Tsiolkovsky rocket equation

Consider the following system:

Tsiolkovsky's theoretical rocket from t = 0 to t = delta_t

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 (${\displaystyle {\vec {F_{i}}}}$) to the change in linear momentum of the whole system (including rocket and exhaust) as follows:

$${\displaystyle \sum _{i}{\vec {F_{i}}}=\lim _{\Delta t\to 0}{\frac {{\vec {P_{2}}}-{\vec {P_{1}}}}{\Delta t}}}$$

where ${\displaystyle {\vec {P_{1}}}}$ is the momentum of the rocket at time ${\displaystyle t=0}$:

$${\displaystyle {\vec {P_{1}}}=m{\vec {V}}}$$

and ${\displaystyle {\vec {P_{2}}}}$ is the momentum of the rocket and exhausted mass at time ${\displaystyle t=\Delta t}$:

$${\displaystyle {\vec {P_{2}}}=\left(m-\Delta m\right)\left({\vec {V}}+\Delta {\vec {V}}\right)+\Delta m{\vec {V_{\text{e}}}}}$$

and where, with respect to the observer:

• ${\displaystyle {\vec {V}}}$ is the velocity of the rocket at time ${\displaystyle t=0}$
• ${\displaystyle {\vec {V}}+\Delta {\vec {V}}}$ is the velocity of the rocket at time ${\displaystyle t=\Delta t}$
• ${\displaystyle {\vec {V_{\text{e}}}}}$ is the velocity of the mass added to the exhaust (and lost by the rocket) during time ${\displaystyle \Delta t}$
• ${\displaystyle m}$ is the mass of the rocket at time ${\displaystyle t=0}$
• ${\displaystyle \left(m-\Delta m\right)}$ is the mass of the rocket at time ${\displaystyle t=\Delta t}$

The velocity of the exhaust ${\displaystyle {\vec {V_{\text{e}}}}}$ in the observer frame is related to the velocity of the exhaust in the rocket frame ${\displaystyle v_{\text{e}}}$ by:

$${\displaystyle {\vec {v_{\text{e}}}}={\vec {V_{\text{e}}}}-{\vec {V}}}$$

thus,

$${\displaystyle {\vec {V_{\text{e}}}}={\vec {V}}+{\vec {v_{\text{e}}}}}$$

Solving this yields:

$${\displaystyle {\vec {P_{2}}}-{\vec {P_{1}}}=m\Delta {\vec {V}}+{\vec {v_{\text{e}}}}\Delta m-\Delta m\Delta {\vec {V}}}$$

If ${\displaystyle {\vec {V}}}$ and ${\displaystyle {\vec {v_{\text{e}}}}}$ are opposite, ${\displaystyle {\vec {F_{\text{i}}}}}$ have the same direction as ${\displaystyle {\vec {V}}}$, ${\displaystyle \Delta m\Delta {\vec {V}}}$ are negligible (since ${\displaystyle dmd{\vec {v}}\rightarrow 0}$), and using ${\displaystyle dm=-\Delta m}$ (since ejecting a positive ${\displaystyle \Delta m}$ results in a decrease in rocket mass in time),

$${\displaystyle \sum _{i}F_{i}=m{\frac {dV}{dt}}+v_{\text{e}}{\frac {dm}{dt}}}$$

If there are no external forces then ${\textstyle \sum _{i}F_{i}=0}$ (conservation of linear momentum) and

$${\displaystyle -m{\frac {dV}{dt}}=v_{\text{e}}{\frac {dm}{dt}}}$$

Assuming that ${\displaystyle v_{\text{e}}}$ is constant (known as Tsiolkovsky's hypothesis[2]), so it is not subject to integration, then the above equation may be integrated as follows:

$${\displaystyle -\int _{V}^{V+\Delta V}\,dV={v_{e}}\int _{m_{0}}^{m_{f}}{\frac {dm}{m}}}$$

This then yields

$${\displaystyle \Delta V=v_{\text{e}}\ln {\frac {m_{0}}{m_{f}}}}$$

or equivalently

$${\displaystyle m_{f}=m_{0}e^{-\Delta V\ /v_{\text{e}}}}$$

or

$${\displaystyle m_{0}=m_{f}e^{\Delta V/v_{\text{e}}}}$$

or

$${\displaystyle m_{0}-m_{f}=m_{f}\left(e^{\Delta V/v_{\text{e}}}-1\right)}$$

where ${\displaystyle m_{0}}$ is the initial total mass including propellant, ${\displaystyle m_{f}}$ the final mass, and ${\displaystyle v_{\text{e}}}$ the velocity of the rocket exhaust with respect to the rocket (the specific impulse, or, if measured in time, that multiplied by gravity-on-Earth acceleration). If ${\displaystyle v_{\text{e}}}$ is NOT constant, we might not have rocket equations that are as simple as the above forms. Many rocket dynamics researches were based on the Tsiolkovsky's constant ${\displaystyle v_{\text{e}}}$ hypothesis.

The value ${\displaystyle m_{0}-m_{f}}$ is the total working mass of propellant expended.

${\displaystyle \Delta V}$ (delta v) is the integration over time of the magnitude of the acceleration produced by using the rocket engine (what would be the actual acceleration if external forces were absent). In free space, for the case of acceleration in the direction of the velocity, this is the increase of the speed. In the case of an acceleration in opposite direction (deceleration) it is the decrease of the speed. Of course gravity and drag also accelerate the vehicle, and they can add or subtract to the change in velocity experienced by the vehicle. Hence delta-v may not always be the actual change in speed or velocity of the vehicle.

The equation can also be derived from the basic integral of acceleration in the form of force (thrust) over mass. By representing the delta-v equation as the following:

$${\displaystyle \Delta v=\int _{t_{0}}^{t_{f}}{\frac {|T|}{{m_{0}}-{t}\Delta {m}}}~dt}$$

where T is thrust, ${\displaystyle m_{0}}$ is the initial (wet) mass and ${\displaystyle \Delta m}$ is the initial mass minus the final (dry) mass,

and realising that the integral of a resultant force over time is total impulse, assuming thrust is the only force involved,

$${\displaystyle \int _{t_{0}}^{t_{f}}F~dt=J}$$

The integral is found to be:

$${\displaystyle J~{\frac {\ln({m_{0}})-\ln({m_{f}})}{\Delta m}}}$$

Realising that impulse over the change in mass is equivalent to force over propellant mass flow rate (p), which is itself equivalent to exhaust velocity,

$${\displaystyle {\frac {J}{\Delta m}}={\frac {F}{p}}=V_{\text{exh}}}$$

the integral can be equated to

$${\displaystyle \Delta v=V_{\text{exh}}~\ln \left({\frac {m_{0}}{m_{f}}}\right)}$$

Imagine a rocket at rest in space with no forces exerted on it (Newton's First Law of Motion). From the moment its engine is started (clock set to 0) the rocket expels gas mass at a constant mass flow rate R (kg/s) and at exhaust velocity relative to the rocket v_e (m/s). This creates a constant force F propelling the rocket that is equal to R × v_e. The rocket is subject to a constant force, but its total mass is decreasing steadily because it is expelling gas. According to Newton's Second Law of Motion, its acceleration at any time t is its propelling force F divided by its current mass m:

$${\displaystyle ~a={\frac {dv}{dt}}=-{\frac {F}{m(t)}}=-{\frac {Rv_{\text{e}}}{m(t)}}}$$

Now, the mass of fuel the rocket initially has on board is equal to m₀ – m_f. For the constant mass flow rate R it will therefore take a time T = (m₀ – m_f)/R to burn all this fuel. Integrating both sides of the equation with respect to time from 0 to T (and noting that R = dm/dt allows a substitution on the right) obtains:

$${\displaystyle ~\Delta v=v_{f}-v_{0}=-v_{\text{e}}\left[\ln m_{f}-\ln m_{0}\right]=~v_{\text{e}}\ln \left({\frac {m_{0}}{m_{f}}}\right).}$$

The rocket equation can also be derived as the limiting case of the speed change for a rocket that expels its fuel in the form of ${\displaystyle N}$ pellets consecutively, as ${\displaystyle N\to \infty }$, with an effective exhaust speed ${\displaystyle v_{\text{eff}}}$ such that the mechanical energy gained per unit fuel mass is given by ${\textstyle {\tfrac {1}{2}}v_{\text{eff}}^{2}}$.

In the rocket's center-of-mass frame, if a pellet of mass ${\displaystyle m_{p}}$ is ejected at speed ${\displaystyle u}$ and the remaining mass of the rocket is ${\displaystyle m}$, the amount of energy converted to increase the rocket's and pellet's kinetic energy is

$${\displaystyle {\tfrac {1}{2}}m_{p}v_{\text{eff}}^{2}={\tfrac {1}{2}}m_{p}u^{2}+{\tfrac {1}{2}}m(\Delta v)^{2}.}$$

Using momentum conservation in the rocket's frame just prior to ejection, ${\textstyle u=\Delta v{\tfrac {m}{m_{p}}}}$, from which we find

$${\displaystyle \Delta v=v_{\text{eff}}{\frac {m_{p}}{\sqrt {m(m+m_{p})}}}.}$$

Let ${\displaystyle \phi }$ be the initial fuel mass fraction on board and ${\displaystyle m_{0}}$ the initial fueled-up mass of the rocket. Divide the total mass of fuel ${\displaystyle \phi m_{0}}$ into ${\displaystyle N}$ discrete pellets each of mass ${\displaystyle m_{p}=\phi m_{0}/N}$. The remaining mass of the rocket after ejecting ${\displaystyle j}$ pellets is then ${\displaystyle m=m_{0}(1-j\phi /N)}$. The overall speed change after ejecting ${\displaystyle j}$ pellets is the sum [6]

$${\displaystyle \Delta v=v_{\text{eff}}\sum _{j=1}^{j=N}{\frac {\phi /N}{\sqrt {(1-j\phi /N)(1-j\phi /N+\phi /N)}}}}$$

Notice that for large ${\displaystyle N}$ the last term in the denominator ${\displaystyle \phi /N\ll 1}$ and can be neglected to give

$${\displaystyle \Delta v\approx v_{\text{eff}}\sum _{j=1}^{j=N}{\frac {\phi /N}{1-j\phi /N}}=v_{\text{eff}}\sum _{j=1}^{j=N}{\frac {\Delta x}{1-x_{j}}}}$$

where ${\textstyle \Delta x={\frac {\phi }{N}}}$ and ${\textstyle x_{j}={\frac {j\phi }{N}}}$.

As ${\displaystyle N\rightarrow \infty }$ this Riemann sum becomes the definite integral

$${\displaystyle \lim _{N\to \infty }\Delta v=v_{\text{eff}}\int _{0}^{\phi }{\frac {dx}{1-x}}=v_{\text{eff}}\ln {\frac {1}{1-\phi }}=v_{\text{eff}}\ln {\frac {m_{0}}{m_{f}}},}$$

since the final remaining mass of the rocket is ${\displaystyle m_{f}=m_{0}(1-\phi )}$.

If special relativity is taken into account, the following equation can be derived for a relativistic rocket,[7] with ${\displaystyle \Delta v}$ again standing for the rocket's final velocity (after expelling all its reaction mass and being reduced to a rest mass of ${\displaystyle m_{1}}$) in the inertial frame of reference where the rocket started at rest (with the rest mass including fuel being ${\displaystyle m_{0}}$ initially), and ${\displaystyle c}$ standing for the speed of light in vacuum:

$${\displaystyle {\frac {m_{0}}{m_{1}}}=\left[{\frac {1+{\frac {\Delta v}{c}}}{1-{\frac {\Delta v}{c}}}}\right]^{\frac {c}{2v_{\text{e}}}}}$$

Writing ${\textstyle {\frac {m_{0}}{m_{1}}}}$ as ${\displaystyle R}$ allows this equation to be rearranged as

$${\displaystyle {\frac {\Delta v}{c}}={\frac {R^{\frac {2v_{\text{e}}}{c}}-1}{R^{\frac {2v_{\text{e}}}{c}}+1}}}$$

Then, using the identity ${\textstyle R^{\frac {2v_{\text{e}}}{c}}=\exp \left[{\frac {2v_{\text{e}}}{c}}\ln R\right]}$ (here "exp" denotes the exponential function; see also Natural logarithm as well as the "power" identity at Logarithmic identities) and the identity ${\textstyle \tanh x={\frac {e^{2x}-1}{e^{2x}+1}}}$ (see Hyperbolic function), this is equivalent to

$${\displaystyle \Delta v=c\tanh \left({\frac {v_{\text{e}}}{c}}\ln {\frac {m_{0}}{m_{1}}}\right)}$$

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 and 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.

In aerospace engineering, the propellant mass fraction is the portion of a vehicle's mass which does not reach the destination, 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.

The effective exhaust velocity is often specified as a specific impulse and they are related to each other by:

$${\displaystyle v_{\text{e}}=g_{0}I_{\text{sp}},}$$

where

• ${\displaystyle I_{\text{sp}}}$ is the specific impulse in seconds,
• ${\displaystyle v_{\text{e}}}$ 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),
• ${\displaystyle g_{0}}$ is the standard gravity, 9.80665 m/s² (in Imperial units 32.174 ft/s²).