Tsiolkovsky rocket equation

The equation is named after Russian scientist Konstantin Tsiolkovsky who independently derived it and published it in his 1903 work. The equation had been derived earlier by the British mathematician William Moore in 1810, and later published in a separate book in 1813. American Robert Goddard independently developed the equation in 1912 when he began his research to improve rocket engines for possible space flight. German engineer Hermann Oberth independently derived the equation about 1920 as he studied the feasibility of space travel. While the derivation of the rocket equation is a straightforward calculus exercise, Tsiolkovsky is honored as being the first to apply it to the question of whether rockets could achieve speeds necessary for space travel. ==Derivation==

Most popular derivation Consider the following system: In the following derivation, "the rocket" is taken to mean "the rocket and all of its unexpended propellant". Newton's second law of motion relates external forces (\vec{F}_i) to the change in linear momentum of the whole system (including rocket and exhaust) as follows: \sum_i \vec{F}_i = \lim_{\Delta t \to 0} \frac{\vec{P}_{\Delta t} - \vec{P}_0}{\Delta t} where \vec{P}_0 is the momentum of the rocket at time t = 0: \vec{P}_0 = m \vec{V} and \vec{P}_{\Delta t} is the momentum of the rocket and exhausted mass at time t = \Delta t: \vec{P}_{\Delta t} = \left(m - \Delta m \right) \left(\vec{V} + \Delta \vec{V} \right) + \Delta m \vec{V}_\text{e} and where, with respect to the observer: • \vec{V} is the velocity of the rocket at time t = 0 • \vec{V} + \Delta \vec{V} is the velocity of the rocket at time t = \Delta t • \vec{V}_\text{e} is the velocity of the mass added to the exhaust (and lost by the rocket) during time \Delta t • m is the mass of the rocket at time t = 0 • \left( m - \Delta m \right) is the mass of the rocket at time t = \Delta t The velocity of the exhaust \vec{V}_\text{e} in the observer frame is related to the velocity of the exhaust in the rocket frame v_\text{e} by: \vec {v}_\text{e} = \vec{V}_\text{e} - \vec{V} thus, \vec {V}_\text{e} = \vec{V} + \vec{v}_\text{e} Solving this yields: \vec{P}_{\Delta t} - \vec{P}_0 = m\Delta \vec{V} + \vec{v}_\text{e} \Delta m - \Delta m \Delta \vec{V} If \vec{V} and \vec{v}_\text{e} are opposite, \vec{F}_\text{i} have the same direction as \vec{V}, \Delta m \Delta \vec{V} are negligible (since dm \, d\vec{v} \to 0), and using dm = -\Delta m (since ejecting a positive \Delta m results in a decrease in rocket mass in time), \sum_i F_i = m \frac{dV}{dt} + v_\text{e} \frac{dm}{dt} If there are no external forces then \sum_i F_i = 0 (conservation of linear momentum) and -m\frac{dV}{dt} = v_\text{e}\frac{dm}{dt} Assuming that v_\text{e} is constant (known as Tsiolkovsky's hypothesis \Delta v = v_\text{eff} \sum ^{j=N}_{j=1} \frac{\phi/N}{\sqrt{(1-j\phi/N)(1-j\phi/N+\phi/N)}} Notice that for large N the last term in the denominator \phi/N\ll 1 and can be neglected to give \Delta v \approx v_\text{eff} \sum^{j=N}_{j=1}\frac{\phi/N}{1-j\phi/N} = v_\text{eff} \sum ^{j=N}_{j=1} \frac{\Delta x}{1-x_j} where \Delta x = \frac{\phi}{N} and x_j = \frac{j\phi}{N} . As N\rightarrow \infty this Riemann sum becomes the definite integral \lim_{N\to\infty}\Delta v = v_\text{eff} \int_{0}^{\phi} \frac{dx}{1-x} = v_\text{eff}\ln \frac{1}{1-\phi} = v_\text{eff} \ln \frac{m_0}{m_f} , since the final remaining mass of the rocket is m_f = m_0(1-\phi). Special relativity If special relativity is taken into account, the following equation can be derived for a relativistic rocket, with \Delta v again standing for the rocket's final velocity (after expelling all its reaction mass and being reduced to a rest mass of m_1) in the inertial frame of reference where the rocket started at rest (with the rest mass including fuel being m_0 initially), and c standing for the speed of light in vacuum: \frac{m_0}{m_1} = \left[\frac{1 + {\frac{\Delta v}{c}}}{1 - {\frac{\Delta v}{c}}}\right]^{\frac{c}{2v_\text{e}}} Writing \frac{m_0}{m_1} as R allows this equation to be rearranged as \frac{\Delta v}{c} = \frac{R^{\frac{2v_\text{e}}{c}} - 1}{R^{\frac{2v_\text{e}}{c}} + 1} Then, using the identity R^{\frac{2v_\text{e}}{c}} = \exp \left[ \frac{2v_\text{e}}{c} \ln R \right] (here "exp" denotes the exponential function; see also Natural logarithm as well as the "power" identity at logarithmic identities) and the identity \tanh x = \frac{e^{2x} - 1} {e^{2x} + 1} (see Hyperbolic function), this is equivalent to \Delta v = c \tanh\left(\frac {v_\text{e}}{c} \ln \frac{m_0}{m_1} \right) ==Terms of the equation==

Delta-v Delta-v (literally "change in velocity"), symbolised as 'Δv''' and pronounced delta-vee, as used in spacecraft flight dynamics, is a measure of the impulse that is needed to perform a maneuver such as launching from, or landing on a planet or moon, or an in-space orbital maneuver. It is a scalar that has the units of speed. As used in this context, it is not'' the same as the physical change in velocity of the vehicle. Delta-v is produced by reaction engines, such as rocket engines, is proportional to the thrust per unit mass and burn time, and is used to determine the mass of propellant required for the given manoeuvre through the rocket equation. For multiple manoeuvres, delta-v sums linearly. For interplanetary missions delta-v is often plotted on a porkchop plot which displays the required mission delta-v as a function of launch date. Mass fraction In aerospace engineering, the propellant mass fraction is the portion of a vehicle's mass which does not reach the destination and is instead burned as propellant, usually used as a measure of the vehicle's performance. In other words, the propellant mass fraction is the ratio between the propellant mass and the initial mass of the vehicle. In a spacecraft, the destination is usually an orbit, while for aircraft it is their landing location. A higher mass fraction represents less weight in a design. Another related measure is the payload fraction, which is the fraction of initial weight that is payload. While the original wording of the Tsiolkovsky rocket equation does not directly use the mass fraction, the mass fraction can be derived from the used ratio of initial to final mass, or \frac{m_0}{m_f}=\frac{m_f+m_p}{m_f}=\frac{m_p}{m_f}+1. Effective exhaust velocity The effective exhaust velocity is often specified as a specific impulse and they are related to each other by: v_\text{e} = g_0 I_\text{sp}, where • I_\text{sp} is the specific impulse in seconds, • v_\text{e} is the specific impulse measured in m/s, which is the same as the effective exhaust velocity measured in m/s (or ft/s if g is in ft/s2), • g_0 is the standard gravity, 9.80665m/s2 (in Imperial units 32.174ft/s2). ==Applicability==

The rocket equation captures the essentials of rocket flight physics in a single short equation. It also holds true for rocket-like reaction vehicles whenever the effective exhaust velocity is constant, and can be summed or integrated when the effective exhaust velocity varies. The rocket equation only accounts for the reaction force from the rocket engine; it does not include other forces that may act on a rocket, such as aerodynamic or gravitational forces. As such, when using it to calculate the propellant requirement for launch from (or powered descent to) a planet with an atmosphere, the effects of these forces must be included in the delta-V requirement (see Examples below). In what has been called "the tyranny of the rocket equation", there is a limit to the amount of payload that the rocket can carry, as higher amounts of propellant increment the overall weight, and thus also increase the fuel consumption. The equation does not apply to non-rocket systems such as aerobraking, gun launches, space elevators, launch loops, tether propulsion or light sails. The rocket equation can be applied to orbital maneuvers in order to determine how much propellant is needed to change to a particular new orbit, or to find the new orbit as the result of a particular propellant burn. When applying to orbital maneuvers, one assumes an impulsive maneuver, in which the propellant is discharged and delta-v applied instantaneously. This assumption is relatively accurate for short-duration burns such as for mid-course corrections and orbital insertion maneuvers. As the burn duration increases, the result is less accurate due to the effect of gravity on the vehicle over the duration of the maneuver. For low-thrust, long duration propulsion, such as electric propulsion, more complicated analysis based on the propagation of the spacecraft's state vector and the integration of thrust are used to predict orbital motion. ==Examples==

Assume an exhaust velocity of and a \Delta v of (Earth to LEO, including \Delta v to overcome gravity and aerodynamic drag). • Single-stage-to-orbit rocket: 1-e^{-9.7/4.5} = 0.884, therefore 88.4% of the initial total mass has to be propellant. The remaining 11.6% is for the engines, the tank, and the payload. • Two-stage-to-orbit: suppose that the first stage should provide a \Delta v of ; 1-e^{-5.0/4.5} = 0.671, therefore 67.1% of the initial total mass has to be propellant to the first stage. The remaining mass is 32.9%. After disposing of the first stage, a mass remains equal to this 32.9%, minus the mass of the tank and engines of the first stage. Assume that this is 8% of the initial total mass, then 24.9% remains. The second stage should provide a \Delta v of ; 1-e^{-4.7/4.5} = 0.648, therefore 64.8% of the remaining mass has to be propellant, which is 16.2% of the original total mass, and 8.7% remains for the tank and engines of the second stage, the payload, and in the case of a space shuttle, also the orbiter. Thus together 16.7% of the original launch mass is available for all engines, the tanks, and payload. ==Stages==

In the case of sequentially thrusting rocket stages, the equation applies for each stage, where for each stage the initial mass in the equation is the total mass of the rocket after discarding the previous stage, and the final mass in the equation is the total mass of the rocket just before discarding the stage concerned. For each stage the specific impulse may be different. For example, if 80% of the mass of a rocket is the fuel of the first stage, and 10% is the dry mass of the first stage, and 10% is the remaining rocket, then \begin{align} \Delta v \ & = v_\text{e} \ln { 100 \over 100 - 80 }\\ & = v_\text{e} \ln 5 \\ & = 1.61 v_\text{e}. \\ \end{align} With three similar, subsequently smaller stages with the same v_\text{e} for each stage, gives: \Delta v \ = 3 v_\text{e} \ln 5 \ = 4.83 v_\text{e} and the payload is 10% × 10% × 10% = 0.1% of the initial mass. A comparable SSTO rocket, also with a 0.1% payload, could have a mass of 11.1% for fuel tanks and engines, and 88.8% for fuel. This would give \Delta v \ = v_\text{e} \ln(100/11.2) \ = 2.19 v_\text{e}. If the motor of a new stage is ignited before the previous stage has been discarded and the simultaneously working motors have a different specific impulse (as is often the case with solid rocket boosters and a liquid-fuel stage), the situation is more complicated. ==See also==