Basic rocket equation

The basic rocket equation in space physics is the equation of motion of an otherwise force-free rocket , which accelerates through continuous ejection of supporting mass . In particular, it indicates the maximum achievable speed of the rocket.

The basic principle of rocket propulsion is to eject a limited amount of fuel at a certain exit speed and, according to Newton's 3rd law (Actio = Reactio), to increase the momentum and thus the speed of the rocket with its payload in the opposite direction.

If a single-stage rocket with an initial mass and initial speed of zero is considered, whose engine ejects the supporting mass continuously and at constant speed , then the basic rocket equation applies (under idealized conditions) for the rocket's speed as a function of the residual mass ( i.e. the amount of fuel used reduced initial dimensions): ${\displaystyle \,m_{0}}$ ${\displaystyle \,v_{\mathrm {g} }}$ ${\displaystyle \,v}$ ${\displaystyle \,m}$

${\displaystyle v(m)=v_{\mathrm {g} }\cdot \ln {\frac {m_{0}}{m}}}$

history

The first documented derivation of this equation comes from the British mathematician William Moore and was first in 1810 in a journal and then in 1813 in the book A Treatise on the Motion of Rockets (A Treatise on the movement of missiles) published. In 1862, William Leitch published God's glory in the Heavens , where he argued that rockets were the most effective method of traveling in space. In 1903 Konstantin Ziolkowski published his derivation independently and also wondered whether rockets could reach the speeds required for space travel, which is why the first derivation is often attributed to him. Hermann Oberth and Robert Goddard , who are often referred to as pioneers of modern space travel , were later also derived independently .

Delimitation from the processes with intermittent acceleration

A fundamental aspect of rocket acceleration is that not only the payload but also the propellant has to be accelerated. Its mass decreases over time due to the ejection. If the influence of the decreasing mass to be accelerated during the rocket launch is not correctly taken into account, incorrect values ​​result for the speed achieved.

An example of understanding is given below for illustration. The fuel of a rocket is released in discrete portions and the change in speed (based on the method of small steps ) is calculated at the same time intervals. If all the fuel could be ejected in one fell swoop, the payload would be at a speed of 90 m / s.

Basic scenario

Initial conditions:

• The missile is at rest. → Step 0
• Rocket has a total mass of 100 kg (10 kg payload and 90 kg fuel). → Step 0
• The fuel is expelled in 9 portions of 10 kg each at a speed of 10 m / s. → Steps 1 to 9

step	Dimensions	speed	pulse	Additional impulse	(Residual) mass	Speed ​​increase	Overall speed
	the fuel portion				the rocket
0	0 kg	0 m / s	0 kg	0 kg m / s	100 kg	0 m / s	0 m / s
1	10 kg	A1 −10 m / s	A1 −100 kg m / s	100 kg m / s	90 kg	1.11 m / s	1.11 m / s
2	10 kg				80 kg	1.25 m / s	2.36 m / s
3	10 kg				70 kg	1.43 m / s	3.79 m / s
4th	10 kg				60 kg	1.67 m / s	5.46 m / s
5	10 kg				50 kg	2.00 m / s	7.46 m / s
6th	10 kg				40 kg	2.50 m / s	9.96 m / s
7th	10 kg				30 kg	3.33 m / s	13.29 m / s
8th	10 kg				20 kg	5.00 m / s	18.29 m / s
9	10 kg				10 kg	10.00 m / s	A2 28.29 m / s

^A1 The speed and momentum of the propellant are negative because they are directed opposite to the positive direction of flight of the rocket.

^A2 The payload has reached its final speed.

Refined consideration

• If 90 steps with 1 kg of fuel are used for the calculation, the final speed is 23.48 m / s.
• If 900 steps with 0.1 kg of fuel are used for the calculation, the final speed is 23.07 m / s.
• According to the basic rocket equation, the final speed of the payload is calculated to be 23.03 m / s after the fuel has been completely ejected.

The basic rocket equation thus describes the limit value in the event that the supporting mass is ejected in infinitely many partial steps in infinitesimally small portions.

Mathematical derivation via conservation of momentum

The mass of the rocket has already decreased and is now changing as a small unit . The support composition is in the reference system of the missile with the speed , in the system of the observer, that is, with ejected and thus transmits the pulse . Since there are no external forces acting, the total momentum of the rocket and supporting mass is preserved: ${\displaystyle m}$ ${\displaystyle \mathrm {d} m<0}$ ${\displaystyle \mathrm {d} m}$ ${\displaystyle -v_{\mathrm {g} }}$ ${\displaystyle v-v_{\mathrm {g} }}$ ${\displaystyle -\mathrm {d} m(v-v_{\mathrm {g} })}$

${\displaystyle \mathrm {d} p=\underbrace {\mathrm {d} (mv)} _{\text{Rakete}}+\underbrace {(-\mathrm {d} m)(v-v_{\mathrm {g} })} _{\text{Stützmasse}}=\mathrm {d} m\cdot v+m\cdot \mathrm {d} v-\mathrm {d} m\cdot v+\mathrm {d} m\cdot v_{\mathrm {g} }=m\cdot \mathrm {d} v+\mathrm {d} m\cdot v_{\mathrm {g} }=0}$

and thus

${\displaystyle \mathrm {d} v=-v_{\mathrm {g} }{\frac {\mathrm {d} m}{m}}}$.

This differential equation is now integrated from to . Integration of the left gives (an antiderivative of ). On the right side only has to be integrated via , since it was assumed to be constant: ${\displaystyle m_{0}}$ ${\displaystyle m}$ ${\displaystyle \,v(m)-v(m_{0})}$ ${\displaystyle \,f(v(m))=1}$ ${\displaystyle -{\frac {\mathrm {d} m}{m}}}$ ${\displaystyle \,v_{\mathrm {g} }}$

${\displaystyle v(m)=-v_{\mathrm {g} }\int _{m_{0}}^{m}{\frac {\mathrm {d} m}{m}}+v(m_{0})=-v_{\mathrm {g} }{\biggl (}\ln(m)-\ln(m_{0}){\biggl )}+v(m_{0})=v_{\mathrm {g} }\ln \left({\frac {m_{0}}{m}}\right)+v(m_{0})}$.

Evaluation of the equation

The final speed, when all the fuel is expelled, is ${\displaystyle \,m_{\mathrm {T} }}$

${\displaystyle v_{\mathrm {End} }=v(m_{\mathrm {End} })=v_{\mathrm {g} }\ln {\frac {m_{0}}{m_{\mathrm {End} }}}+v(m_{0})}$,

is therefore greater, the greater the exit speed and the smaller the residual mass , which consists of the payload, the engine and structural material. ${\displaystyle \,v_{\mathrm {g} }}$ ${\displaystyle \,m_{\mathrm {End} }}$

It is remarkable that top speeds are greater than achievable. However, in order to reach speeds far beyond that, parts of the structure (empty tanks) or the engine (booster) are left behind on the way, see multi-stage rocket . The case of successive stages is clear, with the upper stages representing the payload of the lower stages. ${\displaystyle \,v_{\mathrm {g} }}$ ${\displaystyle \,v_{\mathrm {g} }}$

example

Let us assume a two-stage rocket, the stages of which have a mass of 100 or 20 (in arbitrary units) and each consist of 90% propellant, i.e. have structural masses of 10 or 2. The payload is also 2 units. The basic rocket equation is applied twice, whereby the contributions of both stages add up (you can see this when you switch to the reference system in which the second stage initially rests when the first stage burns out):

${\displaystyle {\frac {v_{\mathrm {End} }}{v_{\mathrm {g} }}}=\ln {\frac {100+20+2}{10+20+2}}+\ln {\frac {20+2}{2+2}}\approx 1{,}34+1{,}70=3{,}04}$.

For comparison, the single-stage rocket with the same fuel and structural mass:

${\displaystyle {\frac {v_{\mathrm {End} }}{v_{\mathrm {g} }}}=\ln {\frac {100+20+2}{10+2+2}}\approx 2{,}16}$.

Limitations or idealizations

In particular, the following was idealized:

• The exhaust velocity of real engines is not constant, but varies due to various technical factors.
• The influence of gravity is not taken into account in the basic rocket equation.
• The influence of air resistance is also not taken into account. The air resistance is not constant, but depends on the current flight speed and, due to the decreasing density and changing composition of the atmosphere, also depends on the flight altitude.

The following are irrelevant for chemical drives:

• The derivation of the basic rocket equation only applies to non- relativistic speeds.
• Real fuel is not emitted in infinitesimally small packages, but in discrete portions (particles).

The following applies to vertical rocket launches, low climbing heights and neglecting air resistance

${\displaystyle v_{\mathrm {End} }=v_{\mathrm {g} }\ln {\frac {m_{0}}{m_{\mathrm {End} }}}-g\,\Delta t}$

with the gravitational acceleration and the burning time . However, this formula is unsuitable for optimizing the reaching of the earth's orbit, because besides the acceleration due to gravity, the thrust vector also changes continuously. ${\displaystyle \!\,g}$ ${\displaystyle \,\Delta t}$

literature

• Ernst Messerschmid, Stefanos Fasoulas: Space systems. An introduction with exercises and solutions, Springer Verlag, Berlin / Heidelberg 2000, ISBN 978-3-662-09674-1 .
• Wolfgang Steiner, Martin Schagerl: Space flight mechanics. Dynamics and control of space vehicles, Springer Verlag, Berlin / Heidelberg 2004, ISBN 3-540-20761-9
• Armin Dadieu, Ralf Damm, Eckart W. Schmidt: Rocket fuels. Springer Verlag, Vienna / New York 1968.
• Friedrich U. Mathiak: Technical Mechanics 3. Kinematics and Kinetics with Maple and MapleSim applications, De Gruyter Verlag, Berlin 2015, ISBN 978-3-1104-3804-8 .
• HG Münzberg: Flight drives. Basics - systematics and technology of aerospace propulsion systems, Springer Verlag, Berlin / Heidelberg 1972, ISBN 978-3-662-11758-3 .

Web links

• Bernd Leitenberger: The basic rocket equation
• Student work on rocket physics (accessed on June 9, 2018)
• Lecture notes for the space systems lecture (accessed on June 9, 2018)

Individual evidence

1. ^ William Moore: A Journal of Natural Philosophy, Chemistry and the Arts Vol. XXVII, December 1810, Article IV: Theory on the motion of Rockets . W. Nichelson, London 1810.
2. ^ William Moore: A Treatise on the Motion of Rockets. To which is added, An Essay on Naval Gunnery . G. and S. Robinson, London 1813.
3. ^ William Leitch: God's Glory in the Heaves . Ed .: Alexander Strahan. 1862.