The Change in the Motions of the Earth and Spacecraft Launching

Revista Brasileira de Ensino de F´ısica, v. 35, n. 3, 3315 (2013) www.sbﬁsica.org.br

The change in the motions of the Earth and spacecraft launching - a college physics level analysis (As mudan¸casno movimento da terra e o lan¸camentode ve´ıculosespaciais - uma an´aliseno n´ıvelda gradua¸c˜aoem f´ısica)

Gengmin Zhang1

Key Laboratory for the Physics and Chemistry of Nanodevices and Department of Electronics, Peking University, Beijing, China Recebido em 25/1/2013; Aceito em 14/3/2013; Publicado em 26/9/2013

Both the translational velocity and the angular velocity of the Earth change during a spacecraft launching process, in which a spacecraft is accelerated from the ground and eventually sent into space. This article presents a systematic study of the role played by the changes in the translation and rotation of the Earth in spacecraft launching. Neglecting these changes, which inevitably arise in the interaction between the Earth and the spacecraft, there is an obvious conflict with the conservation laws of momentum and angular momentum. Nevertheless, this flaw in principle is not accompanied by any technically erroneous answers when college students solve the often-encountered exercise problems, thanks to the special reference frames students use. It is pointed out that the technical validity of the Earth-in-constant-motion approximation cannot be generalized to arbitrary reference frames. For example, the correct values of the second and third cosmic velocities cannot be found in an arbitrary reference frame if the velocity of the Earth is treated as a constant. In an arbitrary reference frame, the increase in the translational kinetic energy of the Earth, which is caused by the work done by the gravitational pull by the spacecraft, is not negligible if compared with the increase in the kinetic energy of the spacecraft. It is also demonstrated that the disparity in the energy consumed in launching a spacecraft from the ground along different directions cannot be well interpreted if the angular velocity of the Earth is treated as a constant. When the spacecraft is launched eastwards, the increase in its kinetic energy is partly gained, either directly or indirectly, at the expense of a decrease in the rotational kinetic energy of the Earth. Keywords: second cosmic velocity, third cosmic velocity, Earth motion, conservation of momentum, conservation of angular momentum, kinetic energy of the Earth.

Tanto a velocidade de transla¸cãoquanto a velocidade angular de rota¸cãoda Terra mudam durante o processo de um lan¸camento espacial, em que o ve´ıculoéacelerada a partir do solo e, finalmente, enviado para o espa¸co.Este artigo apresenta um estudo sistemáticoda influênciadas mudan¸casna transla¸cãoe na rota¸cãoda Terra sobre o processo de lan¸camento de um ve´ıculoespacial. Negligenciando essas mudan¸cas,háum conflito evidente com as leis de conserva¸cãode momento e momento angular. No entanto, esta falha, em princ´ıpio, nãoéacompanhada por respostas tecnicamente erradas quando os estudantes universitáriosresolvem esse tipo comum de exerc´ıcio,devido àutiliza¸cãode sistemas de referênciamuito particulares. E´ importante observar que a validade técnicada aproxima¸cãoda ”Terra em movimento constante” nãopode ser generalizada para sistemas de referênciaarbitrários. Por exemplo, os valores corretos da segunda e da terceira velocidades cósmicasnão podem ser encontrados em um sistema de referênciaarbitráriose a velocidade da Terra for tratada como uma constante. Em um sistema de referênciaarbitrárioo aumento da energia cinéticade transla¸cãoda Terra, cau- sado pelo trabalho realizado pela for¸cagravitacional da ve´ıculoespacial, nãoéinsignificante quando comparado com o aumento da energia cinéticado própriove´ıculo. Tambémfica demonstrado que a disparidade na energia consumida no lan¸camento de um ve´ıculoespacial a partir do solo ao longo de dire¸cõesdiferentes nãopode ser explicada se a velocidade angular da Terra for tratada como uma constante. Quando o ve´ıculoélan¸cadopara leste, o aumento da sua energia cinéticaéparcialmente ganho, quer direta ou indiretamente, àscustas de uma diminui¸cãoda energia cinéticade rota¸cãoda Terra. Palavras-chave: segunda velocidade cósmica,terceira velocidade cósmica,movimento da Terra, conserva¸cão do momento, conserva¸cãodo momento angular, energia cinéticada Terra.

1E-mail: [email protected].

1. Introduction from the Sun and the Moon, omitted, a reference frame in which the center of mass of the spacecraft-Earth As is well known, neglecting the motion of the Earth in system is either at rest or in uniform rectilinear mo- studying the motions of particles in the vicinity of the tion should be rigorously speaking an inertial reference Earth leads to contradictions in principle. The most frame. Hereafter, a frame in which the center of mass cited example seems to be that the momentum of the of the spacecraft-Earth system is at rest is termed the particle-Earth system is not conserved when the par- “S frame” and a frame in which the center of mass un- ticle is supposed to fall to a “motionless” ground [1]. dergoes uniform rectilinear motion is termed the “S’ Generally, however, this conceptual deficiency is not frame”. accompanied by technically detectable mistakes. This Of course, one can also opt to use a reference frame article systematically re-examines this topic from the in which the center of the Earth is either at rest or perspective of spacecraft launching. The changes in in uniform rectilinear motion. In this case, the refer- the motion of the Earth are studied during both the ence frame is non-inertial because of the gravitational accelerating process and escaping process of a space- pull received by the Earth from the spacecraft and iner- craft. To maintain a rigorous application of Newton’s tia forces should be introduced to guarantee conceptual second law, inertial reference frames, in which the cen- rigor [2]. The discussion in this article is restricted to ter of mass of the spacecraft-Earth system is either at inertial frames only. rest or in a uniform rectilinear motion, are employed. Changes in the motion of the Earth play arguably an 2.3. The exact meaning of the term “− GMem ” indispensable role in the process of spacecraft launch- Re ing. Omitting these changes could lead to erroneous In some textbooks, it is explicitly stated and even em- results. phasized that the gravitational potential energy is as- For clarity, the well-known definitions of the sec- sociated with the particle-Earth system instead of the ond and third cosmic velocities are repeated here. The particle alone [3]. Nonetheless, a student might still minimum velocity (or more unambiguously, speed) re- be dimly impressed that a particle alone has a gravi- quired for a spacecraft at the surface of the Earth to tational potential energy near the Earth. For example, escape the gravitational pull of the Earth is defined as according to Eq. (1), during the escape of the space- the second cosmic velocity (or the “escape velocity”). craft from the Earth, the gravitational potential energy increases from − GMem to zero at the expense of a loss The minimum velocity for a spacecraft at the surface Re of the Earth to escape the gravitational pull of the Sun in the kinetic energy of the spacecraft alone. That is, is defined as the third cosmic velocity. In these defini- the increase in the kinetic energy of the Earth does not tions, it has to be emphasized, the two cosmic velocities appear to play any role in the increase in the potential are both measured with respect to the Earth center. energy. (Hereafter, the word “increase” means the difference between the kinetic energies at two states, with 2. Some ambiguities in determining the the kinetic energy of the final state as the minuend and that of the initial state as the subtrahend. An “in- second cosmic velocity and their clar- crease” can be either positive or negative.) Therefore, ifications it is quite natural for a student to relate the potential energy − GMem solely to the spacecraft. Re 2.1. Possible ambiguities As common knowledge, the introduction of poten- The mass of the spacecraft, the mass and radius of the tial energy arises from the work done by conservative Earth are denoted by m, Me and Re, respectively. As forces. It should be emphasized that work is actually is usually taught in textbooks, the second cosmic veloc- done by a pair of forces in the above example. That ity (denoted here by v2) can be found by invoking the is, the gravitational pull exerted on the Earth by the theorem of kinetic energy: spacecraft also plays an indispensable role. The following discussion is restricted to the assumption that the − 1 2 −GMem force and counter force between two arbitrary particles 0 mv2 = . (1) 2 Re lie along the straight line that joins the two particles. Despite the extreme simplicity of this problem, sev- The work done by such a pair of force and counter force eral ambiguities could arise, including: (i) the reference between two particles is frame in which the problem is treated; (ii) the precise meaning of the term “− GMem ”; and (iii) the reason- Re · · · − ableness in neglecting the kinetic energy of the Earth. dW = F12 dr1 + F21 dr2 = F12 d (r1 r2) , (2) where dr and dr are respectively the elementary dis- 2.2. The appropriate reference frames 1 2 placements of particles 1 and 2 in an arbitrary reference Newton’s second law is valid only in an inertial refer- frame, F12 is the force exerted on particle 1 by particle ence frame. With all the external forces, e.g. those 2, and F21 that on particle 2 by particle 1. cabe¸calhodo t´ıtulo 3315-3

Let F12 = |F12|, ρ = r1 − r2, and ρ = |ρ|. Then [4] when measured in the S frame, the velocity of a spacecraft with the second cosmic velocity is Ve + v2, not dW = ±F12dρ. (3) v2.) Moreover, another equation derived from the con- Importantly, dρ is actually the increase in the dis- servation of momentum should be added tance between particles 1 and 2 and is independent of the choice of reference frame in Newtonian mechanics. Therefore, the work done by this force pair is indepen- m (Ve + v2) + MeVe = 0. (6) dent of reference frame and consequently, can be calcu- The process involving a spacecraft escaping the lated in any reference frame. Generally, the most con- Earth, as observed in the S frame, is illustrated in venient frame for the calculation is the frame in which Fig. 1. Two cases are described. In Fig. 1(a), both one of the two particles, e.g. particle 2, is at rest. In the initial velocity of the spacecraft and that of the this case, dr2 = 0 and Earth are along the straight line that joins the spacecraft and the Earth’s center. In Fig. 1(b), the spacecraft is launched along a tangent to the Earth’s surface. · · · dW = F12 dr1 + F21 dr2 = F12 dr1. (4) Fig. 1(a) provides the most direct visualization of this That is, calculating the work done by the force on problem whereas Fig. 1(b) illustrates the use of the particle 1 only in this particle-2-at-rest frame is equiv- Earth’s rotation in the actual launch, as will be dis- alent to calculating the work done by the force pair in cussed in section 4. an arbitrary reference frame. As commonly practiced in most textbooks, in the reference frame in which the Earth’s center is at rest, the work done by the Earth’s gravitational pull on the spacecraft is easily calculated to be “− GMem ”. From Re the above discussion, it is readily known that in the S frame the sum of the work done by the gravitational pull exerted on the spacecraft and that on the Earth is also “− GMem ”. Equivalently, the term “− GMem ” can Re Re be understood as the potential energy of the spacecraft- Earth system instead of that of the spacecraft alone. Consequently, in the S frame, the term − GMem Re should be equal to the increase in the kinetic energy of the spacecraft-Earth system instead of the kinetic en- Figure 1 - Launching a spacecraft with the second cosmic veloc- ergy of the spacecraft alone. According to K¨onig’sthe- ity as observed in the S frame. (The initial velocity of the Earth orem, the kinetic energy of the Earth, which is treated and the distance from the center of mass of the spacecraft-Earth as a rigid body here, is the sum of two parts, namely, system to the Earth’s center are both greatly exaggerated.) (a) its “translational kinetic energy” and its “rotational The spacecraft is launched along the straight line that joins the spacecraft and the Earth’s center; (b) The spacecraft is launched kinetic energy”. During the escape of the spacecraft, along a tangent to the Earth’s surface. the rotational kinetic energy of the Earth remains un- changed and is not taken into consideration when the 2.4. On the technical validity of omitting the theorem of kinetic energy is used. increase in the Earth’s translational ki- The escape of the spacecraft from the Earth is netic energy reached if its relative velocity with respect to the Earth is at least zero at inﬁnity. Because the center of mass Using Eqs. (5) and (6), the second cosmic velocity can of the spacecraft-Earth system is motionless in the S be found frame, after escape both the spacecraft and the Earth √ have (at least) zero velocity in the S frame. Thus 2G (Me + m) Eq. (1) should be replaced by v2 = , (7) Re [ ] from which the often-used approximate result can be − GMem − 1 2 1 2 comfortably obtained after dropping m = 0 m (Ve + v2) + MeVe , (5) Re 2 2 √ 2GMe where Ve denotes the initial velocity of the Earth in the v2 = . (8) S frame when escape has just commenced. (It might Re be necessary to remind possible freshman readers again As common practice, Eq. (8) is usually obtained that both the second and the third cosmic velocities are directly from Eq. (1), thus it should be interesting to measured with respect to the Earth center. Therefore, study the connection between Eqs. (5), (6) and Eq. (1). 3315-4 Zhang

In fact, after eliminating Ve using Eq. (6), Eq. (5) is speed of the spacecraft with respect to the Earth and rewritten as is independent of the reference frame. The right side of Eq. (10) separates into two parts: [ ] the increase in the kinetic energy of the spacecraft, de- ( )2 GMem 1 Me noted by ∆Ekm, and that of the Earth, denoted by − = 0 − m v2 + Re 2 m + Me ∆E . Taking into consideration m << M , their ex- [ ] ke e ( )2 pressions reduce to − 1 − mv2 0 Me . (9) [ ] 2 m + Me 1 ∆E = m u2 − (U + v )2 = km 2 c e 2 Now, Eq. (1) can be obtained from Eqs. (5) and (6) [ ] 1 2 using two approximations. First, the initial speed of the − m 2uc · v2 + v , (12) 2 2 spacecraft in the S frame, Mev2 , is approximated by m+Me and v2. Second, and more importantly,( the increase) in the 2 ( ) − 1 − mv2 1 Earth’s kinetic energy, 0 2 Me m+M , is omit- 2 − 2 · e ∆Eke = Me uc Ue = muc v2. (13) ted directly. Both approximations can be well justified 2 when the condition m << M is taken into considera- Apparently, ∆Eke is generally not negligible in com- e | · | 2 tion. parison with ∆Ekm unless uc v2 << v2. Conse- However, the situation is different when a reference quently, when solving the problem in the S’ frame, ne- frame in which the spacecraft-Earth system is in motion glecting the increase in the Earth’s kinetic energy during the escape of the spacecraft, i.e. treating the veloc- is used. Let uc denote the constant non-zero velocity of the center of mass of this system in the S′ frame. The ity of the Earth as constant, generally leads to a wrong velocity of the Earth at the beginning of the escape in result. The physical origin of this disparity in treating the the S’ frame is denoted by Ue. Apparently, the defini- Earth’s kinetic energy in the two frames can be read- tion of v2 demands both the velocity of the spacecraft ily understood by comparing the power applied to the and that of the Earth should be uc after the escape of the spacecraft. The process of escape is illustrated in Earth by the gravitational field with that applied to Fig. 2. Accordingly, Eqs. (5) and (6) should be revised the spacecraft, here denoted by Pe and Pm, respec- to state tively. The two gravitational forces here are identical in magnitude but opposite in direction. Furthermore, the velocity of the Earth in the S frame at a certain GM m 1 − e = (m + M ) u2 − moment is denoted by V and that of the spacecraft by e c | | | | [ Re 2 ] v. Obviously, V = V << v = v and

1 2 1 2 m (Ue + v2) + MeUe , (10) Pe V 2 2 = << 1. (14) Pm v and Therefore, the increase of the kinetic energy of the Earth is negligible. However, when one moves to the S’ m (Ue + v2) + MeUe = (m + Me) uc. (11) frame, the corresponding ratio becomes:

′ | | Pe uc + V ′ = γ | | , (15) Pm uc + v where γ is a factor determined by the velocity directions V of the spacecraft and the Earth. Although v << 1, | | uc+V is not necessarily a small quantity. That is, the |uc+v| increase in the kinetic energy of the Earth is not generally negligible in comparison with the increase in the kinetic energy of the spacecraft. Of course, when trying to determine the second cos- Figure 2 - Launching a spacecraft with the second cosmic veloc- mic velocity, few students (and teachers) would use ity as observed in the S’ frame. (For simplicity, uc and Ue are assumed to be along the same direction.) (a) The spacecraft is the S’ frame, because using such a frame only makes launched along the straight line joining the spacecraft and the the issue unnecessarily more troublesome. However, as Earth’s center; (b) The spacecraft is launched along a special shown in the next section, in order to determine the tangent to the Earth’s surface. third cosmic velocity, the S’ frame could be used, per- Equation (7), and thus Eq. (8), can still be obtained haps unconsciously. In this case, the correct answer from Eqs. (10) and (11). That uc does not appear in is not obtainable if the increase in the Earth’s kinetic the ﬁnal result is expected because v2 is the relative energy is erroneously omitted. cabe¸calhodo t´ıtulo 3315-5

3. Calculating the third cosmic velocity directly in the Solar System ( ) GM m 1 1 − e = mξ2 + M U2 − R 2 2 e e1 The velocity and speed of the Earth in the reference e[ ] frame ﬁxed to the Sun are: 1 1 m (U + v )2 + M U2 . (21) √ 2 e 3 2 e e GMs Ue = |Ue| = , (16) Rse Assuming the spacecraft escape time from the Earth

where Ms and Rse are the mass of the Sun and the is very short, the impulse due to the gravitational pull radius of the orbit of the Earth around the Sun, re- from the Sun during this time is negligible. Therefore, spectively. (The orbit is approximated by a circle.) the momentum of the spacecraft-Earth system is ap- The third cosmic velocity is denoted by v3 here. One proximately conserved ostensibly plausible approach to determine v3 directly in the ﬁxed-to-the-Sun reference frame is to resort to the theorem of kinetic energy for the spacecraft MeUe1 + mξ = MeUe + m (Ue + v3) . (22)

Taking m << Me and using Eqs. (16), (19), (21), GMem GMsm 1 2 − − = 0 − m (Ue + v3) , (17) and (22), v3 is found to be Re Rse 2 from which the wrong result √ 2GM v = ξ2 + U 2 − 2U ξ + e = 4 3 e e v = 1.38 × 10 m/s (18) Re 3 √( ) √ GM 2GM is obtained. 3 − 2 2 s + e = 1.66 × 104 m/s. (23) The introduction of a critical speed, ξ = |ξ|, defined Rse Re as the minimal speed required for the spacecraft in the Earth’s orbit to escape the Solar System, can establish 4. The change in Earth’s rotation in ac- a connection between Eq. (17) and the discussion in the previous section: given that celerating a spacecraft So far, the S frame has proved to be an appropriate 1 2 − GMsm mξ = 0, (19) reference frame in studying the spacecraft escape pro- 2 Rse cess from the Earth after attaining the necessary high Eq. (17) can be re-written as speed, e.g. v2, because the omission of the increase in the Earth’s translational kinetic energy in this refer- GMem 1 2 1 2 − = mξ − m (Ue + v3) . (20) ence frame makes no technical difference to the final Re 2 2 result. However, if the S frame is further used in de- In this treatment, the left side of Eq. (20) is un- termining the energy required to accelerate an initially derstood as the work done by the Earth’s gravitation grounded spacecraft to the same high speed, caution field on the spacecraft during the escape process. The must be taken again not to neglect the change in the right side is the increase in the kinetic energy of the motion of the Earth. spacecraft during this process. Right after the space- As is well known, in terms of energy saving, a spacecraft escapes from the Earth, i.e., when it is sufficiently craft should be launched eastwards, so that the speed far away from the Earth but still near the Earth’s or- of the Earth surface can be used [6]. However, to es- bit around the Sun, the residual speed needs to be at cape the Earth, irrespective of launch direction, the fi- least “ξ” for the spacecraft to further escape the Sun’s nal speed of the spacecraft after the acceleration stage gravitational pull. From the discussion in the preceding has finished is always (almost) v2 as observed in the S section, the reason behind the mistake can be readily frame. If the increase in the kinetic energy of the space- seen: − GMem should be understood as the total work craft alone is considered, the energy expended would Re done by the gravitational pull between the Earth and be independent of launch direction. That is, there is the spacecraft and is equal to the increase in the kinetic no difference in the energy consumed in launching east- energy of the spacecraft-Earth system instead of that of wards or westwards. As is discussed in this section, the spacecraft alone. That is, the erroneous omission of when studied in the S frame, this difference is actually the increase in the Earth’s kinetic energy leads to the incorporated into the change in the Earth’s rotation. wrong answer [5]. The argument is demonstrated in two scenarios. The Denoting the velocity of the Earth after the escape first is directly perceivable but unrealistic; the second of the spacecraft by Ue1, Eq. (20) is revised as has a more factual basis in modern astronautics. 3315-6 Zhang

4.1. Verne’s gun

Mev2 mv2 Verne envisaged a powerful gun that could send a space- vf = ± eˆ and Vf = ∓ e.ˆ (28) craft to the Moon. Impractical as this approach is, it Me + m Me + m is nonetheless interesting to study the process where For both eastward and westward launches, the in- a spacecraft is accelerated to the second cosmic veloc- crease in the spacecraft’s kinetic energy is ity by such a gun on the surface of the Earth in the S frame. For simplicity, it is assumed that the gun is in 1 2 − 1 2 rigid joint with the Earth. That is, the relative motion ∆Ekm = mvf mvi = ( 2) 2 between the gun and the Earth caused by the spacecraft 2 ( ) launching is neglected. Instead, it is imagined that the 1 Me 2 − 2 2 m v2 Ωi Re . (29) Earth and the gun jointly receives the recoil. 2 Me + m Let vi = vieˆ (vi = |vi|) and Vi = −Vieˆ (Vi = |Vi|), Similarly, in both eastward and westward launch- wheree ˆ is the eastward unit vector, denote the respec- ings, the increase in the Earth’s translational kinetic tive velocities of the spacecraft grounded at point “A” energy is on the Earth surface and the center of the Earth “O” in the S frame before the launching. In the S frame, 1 2 − 1 2 the center of mass of the spacecraft-Earth system “C” ∆Ekt = MeVf MeVi = ( ) 2 2 is at rest, thus 2 ( ) 1 m 2 − 2 2 Me v2 Ωi Re << ∆Ekm. (30) 2 Me + m mvi + MeVi = 0. (24) Furthermore, the recoil of the powerful gun in- Before launching, as depicted in Fig. 3, the space- evitably changes not only the velocity of the Earth’s craft is motionless with respect to the surface of the center, but also, more importantly, the angular veloc- Earth and its velocity in the S frame is purely the “ve- ity of the Earth. Neglecting external forces, the angu- locity of following”. For simplicity, the spacecraft is lar momentum of the spacecraft-Earth system is con- assumed to be on the equator. served. Also, the period of acceleration is approximated as being very short, so that the change in the spacecraft’s position during acceleration can be neglected. v = V + Ω × OA, v = −V + Ω R , (25) i i i i i i e The most straightforward reference point (not necessarily the most convenient one) for calculating angular where Ωi is the angular velocity of the Earth before momenta is the center of mass, i.e. point C. The axis launch, and Ωi = |Ωi|. Thus, that passes the Earth’s center O and is perpendicular to the equator plane is approximated as a principal axis mΩiRe MeΩiRe of inertia and the corresponding moment of inertia is Vi = and vi = . (26) Me + m Me + m denoted by I. Then,

CA × mvi + CO × MeVi + IΩi =

CA × mvf + CO × MeVf + IΩf , (31)

where Ωf is the angular velocity of the Earth after the accelerating process. Using conservation of momentum, Eq. (31) can be rewritten as

Figure 3 - Launch of an initially grounded spacecraft using · ± · Verne’s gun, as observed in the S frame. (a) Before launch; (b) Re mvi + IΩi = Re mvf + IΩf . (32) after an eastward launch; (c) after a westward launch. Equation (32) demonstrates that the most conve- When the accelerating process concludes, the space- nient reference point to calculate the angular momen- craft gains a relative velocity v2 with respect to the tum of the spacecraft-Earth system is actually the spa- Earth’s center. That is, tial point in the S frame that momentarily coincides with the Earth’s center O during the accelerating of

vf − Vf = v2 = ±v2e,ˆ (27) the spacecraft. Rearrangement of Eq. (32) yields where the upper sign corresponds to an eastward launch Rem Ω = Ω + (v ∓ v ) . (33) and the lower sign to a westward launch. Again, us- f i I i f ing momentum conservation, the ﬁnal velocities of the Now the increase in the rotational kinetic energy of spacecraft and the Earth’s center can be calculated the Earth can be calculated cabe¸calhodo t´ıtulo 3315-7

mass due to the expended fuel over an inﬁnitesimal time interval is dm < 0. Correspondingly, the increase in the 1 2 − 1 2 ∆Ekr = IΩf IΩi = velocity of the rocket is dv. [ 2 2 ] The conservation of the momentum of the rocket Rem Rem (vi ∓ vf ) · (vi ∓ vf ) + Ωi . (34) system requires: 2I

As a good approximation, it is assumed that I = 2 Rem ∓ kMeRe. Obviously, in Eq. (34), 2I (vi vf ) = − ∓ mv = (m + dm)(v + dv) + ( dm)(v + dv ηeˆ) . m (vi∓vf ) << Ωi. Hence, by using Eqs. (26), (28), 2kMe Re (36) (29), (30), (34) and approximating Me and m by Therefore, Me+m Me+m 1 and 0, respectively, the increase in the kinetic energy of the spacecraft-Earth system is calculated to be ∫ ± ∫ vf eˆ m dm dv = ∓ηeˆ , (37) vieˆ m0 m ∆Ek = ∆Ekm + ∆Ekt + ∆Ekr = 1 2 ∆Ekm + ∆Ekr = m (v2 ∓ ΩiRe) . (35) from which the relation between the mass of the space- 2 craft and the initial mass of the rocket can be estab- This result is in agreement with that obtained in lished: the reference frame ﬁxed to the ground. Irrespective of

whether observed in the S frame or the frame fixed to ∓ vf vi the ground, the conclusion is the same: launching the m0 = me η . (38) spacecraft eastwards consumes less energy than launching westwards. Nonetheless, the interpretations in the That is, for identical final speed v (> v ), a rocket two frames differ. In the frame fixed to the ground, f i can have a smaller mass when launched eastwards than the initial spacecraft speed is zero. The final spacecraft launched westwards. Unlike Verne’s gun, here the ad- speed in an eastwards launch, v −Ω R , is smaller than 2 i e vantage in launching eastwards is apparent without the that in a westwards launch, v + Ω R . In the S frame, 2 i e need to consider the increase in the rotational kinetic when the spacecraft is launched eastwards, the gun’s re- energy of the Earth. The reason is very simple. In coil slows the rotation of the Earth and ∆E < 0. That kr obtaining the ideal rocket equation, the rocket, includ- is, a certain amount of the Earth’s rotational kinetic en- ing payload and fuel, is approximated as an isolated ergy is converted into kinetic energy of the spacecraft; system. For Verne’s gun, the accelerating of the space- thus less chemical energy from the gunpowder is con- craft requires accelerating the ground in the opposite sumed. When the spacecraft is launched westwards, the direction. For Tsiolkovsky’s rocket, the accelerating of recoil pushes the Earth to rotate faster and ∆E > 0. kr the spacecraft is negated by expelling spent fuel in the In this case, the expended chemical energy from the opposite direction. gunpowder is used to increase both the kinetic energy of the spacecraft and the rotational kinetic energy of However, as illustrated in Fig. 4, because of the in- the Earth. teraction between the spent fuel and the Earth, launching of a spacecraft still inevitably changes the motion of the Earth. To obtain analytical expressions appropriate 4.2. Tsiolkovsky’s rocket for college-level physics, a number of approximations, The derivation of Tsiolkovsky’s ideal rocket equation some of which could not be very well justified in the can be routinely found in college textbooks. It is re- actual situation, have to be employed. For example, peated here with all velocities written in vector form. the accelerating process is assumed to take place over a The total mass of the rocket before launching is m0. short duration, so that the displacements of the space- The mass of the spacecraft, which is the payload of the craft, fuel, and the Earth from their original positions rocket, is m. The initial velocity of the grounded rocket can be neglected. (For clarity, in Fig. 4, spacecraft observed in the S frame is vi = vieˆ, wheree ˆ is still the and fuel are drawn as being separated by a large dis- eastward-pointing unit vector, and the final velocity of tance after launching. In the calculation, this separa- the spacecraft after depletion of all fuel is vf = ±vf eˆ. tion is neglected.) Furthermore, Earth’s atmosphere is The effective exhaust velocity of the fuel is η = ∓ηeˆ. disregarded. In reality, the spent fuel leads directly to Here vf = |vf | > 0 and η = |η| > 0. Again, upper changes in the motion of the surrounding atmosphere signs refer to eastward launchings and lower signs to and eventually changes the Earth’s motion. In this ar- westward launchings. The velocity of the rocket at a ticle, for simplicity, it is assumed that spent fuel falls given moment is v. The mass of the rocket at this mo- to the ground and adheres to the Earth in a perfectly ment is denoted also by m. The increase in the rocket’s inelastic collision. 3315-8 Zhang

ground. In this sense, it is still reasonable to argue that the spacecraft gains its kinetic energy at the expense of the rotational kinetic energy of the Earth. In a westward launch, the corresponding results are

m (v + R Ω ) − mV ∆Ω = f e i i = Figure 4 - Launching of an initially grounded spacecraft using Re [kMe + (k + 1) (m0 − m)] Tsiolkovsky’s rocket as observed in the S frame. (a) Before the m (vf + ReΩi) launching; (b) after an eastward launching; (c) after a westward > 0, (43) launching. kMeRe For an eastward launching, the rocket’s velocity be- and fore launching is vi = (−Vi + ReΩi)e ˆ, and the velocity of the fuel after the launching, which is ﬁrst ex- [ ] pelled from the rocket and eventually falls to ground, 1 2 2 − 2 ∆Ekr = kMeRe (Ωi + ∆Ω) Ωi = is [− (Vi + ∆V ) + Re (Ωi + ∆Ω)]e ˆ, where ∆V and ∆Ω 2 are the increase in the speed and angular velocity of mReΩi (vf + ReΩi) > 0. (44) the Earth caused by the adhesion of the fuel. Momen- tum conservation of the spacecraft-fuel-Earth system 5. Conclusion expressed in the S frame requires When one tries to solve an astronautics problem, such m (−V + R Ω ) − M V = mv + (m − m) × as ﬁnding the second and third cosmic velocities, in an 0 i e i e i f 0 inertial reference frame in which the spacecraft-Earth − − − [ Vi ∆V + Re (Ωi + ∆Ω)] Me (Vi + ∆V ) = 0. (39) system is moving, the increase in the kinetic energy of The angular momentum of the system is also con- the Earth should be considered, because the power of served. Here the spatial point that coincides with the the gravitational pull from the spacecraft doing work Earth’s center at the moment of launch (not the Earth’s on the Earth is not negligible in comparison with the center itself) is taken as the reference point. Therefore, power of the gravitational pull from the Earth doing work on the spacecraft. The launching of an initially grounded spacecraft − 2 Rem0 ( Vi + ReΩi) + kMeReΩi = into space inevitably changes the Earth’s rotation. Remvf + Re (m0 − m) Launching the spacecraft eastwards is more economi- cal in terms of energy consumption because part of the [− (Vi + ∆V ) + Re (Ωi + ∆Ω)] + 2 rotational kinetic energy of the Earth is converted into kMeRe (Ωi + ∆Ω) . (40) kinetic energy of the spacecraft. For Verne’s gun, the As in the preceding section, it is easy to verify that conversion is direct; for Tsiolkovsky’s rocket, the con- the increase in the Earth’s translational kinetic energy version is indirect. is negligible in comparison with the increase in the spacecraft’s kinetic energy. References From Eqs. (39) and (40), ∆Ω can be found [1] T. Moore, Six Ideas that Shaped Physics, Unit C: Con- servation Laws Constrain Interaction (WCB McGram- −m (Vi + vf − ReΩi) ∆Ω = = Hill, New York, 2003), 2nd ed, p. 60. Re [kMe + (k + 1) (m0 − m)] [2] B.K. Gao and T.Z. Xie, College Physics 11, 46 (1991) m (vf − ReΩi) − < 0. (41) (in Chinese). kMeRe [3] R.A. Serway and J.W. Jewett, Principle of Physics The increase in the rotational kinetic energy of the (Qinghua University Press, Beijing, 2004), 3rd ed., p. Earth is thus calculated to be 211. [ ] [4] A.S. Qi and C.Y. Du, Mechanics (Higher Education 1 nd ∆E = kM R2 (Ω + ∆Ω)2 − Ω2 = Press, Beijing, 2005) 2 ed., p. 127 (in Chinese). kr 2 e e i i [5] Z.H. Hu, College Physics 7, 26 (1984) (in Chinese). −mReΩi (vf − ReΩi) < 0. (42) [6] R.A. Serway and J.W. Jewett, Principle of Physics The spacecraft’s accelerating eventually slows down (Qinghua University Press, Beijing, 2004), 3rd ed., p. the Earth’s rotation when the spent fuel falls to the 320.