of 100km, and the thrust stops at 80km lets say, should the engine be jettisoned? Basically, during the coastal stage, does it matter to jettison the unneeded mass when it is being accelerated up? (I guess yes...right?)
also...is there a general formula for a multistage rockets? for example I want x number of stages...could there be such a formula of summation perhaps or limits/series of sorts which would define a rocket with x number of stages and the delta v from it all? I found 3 stage equations...but what I am driving at here is 20 stage rocket equation...
draqon, your 'no horizontal movement' paradigm, that's relative to the earth's surface - the launch site stays directly below the rocket?? You realise that doesn't account for the rotation of the earth, right? And orbiting implies horizontal 'movement'. You have the wrong curves in your paradigm.
thank you, you answered my question. Please Register or Log in to view the hidden image! Basically one in 100km altitude, the satellite will orbit the Earth because of Earth's orbital velocity.
That is not what Vkothii said at all! The Earth's orbital velocity about the Sun has nothing to do with whether a satellite will orbit the Earth. A satellite 100 km up (too low, actually; that is well inside the atmosphere) needs to have a horizontal velocity of more than 7.6 km/second to be in orbit. SpaceShipOne, for example reached an altitude of 111 km but did not achieve orbit, by design. SpaceShip is a suborbital vehicle.
Yes and no. The rocket enters an "orbit", but this orbit has a perigee that would be inside the Earth. There is no simple equation for this. If you neglect resistance, an object in a direct orbit will orbit indefinitely once placed in an orbit around the Earth. (Excluding orbits that intersect the Earth itself, which would complete less than one orbit) In real life, LEO objects are subject to friction from the upper atmosphere, but this is not easy to calculate because the Earth's atmosphere expands and contracts with time due to various causes. Objects placed in retrograde orbit will slowly "fall" in toward Earth due to tidal interaction, but this is also a fairly complex calculation that relies on a number of factors.
well what you said is exactly what I meant. Sorry to be so un-understanding. Am I correct to understand that besides gaining delta v for a specific altitude for altitude, the rocket needs to accelerate itself horizontally for retrograde orbit? where does that value of 7.6 km/s come from?
Delta_V = Isp * g * ln(Mo/Mf) with a rocket engine of specific ISP and initial mass M0 to final mass Mf (due to propellant loss) The rocket accelerates using that Isp to altitude of 110km straight up from the ground, perpendicular to the ground...or is that not going to happen? Once the rocket is at that altitude, will it just fall to the ground slowly? or will it circle the Earth? (assume perfect scenario) Please bear with me, I need to understand it.
not just for retrograde orbit but for any orbit \(V_o = \sqrt{\frac{GM}{r}}\) Where r is the radius of the orbit as measured from the center of mass M(in this case, the Earth).
