PHYSICS OF THE ROCKET*

Howard S. Seifert, Mark M. Mills
and Martin Summerfield

IV. DYNAMICS OF A LONG-RANGE ROCKET

ATTAINMENT OF GREAT HEIGHTS

AND ESCAPE FROM THE EARTH’S GRAVITATIONAL ATTRACTION

39. History of high-altitude rockets

By “long-range rockets” we shall mean rockets having a flight range of not less than 100 miles, with an engine designed for a burning time of not less than approximately 30 sec, and for which maximum range was the principal design objective. In this they differ from artillery rockets, whose task is to throw the greatest payload over a short distance. A “research rocket” is a special type of long-range rocket intended to raise scientific measuring instruments to the greatest height along a vertical trajectory, for the purpose of determining the physical conditions in the upper layers of the atmosphere or even beyond the atmosphere. As is known, pilot balloons reach a maximum altitude of 40 km, and consequently the field of application of rockets begins at this altitude.

Fig. 42. R. Goddard’s rocket using liquid oxygen and gasoline, which in 1935 reached an altitude of 2300 m.

* Amer. J. of Physics 15, 3 (1947) (conclusion). See UFN, 34, issues 1, 34, issue 3, 334 (1948). Translated by M. L. Antokolsky.

The first technical analysis of the problem of reaching great heights was published by Tsiolkovsky in 1903 in Russian¹. In this work the potential superiority of rockets with a liquid propellant over powder rockets was indicated, and the conclusion was drawn that research should be conducted precisely in this direction.

The first experimental attempts to reach great heights with the aid of a rocket were the works of Goddard² in the United States, begun in 1912 and continued until 1941. After experiments to determine the exhaust velocities obtained with solid propellants—black and smokeless powder—and with the oxygen–gasoline combination, Goddard built and tested several liquid-propellant rockets.

The maximum height—2300 m—was reached in 1935 by a rocket 4.3 m long with an oxygen–gasoline engine (Fig. 42). One of the interesting results of these works was the principle of stabilizing the rocket by the gyroscopic action of blades rotated by the stream of outgoing gases.

In Germany, in the period from 1923 to 1930, Hermann Oberth³ significantly developed the theory of the rocket. Active experimental work was carried out by the Verein für Raumschiffahrt—a society founded in 1927 for research in the field of rocket engines. With its assistance a number of liquid-propellant rockets operating on gasoline with oxygen were built and tested. The maximum height reached in the experiments was about 1800 m. This society ceased to exist in 1932, in connection with the transfer of the work into the hands of the German artillery department⁴.

Fig. 43. WAC-Corporal rocket installed in the launching tower.

A similar voluntary society was organized in the United States in 1931, and at present it bears the name of the American Rocket Society. Although the development of high-altitude research rockets was from the very beginning one of the society’s tasks, it succeeded in developing only one rocket using gasoline and oxygen for

by the time his activity was interrupted by the outbreak of the Second World War⁵.

At the California Institute of Technology, interest in the possibility of reaching altitudes exceeding 30 km with the aid of a rocket was aroused by the work of Malina and his associates, begun in 1936. During the period 1936–1940, a theoretical analysis was carried out of the flight of a high-altitude rocket and of rocket thermodynamics, and the characteristics of various propellants were investigated⁶.

Fig. 44. Flight of a rocket with solid propellant, built during the preliminary development of the WAC-Corporal research rocket. The smoke trail is about 4.8 km long.

Fig. 45. German “V-2” rocket at the moment of takeoff at the White Sands proving ground (New Mexico).

In 1940, the program of work of the California Institute of Technology was expanded and directed mainly toward the military application of rockets. Interest in research rockets was actively revived in 1945, when a research rocket, the so-called WAC-Corporal, intended for lifting

to a great altitude measuring instruments weighing about 11 kg. In August of that year 10 such rockets were launched, reaching a maximum altitude of 70 km⁷ (see Figs. 40 and 43). Figure 44 shows the flight of an experimental rocket with a solid propellant, used in the preliminary experiments connected with the development of the WAC-Corporal.

The highest point to date was reached in 1944, when two German “V-2” rockets, launched vertically upward,

Labels in the figure: combustion chamber and Venturi tube; 4 outer control vanes; 4 inner control vanes; turbine and pump unit; liquid-oxygen tank; alcohol tank; control instruments; warhead; stabilizing fins.

Fig. 46. British official depiction of the “V-2” rocket, published on December 8, 1945.

reached an altitude of 170 km⁸. In the normal flight of this rocket in the horizontal direction, the highest point of the trajectory is at an altitude of about 96 km. In the latest experimental launches of captured “V-2” rockets at the Proving Ground of the Artillery Administration at White Sands, New Mexico, altitudes exceeding 160 km were reached. This design represents the highest technical achievement attained up to the present time (Figs. 45 and 46).

40. General equations of rocket motion

In this section we shall consider the dynamics of the vertical flight of a high-altitude rocket. The study of the influence of the various factors determining the conditions of flight makes it possible to estimate the prospects for the future development of high-altitude rockets.

A rocket vehicle such as the “V-2” or the WAC-Corporal consists of the following principal parts: the motor, the so-called useful load—which in research rockets consists of various measuring instruments—the supply of propellant, the shell, and the load-bearing structures. The propellant generally constitutes the largest—

…part of the total mass. Before launch the rocket is placed in a vertical position, nose upward. If the rocket is not equipped with automatic control, then for its launch a guiding tower is necessary, in which the rocket is accelerated to a speed sufficient for its stabilization by the action of aerodynamic forces.

Below only the conditions of the rocket’s flight will be considered. Other dynamical problems, such as, for example, flight stability, the effect on the flight of the control and stabilization devices, and the stabilization of these devices themselves, are too specialized to be treated in this article.

During operation of the motor the rocket experiences acceleration under the action of the thrust, which must overcome the opposing force of gravity and air resistance (Fig. 47). If the limiting altitude is small in comparison with the earth’s radius, then the change in the force of gravity may be neglected. However, the mass of the rocket, the aerodynamic drag, and the thrust developed change during the flight.

If \(F\) is the thrust of the motor, \(D\) is the air resistance, and \(M\) is the instantaneous value of the rocket mass, then its acceleration is determined by the relation

\[ a = \frac{dv}{dt} = \frac{F - D - Mg}{M}. \tag{94} \]

Fig. 47. Forces acting on a rocket during vertical flight in the atmosphere.

Usually rockets are designed so that the mass of propellant flowing out per unit time is constant. In this case the rocket mass varies linearly with time, according to the equation

\[ M = M_0 \left(1 - \frac{\zeta t}{t_p}\right) \quad \text{for } t \leq t_p, \tag{95} \]

where \(M_0\) is the initial mass, \(t_p\) is the total duration of motor operation, and \(\zeta\) is the ratio of the initial amount of propellant to the total mass of the rocket.

The thrust at any instant of time is expressed by the relation

\[ F = \frac{M_p c}{t_p}, \tag{96} \]

where \(M_p = \zeta M_0\) is the initial mass of propellant, and \(c\) is the effective exhaust velocity in the jet. The dependence of \(c\) on flight altitude was considered in Section 10, but to simplify the calculation in expression (96) and throughout the subsequent analysis a constant mean value of \(c\) is assumed. Consequently, here we shall take for the thrust a constant mean value, despite the fact that in reality at the end of burning it may exceed the thrust at launch by 25%.

Aerodynamic drag is determined by the equation

\[ D=C_d A\cdot \frac{1}{2}\rho v^2, \tag{97} \]

where \(C_d\) is the drag coefficient, \(A\) is the area of the frontal cross section, and \(\rho\) is the density of the atmosphere. The coefficient \(C_d\) depends not only on the dimensions and shape of the rocket and the character of its surface, but also on the velocity \(v\). This latter dependence is expressed by means of the Mach number, which represents the ratio of the flight speed to the speed of sound propagation under the given conditions. In what follows, for simplicity, we shall take for \(C_d\) a certain constant average value.

Combining equations (94) and (97) and integrating the resulting equation of motion, we arrive at the following expression for the velocity \(v\) at any moment of flight with the motor operating:

\[ v=\int_0^t \frac{\dfrac{\zeta C}{t_p}}{1-\zeta \dfrac{t}{t_p}}\,dt -\int_0^t g\,dt -\int_0^t \frac{\dfrac{C_d}{\mu}\cdot \dfrac{1}{2}\rho v^2}{1-\dfrac{\zeta t}{t_p}}\,dt +v_0, \tag{98} \]

where \(v_0\) is the initial velocity at launch. The parameter \(\mu\) denotes the ratio \(M_0/A\), on which the relative weight of the third term in (98) depends. The first two integrals can be evaluated directly, while the third can be evaluated by numerical methods; we shall denote its value by \(\dfrac{q_1 C_d}{\mu}\), where

\[ q_1=\int_0^{t_p} \frac{\dfrac{1}{2}\rho v^2}{1-\zeta \dfrac{t}{t_p}}\,dt. \tag{99} \]

The velocity at the end of motor operation proves to be equal to

\[ v_p=-c\ln(1-\zeta)-gt_p-q_1\frac{C_d}{\mu}+v_0. \tag{100} \]

For flight in a vacuum, the velocity at the end of motor operation is equal to

\[ v_{p_0}=-c\ln(1-\zeta)-gt_p+v_0. \tag{101} \]

A second integration of equation (98) gives the height reached by the end of motor operation:

\[ h_p=ct_p\left[1+\frac{1-\zeta}{\zeta}\ln(1-\zeta)\right] -\frac{1}{2}gt_p^2+v_0t_p+h_0-q_2\frac{C_d}{\mu}, \tag{102} \]

where \(q_2\) is the double integral with respect to time of the quantity standing under

sign of the integral in the expression for \(q_1\). If air resistance may be neglected, then the last term \(q_2 \dfrac{C_d}{\mu}\) vanishes.

After combustion has ended, the rocket continues to gain altitude until its velocity becomes zero. This additional increase in altitude, if it is assumed that the rocket is already outside the atmosphere, is simply equal to

\[ h_c=\frac{\frac{1}{2}v_p^2}{g}. \tag{103} \]

The highest point of the trajectory is equal to the sum of \(R_p\) and \(h_c\), or

\[ \begin{aligned} h={}&\frac{c^2}{2g}\,[\ln(1-\zeta)]^2 +ct_p\left[1+\frac{1}{\zeta}\ln(1-\zeta)\right] +\frac{v_0^2}{2g}+h_0 \\ &-\frac{v_0c}{g}\ln(1-\zeta) -\frac{C_d}{\mu}\left[ q_2+\frac{v_0q_1}{g} -\frac{cq_1}{g}\ln(1-\zeta)\right.\\ &\left.\qquad\qquad\qquad -q_1t_p-\frac{q_1^2}{2g}\frac{C_d}{\mu} \right]. \end{aligned} \tag{104} \]

Let us consider the significance of each term in (104). The first of them, which usually has the predominant significance, is proportional to the square of the exhaust velocity. Owing to the presence of the logarithmic factor, it is very sensitive to small changes in the relative mass of the propellant \(\zeta\), especially for values of \(\zeta\) close to unity. Indeed, however small the exhaust velocity may be, the rocket’s flight altitude tends to infinity when the unexpended mass of the rocket tends to zero. The second term is always negative and, since it is proportional to \(t_p\), the maximum flight altitude is the lower the greater the burning time. If \(t_p\) were the only factor, then the greatest flight altitude would be attained with instantaneous combustion of all the propellant. However, a rocket capable of withstanding the stresses developing in this case would have to possess a large value of the ratio of structural mass to propellant mass, which leads to a considerable reduction of the first term in (104). Moreover, when launching a rocket from the earth’s surface, a very large initial velocity would greatly increase air resistance. Thus, generally speaking, there exists an optimum burning time, which must be determined for each type of rocket. The third, fourth, and fifth terms are corrections for the case in which the rocket already has an initial velocity before its ignition, as for example when launched with the aid of an auxiliary rocket, and for the case in which the launch is made from a point elevated above the earth.

Finally, the last term, negative in sign, takes account of the braking of the rocket by air. Of course it disappears when \(C_d=0\) or \(\rho=0\), i.e. in a vacuum or beyond the limits of the atmosphere. It is noteworthy,

that it becomes negligible for large values of \(\mu\). The parameter \(\mu=M_0/A\) is proportional to the product of the rocket length and its mean density. Therefore, when similar rockets are compared with one another, it turns out that the effect of air resistance becomes negligible for rockets of sufficiently large dimensions, and the operation of large rockets is described with satisfactory approximation by the first five terms, which are independent of the dimensions.

The altitude attained by a sufficiently large rocket launched from rest and with \(h_0=0\) is given simply by the equation

\[ h=\frac{c^2}{2g}\,[\ln(1-\zeta)]^2+ct_p\left[1+\frac{1}{\zeta}\ln(1-\zeta)\right]. \tag{105} \]

The influence on the attainable altitude of the exhaust velocity \(c\), of the relative mass of propellant \(\zeta\), and of the ratio of the total mass of the rocket to the cross section (the transverse loading) \(\mu\) is presented in Figs. 48 and 49, which show the altitude computed from equation (104). In the calculations the air-resistance curve of Fig. 50 was used; it represents results obtained for successfully designed rockets. For the density and temperature distribution of the atmosphere, the standard data of the National Advisory Committee for Aeronautics (NACA)\(^9\) were used.

Fig. 48. Influence of the relative mass of propellant and the effective exhaust velocity on the altitude attained in vertical flight. The influence of air resistance is assumed negligible, and it is taken that \(\mu=\infty\), \(t_p=30\) sec, \(h_0=0\), and \(v_0=0\).

In Table IX data are presented on the WAC-Corporal and “V-2” rockets for comparison of actually obtained data with the quantities predicted by equation (104), using, for determination of air resistance, the curve of Fig. 50. The observed discrepancy is apparently caused by such circumstances as an increase of head resistance due to small transverse oscillations and incomplete use of the propellant during the period

launch and end of combustion, etc. In particular, for WAC-Corporal it is known that its drag exceeds the figures given—

Fig. 49. Effect of the burning duration \(t_p\) and the transverse loading \(\mu\) on the altitude of the vertical flight of a rocket. It is assumed that \(\zeta = 0.70\), \(c = 2350\ \text{m/sec}\), \(h_0 = 0\), \(v_0 = 0\), and \(C_d\) is determined from Fig. 50.

Fig. 50. Dependence of the drag coefficient \(C_d\) on the Mach number, used in calculating rocket trajectories. It is assumed that there is no exhaust jet, that the angle of attack is zero, and that standard NACA tables for atmospheric pressure can be applied.

—by Fig. 50. The exhaust velocity of the “V-2” rocket was corrected taking into account the drag of the stabilizing fins placed in the exhaust jet.

Table IX

Comparative data for the WAC-Corporal and “V-2” rockets

41. Considerations on Flight into Outer Space

The question of flight into outer space includes many different problems, including questions of trajectory control, navigation, communication, and take-off and landing techniques. The fundamental physical question, however, is the imparting to the craft of the energy necessary to break away from the Earth’s attraction. Although considerations have sometimes been expressed about firing a projectile from a tube, like a cannon, all investigators agree that the only suitable method is the rocket[^10a]1.

We shall first consider the question of the energy, or velocity, necessary to overcome the Earth’s attraction, and then considerations on the design of a rocket suitable for producing such velocities. Four cases are of interest.

1. An Earth satellite revolving around the Earth in a circular orbit outside the atmosphere can remain in such an orbit indefinitely, expending no additional energy, except in the event of collision with a meteor[^11]. The distance of the orbit from the Earth must be sufficient that the satellite does not encounter air resistance. The velocity of the satellite must be such as to compensate the Earth’s attraction, and is determined by the following equations:

\[ \frac{v^2}{R+h}=\frac{g_0 R^2}{(R+h)^2}, \tag{106} \]

\[ v=\sqrt{\frac{g_0 R^2}{R+h}}, \tag{107} \]

where \(R\) is the radius of the Earth, \(h\) is the height of the orbit above the Earth’s surface, and \(g_0\) is the acceleration of gravity at the Earth’s surface. The time of one revolution is

\[ T=\frac{2\pi(R+h)}{v} =2\pi\left(1+\frac{h}{R}\right)\sqrt{\frac{R+h}{g_0}} . \tag{108} \]

The total energy \(E\) per unit mass which must be imparted to the craft in order for it to attain this orbit is the sum of the required kinetic and potential energies, i.e.

\[ E=\frac{1}{2}v^2+\int_0^h g_0\frac{R^2}{(R+x)^2}\,dx =\frac{v^2}{2}+g_0h\frac{R}{R+h}, \tag{109} \]

or, substituting the value of \(v\) from equation (107),

\[ E=\frac{1}{2}g_0R\left(1+\frac{h}{R+h}\right). \tag{110} \]

In this equation the possibility of adding to the required total velocity the peripheral velocity of the Earth’s surface has not been taken into account. The change in the required total energy as a function of height is determined—

is the second term of equation (110), from which it is seen that the energy for an orbit at an altitude of 320 km is only 5% greater than for an orbit at the very surface of the Earth. The total energy per unit mass needed to reach such an orbit is \(3.43 \cdot 10^{10}\) joule/kg. The orbital velocity is \(7800\) m/sec and the period of revolution is 1.51 hours.

Projects of this kind raise many interesting questions that lie beyond the scope of this article. Such are the determination of the optimum trajectory for entering orbit, the method of specifying this trajectory or of controlling it during flight, means of communication with the ship during its flight, the probability of a destructive collision with a meteor, and the technique of descent to Earth. A large amount of work and calculation will be required before such a flight can be practically carried out.

1. An interesting special case of a rocket-satellite is the so-called “stationary satellite,” whose angular velocity is equal to the angular velocity of the Earth’s rotation. Such a rocket, moving in an orbit lying in the plane of the equator, will occupy a stationary position with respect to an observer on Earth. It has been pointed out that a stationary satellite may be a convenient relay station for short-wave communication over almost half the Earth’s surface. From equations (107), (108), and (110) one can find that a stationary satellite must be located at a distance of 35,800 km from the Earth’s surface, i.e., about six Earth radii or one tenth of the distance to the Moon. Its orbital velocity is \(3110\) m/sec and the total energy per unit mass required to reach this orbit is \(6.07 \cdot 10^{10}\) joule/kg.

2. Another interesting example of a calculation performed by the same method is the determination of the minimum energy required for complete escape from the Earth’s attraction. With the aid of equation (110), this total energy is found to be \(6.55 \cdot 10^{10}\) joule/kg, only a little greater than for a stationary satellite. This energy corresponds to the well-known “escape velocity,” equal to \(11.2\) km/sec.

3. As a limiting case, one may consider escape from the Earth and from the entire solar system. The gravitational attraction of the Sun is sufficiently large to require a considerable expenditure of energy in order to overcome it. By contrast, the influence of all the planets may be neglected, since their total mass is about one thousandth of the mass of the Sun.

The energy per unit mass required to move a ship from the Earth’s surface to a point remote from the solar system is equal to the sum of the increase in potential energy of the terrestrial field plus the increase in potential energy of the solar field and minus the kinetic energy of the Earth’s orbital velocity. The first

and the second terms are respectively equal to \(\dfrac{GM_e}{R}\) and \(\dfrac{GM_s}{S}\), where \(G\) is the gravitational constant, \(M_e\) is the mass of the Earth, \(R\) is the radius of the Earth, \(M_s\) is the mass of the Sun, and \(S\) is the distance from the Earth to the Sun. The third term is equal to \(\dfrac{1}{2}GM_s/S\). The kinetic energy associated with the rotation of the Earth about its axis may be neglected. Thus, the required energy per unit mass will be

\[ E=\frac{GM_e}{R}+\frac{GM_s}{S}-\frac{\frac{1}{2}GM_e}{S} =\frac{GM_e}{R}\left[1+\frac{M_sR}{2M_eS}\right] \tag{111} \]

or, since \(g_0=\dfrac{GM_e}{R^2}\),

\[ E=g_0R\left[1+\frac{1}{2}\frac{M_s}{M_e}\frac{R}{S}\right]. \tag{112} \]

But \(\dfrac{M_s}{M_e}=3.32\cdot 10^5;\quad \dfrac{R}{S}=4.26\cdot 10^{-5}\) and \(g_0R=6.55\cdot 10^{10}\) erg/kg.

Hence the value of \(E\) is found to be \(52.9\cdot 10^{10}\) erg/kg. In the next two sections the characteristics of rockets required to obtain such energies are considered.

42. General theory of multistage rockets

The idea of a multistage rocket arises naturally in view of the difficulty of producing a single rocket of sufficiently light construction capable of acquiring the velocity necessary for escape from the Earth’s attraction. Of course, this may not apply to rockets using atomic energy. In Section 41 it was shown that the required velocities are of the order of \(7500\) m/sec and higher. Taking this value, one can, on the basis of (101), calculate the required characteristics for an extra-atmospheric rocket.

One of the highest values of the exhaust velocity obtained as a result of a chemical reaction corresponds to the oxygen–hydrogen combination and is about \(3180\) m/sec at sea level (see Table IV in the first article). It may be assumed, in accordance with the considerations developed in Section 10, that the average value of the exhaust velocity during flight beyond the atmosphere may be 20% higher than the value at sea level. This effect depends on the decrease of back pressure at high altitudes, and its magnitude depends on the working pressure in the chamber, the shape of the nozzle, and the flight trajectory.

Substituting in (101) the value \(c=3750\) m/sec and assuming the burning time \(t_p=100\) sec, we find that the rocket velocity of \(7500\) m/sec can be attained only when the relative mass of propellant \(\zeta\) exceeds \(0.895\). This means that the mass of the rocket with payload

payload, after subtracting the propellant, should not exceed 10% of the total mass. Still higher velocities, corresponding to cases 2, 3, and 4 of Section 41, require an even lighter structure. A detailed analysis of the factors determining the structural weight of the rocket is not the purpose of this article; it is sufficient only to point out that for the “V-2” rocket the factor \(\xi\) is equal to 0.70, in order to estimate the difficulty of practically realizing a rocket having \(\xi = 0.90\) or still higher.

Fig. 51. One possible arrangement of the component parts of a four-stage rocket.

Labels in the figure:

• Payload (instruments, radio, automatic pilot, etc.)
• Annular oxygen tank
• Launch tube
• Annular tank
• Oxygen feed line
• Hydrogen tank
• Propellant valves
• Rocket motor
• Nozzle
• Guide vanes
• Fourth stage
• Third stage
• Second stage
• First stage

In a single-stage rocket the maximum attainable velocity is sharply limited by the fact that the energy developed by the motor is expended on accelerating the entire mass of the rocket casing even during the period when the greater part of this casing has been emptied and is already unnecessary. Hence arises the idea of a rocket consisting of several stages, operating independently and containing their own separate supply of propellant1. After this supply has been consumed, the corresponding stage is discarded and the remaining part is set in motion by the next stage. The last, smallest stage carries the payload (Fig. 51).

The analysis of the operation of a multistage rocket can be carried out by the method developed in Section 40. Let the number of stages be \(N\), and let the last stage carry a payload of mass \(M_l\). We shall call the payload coefficient \(\lambda\)2 for each stage the ratio of the load carried

by the rocket, to the mass of the rocket at the moments of the beginning of combustion of the given stage. It can be shown^10b that the optimum result is obtained for values of \(\lambda\) that are the same for all stages. Suppose, for example, that a three-stage rocket carries a payload of \(100\ \mathrm{kg}\) in the third stage and that the initial mass of this stage is \(500\ \mathrm{kg}\). Consequently, the payload coefficient of the third stage is 0.20. This third stage is launched by means of the second stage, which has an initial mass of \(2500\ \mathrm{kg}\). The value of \(\lambda_2\) is also 0.20. The total mass of the first stage and, at the same time, the mass of the whole composite rocket is \(12\,500\ \mathrm{kg}\), so that the mass of the second stage amounts to 0.20 of the total mass. In general form this condition leads to the relation

\[ M_0^{(1)}=\lambda^{-N}M_1, \tag{113} \]

where \(M_0^{(1)}\) is the initial mass of the first stage, in other words, the total mass of the rocket.

According to the definition, the total mass \(M_0^{(1)}\) of the first stage includes the mass \(M_0^{(2)}\) of the second stage, the mass \(M_p^{(1)}\) of the propellant in the first stage, and the mass \(M_e^{(1)}\) of the casing of the first stage. In general, for the \(n\)-th stage the relation is

\[ \left. \begin{aligned} M_0^{(n)}&=M_0^{(n+1)}+M_p^{(n)}+M_e^{(n)},\\ &\quad 1\leq n\leq N,\\ M_0^{(N+1)}&=M_e. \end{aligned} \right\} \tag{114} \]

where

and

It may be assumed that each stage of one and the same rocket can be made with the same degree of effective utilization of the casing (apparently this corresponds closely to reality, except perhaps for small rockets with a mass of less than \(50\ \mathrm{kg}\)). This assumption leads to the definition of the coefficient \(\varepsilon\), which we shall call the structural factor, the same for all stages*)

\[ \varepsilon=\frac{M_e^{(n)}}{M_e^{(n)}+M_p^{(n)}};\quad 1\leq n\leq N; \tag{115} \]

the structural factor for the “V-2” is 0.24, which is the lowest figure achieved up to the present time.

The relative mass of the propellant \(\zeta\), defined in Section 40, is expressed in terms of \(\varepsilon\) and \(\lambda\) by the equation

\[ \zeta=(1-\lambda)(1-\varepsilon). \tag{116} \]

*) The symbol \(\varepsilon\) also had another meaning in Section 9.

Figure 52 schematically shows the various masses entering into the definitions of \(\lambda\), \(\varepsilon\), and \(\zeta\).

According to equation (101), the increase in velocity imparted by each stage of the rocket is determined by the equality:

\[ v_n - v_{n-1} = -c_n \ln [\varepsilon(1-\lambda)+\lambda] - g_0 t_n, \tag{117} \]

where \(v_n\) is the velocity of the rocket at the end of combustion of the \(n\)-th stage, \(c_n\) is the mean exhaust velocity during combustion of the \(n\)-th stage, and \(t_n\) is the duration of this combustion. Since the initial stages are sufficiently large to reduce the drag to a negligible value, and the last stages operate outside the atmosphere, the term depending on air resistance is not included in (117). It is assumed that the flight of the rocket during combustion takes place essentially vertically and that, without significant error, the value \(g_0\) corresponding to sea level may be adopted for all stages. Each stage is ignited immediately after the combustion of the preceding one is completed and is discarded immediately after being expended. A detailed analysis of the significance of these assumptions was carried out by Malina and Summerfield. The velocity of the rocket \(v_N\) at the end of combustion of the last stage is obtained from equation (117) by simple summation:

\[ v_N = -Nc \ln [\varepsilon(1-\lambda)+\lambda] - g_0 t_p, \tag{118} \]

where \(t_p\) is the total burning time and \(c\) is the mean value of the exhaust velocity over the entire burning time.

Fig. 52. Distribution in a multistage rocket of the masses entering into the definition of the relative propellant mass \(\zeta\), the structural factor \(\varepsilon\), and the payload coefficient \(\lambda\).

Two basic equalities (113) and (118) may be written in dimensionless form, by introducing the ratio of total masses \(G\) and the ratio of velocities \(S\), in the following form:

\[ G=\frac{M_l^{(1)}}{M_l}=\lambda^{-N}, \tag{119} \]

\[ S=\frac{v_n+g_0 t_p}{c}=-N\ln[\varepsilon(1-\lambda)+\lambda]. \tag{120} \]

The meaning of these equations becomes clear if one considers the general course of selecting a rocket for specified purposes. The purpose of the rocket determines \(v_N\), and the choice of propellant determines the magnitude \(c\), thereby predetermining the value of the velocity ratio \(S\). Accumulated experience in rocket design usually determines the value of \(\varepsilon\) with sufficient accuracy. The choice of \(S\) and \(\varepsilon\) makes it possible to calculate, from (120), various combinations of \(\lambda\) and \(N\), and their combined influence on the mass ratio is determined by (119). In general, the greater the number of stages, the greater the payload coefficient \(\lambda\) and the smaller the required mass ratio \(G\).

Fig. 53. Graph of the payload coefficient \(\lambda\) and the ratio of total masses \(G\) in a multistage rocket, as a function of the number of stages \(N\). Two typical values of the velocity ratio \(S\), equal to 3 and 6, and a structural-efficiency factor \(\varepsilon=0.25\), are taken.

Quantitatively these effects can be determined with the aid of Figs. 53 and 54, on which equations (119) and (120) are represented graphically for various values of \(S\) and \(\varepsilon\). They make it possible to draw several interesting generalizations regarding multistage rockets.

a) For a rocket of a definite purpose, the required mass of the rocket is always proportional to the mass of the payload, irrespective of how small the latter is in comparison with the former. If a payload of \(100\ \text{kg}\) requires a rocket of \(10\,000\ \text{kg}\), then a payload of \(200\ \text{kg}\) will require \(20\,000\ \text{kg}\).

b) For given values of \(\lambda\) and \(\varepsilon\) the number of stages is directly proportional to the velocity ratio, and the required total mass depends

exponentially on the ratio of velocities or on the number of stages. Thus, for example, for a specified payload, if in order to attain a velocity of 7500 m/sec a three-stage rocket weighing \(1000\,M_1\) is required, then in order to attain 15,000 m/sec a six-stage rocket weighing \(1\,000\,000\,M_1\) is required.

Fig. 54. Graph of the payload coefficient \(\lambda\) and the total-mass ratio \(G\), analogous to Fig. 53, for an improved structural efficiency factor \(\varepsilon = 0.20\).

c) The total mass decreases as the number of stages increases, but this decrease becomes negligible for a large number of stages. It can be shown mathematically that, for constant \(\varepsilon\) and \(S\) in equations (119) and (120), \(G\) tends with increasing \(N\) to an asymptotic value. In reality, the number of stages \(N\) should not be chosen too large, since the complexity and multiplicity of the mechanisms cause an undesirable increase in the structural factor \(\varepsilon\). From an examination of Figs. 53 and 54 it follows that favorable combinations of parameters are obtained for values of \(\lambda\) lying between 0.20 and 0.40.

43. Numerical examples of multistage rockets

In Section 41, a calculation was made of the energy required to break away from the Earth’s attraction for four special cases. In Section 42, a method was set forth that makes it possible to compute the velocity at the end of burning of a staged rocket. Then, using the analysis given in Section 40, one can obtain a general expression for the altitude reached at the end of the rocket’s burning. Further, an expression can be written for the total energy of the rocket at the end of burning. By equating this last expression to the required values of energy obtained in Section 41, one can obtain various characteristics of rockets intended for departure from the Earth.

There is no need to present here the rather complicated complete analysis of this kind. An approximate expression for the total energy per unit mass can be obtained as follows. Suppose that the trajectory during the rocket’s operation is directed vertically upward with an average acceleration \(\nu g_0\), and that the force of gravity is constant during the rocket’s burning time. It is easy to derive that the incre-

increment of kinetic energy is then \(\upsilon\) times greater than the increment of potential energy. The total increment of energy can then be written as

\[ E=\frac{\upsilon+1}{\upsilon}\left[\frac{1}{2}(v_N+v_0)^2-\frac{1}{2}v_0^2\right], \tag{121} \]

where \(v_0\) is the initial velocity, or the launch velocity of the first stage, and \(v_N\) is determined by equation (118). With the aid of equation (121) the value of \(v_N\) can be computed for each of the special cases considered.

In this section numerical examples are given for the case \(\upsilon=5\), although it is obvious that the exact value of the acceleration is not of great importance, provided only that \(\upsilon \gg 1\). In any case, very large accelerations should be avoided, since they may cause damage to the rocket as a result of the high temperatures developing from friction against the air. For the first three cases considered in Section 41, the initial velocity \(v_0\) is taken equal to zero. In the fourth case—complete escape from the solar system—the initial velocity is set equal to the orbital velocity of the Earth about the Sun, \(29.9\ \text{km/sec}\). The values of \(v_N\), calculated from equation (121), and the corresponding burning times are given in Table X.

As a first example let us consider a rocket—an Earth satellite—using oxygen and hydrogen as the propellant. As in Section 42, it may be assumed that the mean exhaust velocity during burning reaches \(3800\ \text{m/sec}\), which exceeds by about 20% the sea-level value given in Table IV and corrected in accordance with the note given there. Owing to the low mean density \((0.24)\) of this mixture, tanks, pipelines, valves, pumps, and so forth of larger dimensions are required than for heavier propellants, such as alcohol—oxygen. Therefore the structural factor \(\varepsilon\) should in all probability be higher than in the “V-2” rocket. A reasonable value is \(0.33\). The values \(v_N\) and \(t_p\) are taken from the first row of Table X. Then \(c=3800\ \text{m/sec}\), \(g_0=9.8\ \text{m/sec.}\), \(t_p=150\ \text{sec}\), \(v_N=7800\ \text{m/sec}\), and from equation (120),

\[ S=\frac{8700+150\cdot 9.8}{3800}=2.44. \]

Let us take \(N=4\) stages. Then, substituting the value \(\varepsilon=0.33\) in equations (119) and (120), we find \(\lambda=0.318\) and \(G=98.0\).

Let us define the payload as consisting of instruments and a radio transmitter of \(45\ \text{kg}\). Let us note again that the required mass of the rocket is proportional to the payload mass. Thus, in imagining such a flight, one must think of microinstruments, microtransmitters, and micro-control devices. The total mass of the rocket, according to (113), is

\[ M_0^{(1)}=98.0\times45=4430\ \text{kg}. \]

The mass of propellant in the first stage of the rocket is obtained from (116), namely

\[ \zeta=(1-0.318)(1-0.330)=0.457, \]

so that

\[ M_{p}^{(1)}=0.457\times 4430=2025\ \text{kg}. \]

The assumption that \(\varepsilon\), \(\lambda\), and \(\nu\) are the same for each stage leads, on the basis of equation (96), to the fact that the burning time

Table X

Four characteristic rocket missions

is also the same for each stage. Therefore the burning time of the first stage is equal to one quarter of the total time of 150 sec, or 37.5 sec. The thrust of the first stage is obtained as

\[ F_1=\frac{M_{p}^{(1)}}{t_1}\,c =\frac{2025\cdot 7800}{9.8\cdot 37.5} =21000\ \text{kg}. \]

The acceleration at launch is equal to

\[ a_{1\,\text{init.}}=\frac{21000-4430}{4430}=3.7\,g_0, \]

*) The larger energy corresponding to case IV is calculated relative to the Sun; the velocity is calculated relative to the Earth.

**) The velocity at the end of burning is calculated on the assumption that the mean acceleration at each stage is approximately \(5g_0\).

***) In escaping from the solar system the rocket velocity must be directed tangentially to the Earth’s orbit in order to make use of the orbital velocity of 29.9 km/sec.

****) The flight trajectory during burning in case I is not vertical, but is curved in order to enter orbit.

PHYSICS OF THE ROCKET

Table XI

Characteristics of eight rockets with a payload of 45 kg

while the acceleration at the end of combustion of the first stage is

\[ a_{1\text{end}}=\frac{21000-2405}{2405}=7.7\,g_0 . \]

The approximate size of the rocket can be estimated if one specifies the average density of the rocket in its loaded state. In view of the very low density of hydrogen, we take the overall density of the rocket to be 0.27. Then, taking the ratio of length to radius to be 10.0, considering the rocket cylindrical, and neglecting the volume and mass of the stabilizer, we find the approximate dimensions of the rocket: length 12.8 m, diameter 1.28 m. Figure 51 is a schematic drawing of such a four-stage rocket.

In the same way the characteristics of other multistage rockets can be calculated. The basic characteristics for eight types of rockets are given in Table XI. The payload is everywhere taken to be 45 kg. The mean density for an oxygen-hydrogen rocket is taken as 0.27, and for an acid-aniline rocket as 0.80. The factor \(\varepsilon\) is taken as 0.25 for acid-aniline rockets and 0.33 for oxygen-hydrogen rockets. In all cases the ratio of length to diameter is taken to be 10. The average exhaust velocity for acid-aniline rockets is taken as 2300 m/sec.

Summing up, we become convinced that the design and construction of a rocket capable of breaking away from the Earth’s attraction are at present entirely possible technically. With the aid of multistage rockets it is possible to attain the required colossal speeds without waiting for the appearance of any fantastically light structures or for the use of atomic energy. The importance of choosing a propellant having a high exhaust velocity and a high density has become clear. The central task is the lightening of the construc-

...and thoughtful economical use of the given magnitude of payload. But even these improvements are not necessary: the present state of rocket technology, as evidenced by the “V-2,” is already sufficient for solving the problem posed, since the matter concerns setting the rocket in motion.

ROCKETS USING ATOMIC ENERGY

44. Atomic Energy

The appearance of atomic energy in a technically usable form and in amounts exceeding by millions of times the energy released in ordinary chemical reactions has aroused special interest in the question of using it in rockets. As was indicated at the beginning, for the effective operation of a rocket it is necessary to strive to reduce the mass of substance expended simultaneously with the release of an enormous amount of energy. A rocket with a thrust of 450 kg and an exhaust velocity of 1900 m/sec, operating with a thermal efficiency of 40% (which corresponds to a chamber pressure of 21 kg/cm² at sea level), expends chemical energy at a rate of approximately 2500 kcal/sec. With the same thrust and an even greater exhaust velocity, the energy expenditure will be still higher and is determined by the relation

\[ p=\frac{Fa}{2\eta_i}, \tag{122} \]

where \(p\) is power, \(F\) is thrust, \(a\) is the exhaust velocity*) and \(\eta_i\) is the thermal efficiency.

This is not the place to examine nuclear reactions in detail. A very clear exposition of the general theory of nuclear reactions has been given by Morrison\(^{12}\), and a concise survey of questions of atomic energy has been made by Tzvin\(^{13}\).

Chemical energy is released during the rearrangement of electrons (usually only the outermost electrons of the atom) when atoms are reunited into molecules. The ionization energy of the hydrogen atom is about 13 eV, and this magnitude may be taken as the order of magnitude of the energy released in any elementary chemical process. Atomic energy is released through the rearrangement of protons and neutrons in the atomic nucleus. But the binding energy of one neutron or proton in the nucleus is about \(7\cdot 10^6\) electron-volts. From this the energetic superiority of nuclear reactions over chemical ones is immediately evident.

The nuclear reaction that first opened the way to the technical utilization of atomic energy was the fission of the nucleus \(^{235}_{92}\mathrm{U}\).

*) We shall now denote it by \(a\), since \(c\) is the generally accepted notation for the speed of light. These symbols should not be confused with \(a\), the speed of sound, and \(c\), the effective exhaust velocity in the preceding sections.

This reaction may be written in the form:

\[ {}^{1}_{0}n+{}^{235}_{92}\mathrm{U}\longrightarrow \text{primary fission products}+180\ \mathrm{MeV}. \]

\[ \text{Primary products}\longrightarrow \text{final products}+20\ \mathrm{MeV}. \]

With regard to these processes we must note two circumstances. First, the neutron causing the fission of uranium need not introduce kinetic energy into the nucleus. The very penetration of the neutron into the nucleus causes an increase in the energy of the nucleus by \(6.4\ \mathrm{MeV}\), whereas for fission to be possible only \(5.3\ \mathrm{MeV}\) is required. This circumstance is of great importance, since only slow neutrons have an appreciable probability of being captured by a uranium nucleus. Second, the decay products are not homogeneous, but are distributed among a wide variety of kinds of atomic nuclei, while the sum of the charges and mass numbers remains unchanged. Thus, for example, the primary fission products of one atomic nucleus may be \({}^{141}_{53}\mathrm{J}+{}^{95}_{39}\mathrm{Y}\), and of another, \(2\,{}^{118}_{46}\mathrm{Pd}\). Further, it turns out that the primary products are not stable, since they have an excess of neutrons in comparison with “normal” nuclei of the same charge, and undergo decay with emission of neutrons or electrons until they become a stable nucleus. Let us write, for definiteness, the reaction which apparently has the greatest probability:

\[ {}^{1}_{0}n+{}^{235}_{92}\mathrm{U}\longrightarrow \left({}^{236}_{92}\mathrm{U}\right)\longrightarrow 2\,{}^{1}_{0}n+{}^{140}_{53}\mathrm{J}+{}^{94}_{39}\mathrm{Y}+179\ \mathrm{MeV}, \]

\[ {}^{140}_{53}\mathrm{J}\longrightarrow{}^{140}_{58}\mathrm{Ce}+5e^{0}_{-1}+4\ \mathrm{MeV}, \]

\[ {}^{94}_{39}\mathrm{Y}\longrightarrow{}^{94}_{40}\mathrm{Zr}+e^{0}_{-1}+4\ \mathrm{MeV}, \]

where the transformation of \(\mathrm{J}\) into \(\mathrm{Ce}\) passes through the intermediate stages \({}^{140}\mathrm{Xe}\), \({}^{140}\mathrm{Cs}\), \({}^{140}\mathrm{Ba}\), \({}^{140}\mathrm{La}\). In calculations we usually assume that

\[ {}^{1}_{0}n+{}^{235}_{92}\mathrm{U}\longrightarrow 2\,{}^{118}_{46}\mathrm{Pd}+170\ \mathrm{MeV} \]

and

\[ {}^{118}_{46}\mathrm{Pd}\longrightarrow{}^{1}_{0}n+{}^{117}_{50}\mathrm{Sn}+4e^{0}_{-1}+15\ \mathrm{MeV}. \]

On this basis the kinetic energy of the palladium nucleus is equal to \(85\ \mathrm{MeV}\) (since, for equal masses, the law of conservation of momentum requires an equal distribution of energy), and the particle has a velocity of \(1.2\cdot 10^{9}\ \mathrm{cm/sec}\).

45. Relativistic mechanics of the rocket

The application of relativistic mechanics to the rocket was considered by Ackeret\(^{14}\). Usually the velocity of the fission products is less than \(3\cdot 10^{9}\ \mathrm{cm/sec}\), and relativistic effects are not yet appreciable. Nevertheless, it is interesting to see how the behavior of the “classical” rocket changes under the influence of these effects. This investigation, incidentally, will indicate the region in which one may freely use the classical theory.

1) Static thrust force. The usual relativistic definition of force as the time derivative of momentum gives for the thrust force (in vacuum)

\[ F=\dot m_0 a, \tag{123} \]

where \(\dot m_0\) is the rate of decrease of the rocket’s rest mass, equal to the mass flux in the outflowing jet with respect to a coordinate system at rest relative to the rocket. Let us note that if \(\dot m_{0e}\) is the rest mass carried away per unit time by the outflowing jet, then

\[ \dot m_0=\frac{\dot m_{0e}}{\sqrt{1-\dfrac{a^2}{c^2}}}. \tag{124} \]

Equation (123) has exactly the same form and the same meaning as the equation for the thrust force of a rocket in classical mechanics.

Fig. 55. Coordinate system and notation used in considering the relativistic motion of a rocket. The axes are at rest relative to the rocket in its initial stationary position.

The decrease of the rocket’s rest mass consists of two terms: (a) the amount of inertial mass given to the outflowing jet, \(\dot m_{0e}\), and (b) the mass “converted” into the energy necessary to impart acceleration to the mass leaving the rocket. Since it is the mass of the rocket, and not the expenditure of rest mass in the outflowing jet, that is of interest, equation (123) has the same meaning as in the classical theory.

2) The rocket equation. Although the static equation for thrust force is retained when relativistic mechanics is applied to a rocket, the general character of its motion changes. Its analysis is rather complicated; however, Ackeret developed an ingenious scheme for computing the final velocity of a rocket (in the coordinate system in which the rocket was initially at rest), expressed as a function of the ratio \(\zeta\) of the weight of the propellant to the total weight of the rocket, and of the exhaust velocity relative to the rocket. We shall present this calculation.

Consider a rocket in empty space in the absence of external forces or fields. We observe the motion in a coordinate system in which the rocket was initially at rest. Suppose that the mass flows out to the right with velocity \(u_2\), while the rocket moves to the left with velocity \(u_1\). The rest mass of the rocket at some instant is \(m_{01}\), and the rest mass of a small part of the outflowing ...

jet \(dm_{02}\), as shown in Fig. 55. For brevity, denote

\[ \sqrt{1-\frac{u_1^2}{c^2}} \quad\text{and}\quad \sqrt{1-\frac{u_2^2}{c^2}} \]

by \(k_1\) and \(k_2\), respectively.

The equations of motion of the system are the following equations: conservation of energy

\[ d(m_{01}k_1c^2)=k_2c^2\,dm_{02} \tag{125} \]

and conservation of momenta

\[ d(m_{01}u_1k_1)=k_2u_2\,dm_{02}. \tag{126} \]

From the formula for relativistic addition of velocities,

\[ u_2=\frac{a-u_1}{1-\dfrac{u_1a}{c^2}}, \tag{127} \]

where \(a\) is the (constant) velocity of outflow relative to the rocket. Eliminating \(k_2dm_{02}\) from (125) and (126), we obtain

\[ u_2\,d(m_{01}k_1)-d(m_{01}k_1u_1)=0. \tag{128} \]

Eliminating, with the aid of (127), \(u_2\) from equation (128), we find

\[ (a-u_1)d(m_{01}k_1)+\left(1-\frac{u_1a}{c}\right)d(m_{01}k_1u_1)=0. \tag{129} \]

The last expression can without great difficulty be put into the simpler form:

\[ \frac{dm_{01}}{m_{01}}=-\frac{1}{a_1}\cdot\frac{du_1}{1-\dfrac{u_1^2}{c^4}}. \tag{130} \]

Let \(v\) be the final velocity of the rocket. If \(W_p+W_0\) is the total weight of the rocket before ignition and \(W_0\) is its final weight after the expenditure of all the propellant, then the ratio of the weight of the propellant to the initial total weight is

\[ \zeta=\frac{W_p}{W_0+W_p}. \tag{131} \]

Integrating equation (130) then gives

\[ (1-\zeta)=\left[\frac{c-v}{c+v}\right]^{\frac{c}{2a}} \tag{132} \]

or

\[ \frac{v}{c}=\frac{1-(1-\zeta)^{\frac{2a}{c}}}{1+(1-\zeta)^{\frac{2a}{c}}}. \tag{133} \]

This equation must be compared with the classical equation for a space in which there is no force field [cf.

equation (101)]

\[ \frac{v}{a}=-\ln(1-\zeta). \tag{134} \]

The general character of this result is shown in Fig. 56. We see that the limiting case \(\zeta=1\) corresponds to the speed of light in the relativistic theory and to infinite velocity in the classical theory.

In Fig. 56 the following circumstances may be noted. For \(a=0.1c\) the difference between the classical and relativistic values is small, even for values of \(\zeta\) close to unity. For an exhaust velocity equal to \(0.5c\), significant differences appear already at \(\zeta=0.5\). The limiting case \(a=c\), representing the maximum exhaust velocity possible according to the relativistic theory, is also shown in Fig. 56, since it is of some interest. In general, the relativistic correction to the classical theory always decreases the rocket velocity after the motor has finished operating.

1. The ratio \(\dfrac{v}{c}\), where \(v\) is the final velocity of the rocket in force-free space, as a function of the relative mass of propellant \(\zeta\).

46. Problems of the Utilization of Atomic Energy

As we have seen, fission of the nucleus \(U^{235}\) gives rise to particles having a velocity of \(1.2\cdot 10^9\ \mathrm{cm/sec}\). This is less than one tenth of the speed of light; as we have established, at such exhaust velocities classical mechanics remains applicable up to mass ratios very close to unity.

1) Ideal rocket

Let us suppose that the rocket consists of a main body, covered at the rear by a layer of fissionable material. To simplify the discussion, let us assume that in the reaction one half of the particles produced rushes backward, forming the exhaust jet, while the other half is directed straight forward and imparts its momentum to the rocket. Since no method is known for “reflecting” the fast particles produced in nuclear fission, we must suppose that these particles are stopped in the material of the rocket. We shall assume that the mass of the rocket is large in comparison with the mass of the propellant, the validity of which we shall soon verify.

The law of conservation of momentum gives

\[ Mv=\frac{1}{2}ma, \tag{135} \]

where \(M\) is the mass of the rocket, \(m\) is the total mass of the material, \(v\) is the final velocity of the rocket, and \(a\) is the exhaust velocity; in our example \(a=1.2\cdot 10^9\ \mathrm{cm/sec}\). Let \(M=1\) ton and \(v=1.12\cdot 10^6\ \mathrm{cm/sec}\) (the velocity necessary to escape the Earth’s attraction). Then we find for \(m\) the value \(1.87\ \mathrm{kg}\)—a negligible quantity, less than \(0.2\%\) of the total mass of the rocket. The use of atomic energy for a single-stage rocket reduces the mass ratio \(\zeta\), required for escape from the Earth’s attraction, from \(0.96\) for hydrogen—oxygen to \(0.00187\) for uranium. However, a difficulty immediately arises.

Let us recall that one half of the fission products, carrying with them momentum as well as energy, remains trapped in the body of the rocket. The amount of energy which the rocket must absorb is equal to

\[ E=\frac{1}{2}\cdot\frac{1}{2}ma^2. \tag{136} \]

For the values of \(m\) and \(a\) found, \(E\) amounts to \(1.39\cdot 10^{10}\ \mathrm{kcal}\). If it is assumed that the rocket has a fairly high heat capacity of \(1\ \mathrm{kcal}/\mathrm{kg}\cdot\mathrm{deg}\), then the body of the rocket reaches a temperature of \(1.56\cdot 10^7\,^\circ\mathrm{C}\). A fantastic cooling problem! The situation is further complicated by the fact that the primary decay products, trapped in the body of the rocket, undergo radioactive decay with the release of approximately another \(0.25\cdot 10^9\ \mathrm{kcal}\). In reality, of course, the temperature developed in such a device would be much lower, since the heat capacity, for example, of steel, normally only \(0.1\ \mathrm{kcal}/\mathrm{kg}\cdot\mathrm{deg}\), would increase considerably after evaporation of the material.

2) Other Possibilities for Using Atomic Energy

We have just seen that the direct application of atomic energy for propelling a rocket is apparently impracticable. This raises the question of a technical device that would nevertheless make it possible to use this new source of energy. A convenient starting point for investigating the problem is the question of what the expelled jet of matter should consist of. Assuming that means have been found for flexible control over atomic energy, one may imagine a rocket using a stream of photons (light), a stream of charged particles, or a stream of neutral particles.

a) Photon rocket. Enormous efforts would have to be expended to create a sufficiently intense source of radiation that would make such a rocket possible. Although a jet with a small thrust can, after the lapse of sufficient time, impart any velocity to a rocket, near the Earth the thrust cannot be less than the weight of the rocket. The difficulty of this problem is explained by a simple example.

Suppose that the rocket “motor” is a layer of luminous material applied to the rear part of the rocket. As will be seen below, if exceptionally enormous temperatures are to be avoided, this layer must have a large area, in all probability forming a disk many times larger in diameter than the rocket itself. We shall assume that the radiation is emitted by the disk uniformly in all directions within a solid angle of \(2\pi\) steradians. Suppose that, in a thin layer near the surface of the disk, black-body radiation conditions exist. The radiation pressure \(p\) is determined by the formula\({}^{15}\)

\[ p=\frac{1}{3}\,bT^{4}, \tag{137} \]

where \(T\) is the absolute temperature, and the constant \(b\) has the value \(7.62\cdot 10^{-15}\ \mathrm{erg}\cdot\mathrm{cm}^{-3}\cdot\mathrm{deg}^{-4}\).

Then the energy density \(\rho\) in this layer is

\[ \rho=bT^{4}, \tag{138} \]

and the energy flows out of the rocket with the speed of light \(c\). Accordingly the power radiated per unit area will be

\[ P=bT^{4}c. \tag{139} \]

The loss of weight due to radiation is obtained by dividing the radiated power by \(\dfrac{c^{2}}{g}\). The results of the calculation performed according to these formulas are given in Table XII.

Table XII

Approximate calculation of a radiating disk serving for the propulsion of a photon rocket

These figures give an idea of the colossal power required to create the specified propulsive force. With a force of 450 kg

*) Atmospheric pressure.

a photon rocket consumes 400,000 times more power than an ordinary rocket with a chemical propellant [see equation (129)]. On the other hand, the mass consumption of the photon rocket is reduced enormously, and the specific impulse is \(10^7\) sec instead of 200 sec for an ordinary rocket. Thus, a given mass of “pure photon propellant” is capable of supporting, in the Earth’s gravitational field, a weight equal to its own initial weight for 115 days, whereas an ordinary chemical propellant can do so for 3.3 minutes.

The difficulty of constructing a photon rocket lies in the high surface temperatures required in order to obtain sufficient thrust per square meter of radiating surface. To obtain a radiation pressure of only \(50 \text{ kg}/\text{m}^2\), a surface temperature of \(37\,000^\circ\text{K}\) is needed. To reach atmospheric pressure, \(141\,000^\circ\text{K}\) is required. A lamp with a tungsten filament operates at a temperature of about \(3000^\circ\text{K}\) (the melting point of tungsten is \(3655^\circ\text{K}\)). Let us make the very optimistic assumption that it will be possible to construct a radiating disk operating at a temperature of \(10\,000^\circ\text{K}\). As is seen from Table XII, at the radiation pressure corresponding to this temperature, in order to support 1 ton of rocket weight in the Earth’s gravitational field, an area of about \(3400 \text{ m}^2\) is needed, or a circular disk 65 m in diameter. It seems unlikely that a disk of such area, operating at a high temperature and weighing less than 1 ton, could be constructed. Moreover, the rocket must, of course, also carry the energy source and the payload. Apparently, the “photon rocket” is not feasible in the near future.

b) Jet of charged particles. In recent years the technique of accelerating charged particles has undergone brilliant development, and a jet of electrons or protons with velocities approaching the speed of light can be obtained. However, such a rocket will acquire an electric charge, which will soon stop the outflow of a jet carrying charge of one sign. It may perhaps be possible to produce two jets of different signs acting jointly. However, energy will be required to separate the initial “hot” ions being accelerated, and this energy will prove lost. At present, beams of charged particles obtained in accelerating devices have too low an intensity to create any appreciable thrust.

c) Jet of neutral particles. From the preceding analysis it follows that, in practice, the possibility of using atomic energy for rocket propulsion is reduced to devices in which a nuclear reaction is used to heat a working fluid, which then, flowing out of an ordinary nozzle, creates thrust. Hydrogen would be a suitable working fluid, owing to its low atomic weight. Devices with a working fluid could be

also make it possible to rid the rocket of some radioactive products of the nuclear reaction, if the jet of liquid is made to carry them away with it.

3) Rocket with working liquid.

There is a fundamental difference between an ordinary rocket using a chemical propellant and a rocket combining atomic energy with a working liquid. An ordinary rocket of high efficiency (for example, a rocket intended to break away from the earth’s attraction) must operate at a mass coefficient \(\zeta\) close to unity. Indeed, a single-stage rocket using the energetically most advantageous reaction of hydrogen with oxygen must have \(\zeta = 0.96\) in order to break away from the earth’s attraction. This coefficient is so high that it makes the construction of such a rocket impracticable, and one must either build multistage rockets or seek a fuel giving a larger amount of energy. The fundamental reason for this difficulty is that the propellant is used both as the source of energy and as the mass whose outflow creates the thrust. It is possible, of course, to add inert material to the outflowing jet and thereby make the expenditure of mass and energy to some extent independent of one another. But since even the greatest exhaust velocities attainable in chemical reactions are insufficient for rockets of high efficiency, this consideration is of a purely academic character, and the indicated procedure finds application only to a very limited extent, for overcoming difficulties with cooling (see section 32).

An entirely different situation obtains in systems using atomic energy and a working liquid. It turns out that in this case there is an optimum amount of working liquid, while too small an amount of liquid is also unfavorable. This can be shown as follows. A nuclear-fission reaction, in view of the enormous amount of energy released, represents, on the scale of chemical reactions, almost pure “energy devoid of mass.” Indeed, the energy released by \(1\ \mathrm{kg}\) of \(U^{235}\), in primary fission, is \(1.7 \cdot 10^{10}\ \mathrm{kcal}\), whereas the reaction of hydrogen with oxygen releases only \(3.8 \cdot 10^{3}\ \mathrm{kcal}/\mathrm{kg}\). We have seen, in the example of the “ideal” rocket, that to break away from the earth’s attraction with 1 ton of load only \(1.9\ \mathrm{kg}\) of \(U^{235}\) is required. Therefore, in the following considerations we have full justification, when considering the expenditure of mass, for completely neglecting the mass of uranium in comparison with the mass of the working liquid. Of course, the quantity of uranium is of importance in considering the energy balance and determining the exhaust velocity.

Let us consider a rocket containing a payload \(W_0\), uranium in the amount \(W_u\), and working substance of weight \(W_p\). With the exception of pre-

of the conversion of part of the mass of uranium into energy*); otherwise nonrelativistic mechanics may be applied.

In a coordinate system fixed relative to the rocket, a small fraction \(\delta\) of the mass of uranium is converted into energy \(\delta W_u c^2\), determined by Einstein’s formula. This energy must be equal to the kinetic energy of the outflowing jet

\[ \delta W_u c^2=\frac{1}{2}W_p a^2 . \tag{140} \]

For the nuclear reaction in uranium \(\delta=0.000731\), i.e., less than \(0.1\%\) of the mass of uranium is converted into energy. We assume that there is no external pressure, so that the outflow velocity \(a\) corresponds to the total energy.

Neglecting the weight of the uranium, we may write the mass or weight coefficient in the form

\[ \zeta=\frac{W_p}{W_0+W_p}. \tag{141} \]

Solving (140) with respect to \(a\) and substituting the result, as well as equality (141), into (134), we obtain the final velocity of the rocket

\[ v=-c\sqrt{\frac{2\delta W_u}{W_p}}\, \ln\left(1-\frac{W_p}{W_0+W_p}\right) \]

or

\[ v=c\sqrt{\frac{2\delta W_u}{W_p}}\, \ln\left(1+\frac{W_p}{W_0}\right). \tag{142} \]

Thus it turns out that, for a fixed payload \(W_0\) and amount of uranium \(W_u\), which determines the energy reserve, \(v\) reaches a maximum at \(W_p=4W_0\). This occurs because \(W_p\) enters both the expression for the outflow velocity (under the square root) and the mass coefficient. Near the maximum the curve has a very flat form, and any value of \(\zeta\) exceeding \(0.5\) gives a value of \(v\) close to the maximum. Taking, as we did in the example for the “ideal rocket,” a payload of 1 ton, an amount of uranium of \(1.87\) kg, and adding to the system 4 tons of working fluid, we obtain a final velocity of \(4.0\cdot10^7\) cm/sec—considerably exceeding the velocity attainable in the absence of a working fluid, \(1.12\cdot10^6\) cm/sec.

It is interesting to note that the value \(\zeta=0.8\), which leads to the maximum final value of the velocity \(v\), is determined independently

*) This formulation is not entirely precise, but it is convenient, since it expresses the basic concepts without any ambiguity. More precisely, one should say that part of the total energy of the uranium is converted from mass, or from a special kind of potential energy, into energy in a more ordinary form. Of course, the liberated energy has the same mass that it had before the transformation. In reality, no mass or energy arises or disappears. See [16].

of the amount of energy delivered by the uranium (that is, of $\delta W_u c^2$), provided only that $W_u$ is small in comparison with $W_0$ or $W_p$. This assumption of the smallness of $W_u$ is contained in expression (141) and is always valid if excessively high temperatures are to be avoided.

The possibility of using this optimal value of $\zeta$ is, evidently, limited to a hypothetical case, since the mass of the boiler carrying out the reaction, of the protective shields, and of the useful load cannot be varied arbitrarily. Thus the only important variable available for choice is the mass of the working substance, and under these conditions the optimal amount of the latter is equal to four times the mass of the uncharged rocket. On the other hand, a decrease in the mass of the uncharged rocket as a result of some improvement in design, leading to a decrease in $\zeta$, increases the velocity. Increasing again the value of $\zeta$ to 0.8 by reducing the amount of working fluid, however, makes it possible to obtain, with the same mass of the uncharged rocket and the same consumption of uranium, an even greater final velocity.

In the preceding calculation it was tacitly assumed that the working substance can be heated to any desired temperature. However, there must exist a maximum permissible temperature from the point of view of overheating the apparatus. In such a case the ratio $\frac{W_u}{W_p}$ in equation (142) proves to be fixed. Then the operating conditions of the rocket turn out to be ordinary, and the final velocity increases continuously with an increase of the mass coefficient $\zeta$ even above the value 0.8.

It is appropriate to indicate here what exhaust velocities can be obtained in a system using nuclear fission plus gaseous hydrogen as the working substance, under the limitations imposed by temperature. Taking into account the dissociation of hydrogen molecules and assuming that the exhaust takes place into a vacuum, we may expect to obtain an exhaust velocity of 6500 m/sec if the gas can be heated to 2700°, and 11,400 m/sec if the heating can be brought to 5700°. The latter exhaust velocity is quite sufficient for escape from the earth’s attraction by a rocket with mass coefficient 0.5.

It should be noted that the exhaust velocity increases faster than $\sqrt{T}$. This favorable result is explained by the dissociation of the gas.

It would be premature to judge whether the problems connected with the enormous temperatures and radiation intensities accompanying nuclear reactions can be solved. In all probability, in the near future it will be precisely the temperature conditions that will hinder the application of atomic energy in rockets. Rockets substantially surpassing ordinary chemical rockets in their effectiveness will be able to appear only after the temperature difficulties in the use of atomic energy have been overcome.

References

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2. In Section 10 the symbol \(\lambda\) had an entirely different meaning. ↩