Rocket Propulsion, Classical Relativity, and the Oberth Effect Philip Rodriguez Blanco, Grossmont College, El Cajon, CA Carl E. Mungan, U.S. Naval Academy, Annapolis, MD

he role of work and mechanical energy in classical relativity has been a subject of renewed interest in this publication.1-4 Here we present a problem that illus- Ttrates the relationship between impulse and kinetic energy for a rocket-powered object that can also change its gravitational potential energy. The same introductory physics principles lead to a remarkable result when applied to the mechanics of spaceflight—the Oberth effect—whereby a small impulse can cause a large change in a rocket’s orbital energy without violat- ing any conservation laws. Problem statement Refer to Fig. 1. A rocket-powered skateboard (hereafter the “rocket”) is about to roll downhill starting from rest at Fig. 1. Statement of the problem. A version without annotations is 16 A. At what position (A, B, or C) should the rocket fire a brief provided in the online supplement. impulse to maximize its final speed on the other side of the (2) valley, or does it not matter? To simplify this problem, assume that g is constant, dissipative forces such as friction and air which increases with v. Thus the rocket gains more kinetic drag are negligible, and all speeds in the planet’s reference energy for the same impulse if it is already moving, and there- frame are such that classical (Galilean) relativity applies. fore the impulse should be applied when the rocket is moving With no burst from the rocket engine, the rocket will roll fastest, at the valley bottom B in Fig. 1. back and forth in the potential well between points A and C. The effect of the impulse on the rocket’s final speed is Astute students will recognize that firing the burst at A or C derived below. But why does the same v cause a larger KR has the same effect on the rocket, since it is at rest at the same when applied at a nonzero speed? The short answer is that height for both locations. But they may be unsure about the the work done on the rocket by the force, F x measured in the effect of firing it at B, where the rocket is moving. planet frame, is greatest at B, since the moving rocket covers In the following analysis, set the zero level of gravitational more distance x there for the same time interval t. This work potential energy at the height of the right side of the valley. is equal to the rocket’s change in kinetic energy KR when 2,4 The rocket starts on the left at a depth dA below this level, and both are calculated in the planet’s reference frame. the valley floor at B has a depth dB > dA (as indicated in Fig. 1). The rocket ejects an exhaust mass mex backwards in a brief Extra kinetic energy for the same fuel burst of duration t. This expelled fuel moves at speed vex expended? relative to its initial velocity (which is also the velocity of the In the planet’s frame, giving the rocket an impulse when it rocket + fuel system’s center of mass). Let the remaining mass is already moving appears to provide it with “extra” kinetic en- of the rocket be MR and its change in speed v. From momen- ergy, compared to firing the impulse when the rocket is at rest. tum conservation for the rocket and expelled fuel, valid in any That seems paradoxical. An observer moving with the rocket inertial frame, always measures the same chemical energy converted, to expel the same fuel mass at the same relative exhaust velocity. So MR v = mexvex = F t (1) where does the rocket’s additional kinetic energy come from? for a constant force F between the rocket and the exhaust. For The resolution to this paradox requires accounting for the a time-varying force, the impulse vector’s magnitude given by change in the kinetic energy of the exhaust mass expelled. In the right-hand side of Eq. (1) has to be replaced by an integral, the planet frame, if the rocket and fuel are already moving for- but that does not affect the analysis. wards at speed v, as shown in Fig. 2, the exhaust speed relative to the planet is |vex – v|. (We take the absolute value here to Short solution: Same impulse but not the include the case v > vex, for which the expelled fuel will then same work move in the same direction as the rocket.) The change in the From Eq. (1), the impulse causes the same v regardless kinetic energy of the expelled mass is of the rocket’s motion. However, if the rocket is moving in the ex ex (3) planet’s frame at speed v just prior to the impulse, the result- ing change in its kinetic energy in that frame is DOI: 10.1119/1.5126818 THE PHYSICS TEACHER ◆ Vol. 57, October 2019 439 Fig. 2. A rocket moving at initial speed v in the planet’s frame of reference expels an exhaust mass at speed vex relative to its initial speed, which increases the rocket’s speed by Δv. By vector addition assuming classical relativity, the post-impulse speeds of the rocket and exhaust in the planet frame are v + Δv and |v – vex|, respectively. which can be positive, negative, or zero. Adding Eqs. (2) and (3), and using Eq. (1), the total change in kinetic energy of the rocket + fuel is Fig. 3. A rocket in an elliptical orbit (solid green path) fires an ex (4) impulse at periapsis (labeled B by analogy with Fig. 1) in order to which is independent of v. Here E is the kinetic energy escape at the maximum possible speed. The dashed blue curve chem shows the post-impulse hyperbolic path of the rocket. The dot- gained from the conversion of the fuel’s chemical energy via ted orange curve shows the path of the expelled fuel. Axes are the engine, a quantity for which observers in all inertial frames labeled in units of the central body radius. Details of the orbital must agree. paths are provided in the online supplement.16 In the center of mass frame, Echem is shared by the rocket and exhaust mass according to Eqs. (1) and (4). In the planet frame, from Eqs. (3) and (4), for rocket speeds v > vex/2 we find (8) Kex < 0 and hence KR > Echem, i.e., the rocket gains more kinetic energy than that provided by combusting the fuel! 5 In that case, the additional energy gained by the rocket is obtained Compared to the final mechanical energies if the impulse from a portion of the fuel’s mechanical energy. were fired at A, Eqs. (7) and (8) have an additional term on the right-hand side. (As a check, set dB = dA, as if the valley Calculating the final speed bottom did not exist, and the third term in each equation If the rocket rolls down to point B and fires its impulse vanishes.) From Eq. (1), those terms are equal and opposite, there, what will be its final speed when it emerges on the right so that the sum of ER,f and Eex,f gives a total final energy for side of the valley? the {rocket+fuel} system The mechanical energy of a system is the sum of its kinet- ic and potential energies, and is constant in the absence of (9) external work.6 In the planet frame before the rocket fires, the initial mechanical energy of the rocket+fuel Etotal,i equals as expected from Eq. (4). their gravitational potential energy at position A in Fig. 1, After the impulse at the valley bottom, use conservation of since they are stationary there, mechanical energy for the rocket alone to find its final speed E = –(M + m )gd . (5) vR,f on reaching the top of the valley on the right, where the total,i R ex A gravitational potential energy is zero. From Eq. (7), The speed vB at the valley bottom B, just before firing the burst, is found from mechanical energy conservation as (10) ,i R,f (6) The final speed when the impulse is fired instead at A or C can be found from Eq. (10) by setting dB = dA. Just after firing the impulse at B, the rocket and expelled fuel have speeds of vB + v and | vB –vex |, respectively (cf. Fig. 2). Rocket fuel carries chemical and mechan- Using Eqs. (2) and (3), with v = vB from Eq. (6), their respec- ical energy tive final mechanical energies ER,f and Eex,f become Equations (7) and (10) provide a remarkable result. If we increase the valley depth dB, the rocket’s final mechanical en- (7) ergy, and hence its final speed, can be as large as one wants! If dB is deep enough, it is even possible for a rocket to start at A and escape on the right with vR,f > v, eventually passing an and identical rocket that starts from rest on the valley’s right side and fires its burst there.

440 THE PHYSICS TEACHER ◆ Vol. 57, October 2019 If one considers the energetics of the rocket only, this tions,” Phys. Teach. 56, 96–99 (Feb. 2018). result seems surprising. However, the total post-impulse me- 2. R. Kaufman, “Work and inertial frames,” Phys. Teach. 55, 561– chanical energy of the rocket plus fuel is conserved, and the 564 (Dec. 2017). exhaust mass expelled at B ends up with less mechanical en- 3. S. Ghanbari, “Mechanical energy change in inertial reference ergy as dB is increased. For example, from Eq. (8) the ejected frames,” Phys. Teach. 54, 360–363 (Sept. 2016). fuel mass will become trapped in the potential well between 4. B. Tefft and J. Tefft, “Galilean relativity and the work-kinetic A and C if Eex,f < –mexgdA. energy theorem,” Phys. Teach. 45, 218–220 (April 2007). Examining the right-hand side of Eq. (7), we see that the 5. We desire both the largest possible initial speed v and exhaust rocket always gains energy from some of the fuel’s converted speed vex (also referred to as the fuel’s specific impulse in m/s or . chemical energy, given by the first term (cf. Eq. 4). By firing N s/kg) in order to provide the largest boost to the rocket, its burst at the lowest point B, the rocket can also “steal” ad- according to Eqs. (1) and (2). Also see M. Iona, “Efficiency in rocket propulsion,” Phys. Teach. 16, 600 (Dec. 1978). ditional energy, given by the last term, from the the original 6. The gravitational potential energy of the rocket and fuel should mechanical energy carried by the fuel. (The remaining mid- properly be assigned to the systems of planet+rocket and plan- dle term equals the initial mechanical energy of the rocket et+fuel, respectively. For brevity, we refer to the mechanical en- body.) ergy of the “rocket” or “fuel” alone. If we assume a near-infinite mass for the planet, we can neglect any changes in its motion as The Oberth effect a result of its interactions with the rocket or fuel. The application of this result to spaceflight was first sug- 7. H. Oberth, “Ways to Spaceflight,” in NASA Technical Transla- gested in 1929 by astrodynamics pioneer Hermann Oberth.7 tion F-622 (1972), pp. 194–217. In a bound elliptical orbit around a central body, the fuel 8. D. Landau and N. Strange, “This way to Mars,” Sci. Am. 305, carries more mechanical energy than the same mass at rest 58–65 (Dec. 2011). on the body’s surface. As with our ground-based example, the 9. Perhaps easier to visualize (by reversing the arrows in Fig. 3), (8) greatest boost to the rocket’s orbital energy is produced by a the time-reverse version of this maneuver is an inelastic head- prograde (forward-acting) impulse made near periapsis, the on collision between an orbiting fuel mass and an incoming lowest and fastest point in its path, indicated as position B in rocket, whereby the combined object ends up with a mechan- Compared to the final mechanical energies if the impulse Fig. 3 by analogy with Fig. 1. ical energy less than that of the colliding objects by an amount ΔE . were fired at A, Eqs. (7) and (8) have an additional term on The Oberth effect8 refers to this increase in the rocket’s chem 10. “Inserting Juno into orbit around Jupiter,” online at the right-hand side. (As a check, set d = d , as if the valley orbital energy, which will either increase the size of its ellip- B A physicsfromplanetearth.wordpress.com/2016/07/10/. bottom did not exist, and the third term in each equation tical orbit, or its hyperbolic excess speed as shown in Fig. 3.9 11. The Oberth effect is distinct from a gravity-assist “slingshot” vanishes.) From Eq. (1), those terms are equal and opposite, Conversely, retro-firing at periapsis is the most effective way maneuver, which describes an elastic momentum and energy so that the sum of ER,f and Eex,f gives a total final energy for to reduce the spacecraft’s mechanical energy (by transferring exchange between an unpowered spacecraft and a planet when the {rocket+fuel} system a large portion of it to the expelled fuel). Such a maneuver is both are under the gravitational influence of a third body, typi- often used to achieve orbital capture, as exemplified by the cally the Sun. See A. Bartlett and C. Hord, “The slingshot effect: 10 (9) Juno mission’s insertion into orbit around Jupiter. Explanation and analogies,” Phys. Teach. 23, 466–473 (Nov. Since orbital motion occurs in a 1/r gravitational potential, 1985). In the Oberth effect, no third body is required, and since as expected from Eq. (4). with angular momentum as an additional constraint, the de- the firing of the impulse does not change the motion of the After the impulse at the valley bottom, use conservation of tails of the orbital Oberth effect differ from the ground-based rocket+fuel’s center of mass, it has no effect on the planet. mechanical energy for the rocket alone to find its final speed problem analyzed here, but the physical principles are the 12. Online at http://orbit.medphys.ucl.ac.uk . 11 13. To obtain free educational licenses for Systems Tool Kit (STK), vR,f on reaching the top of the valley on the right, where the same. We invite users of orbital mechanics software such gravitational potential energy is zero. From Eq. (7), as Orbiter,12 STK,13 or Kerbal Space Program14 to simulate visit http://www.agi.com/education . the effects of an impulse on a spacecraft at different locations 14. Online at http://www.kerbalspaceprogram.com . See also R. in its elliptical orbit. A computational physics simulation of Munroe, “The six words you never say at NASA,” https://xkcd. com/1244 . (10) Fig. 1 (e.g. in Algodoo15), or a laboratory apparatus using a R,f 15. B. Gregorcic and M. Bodin, “Algodoo: A tool for encouraging spring-powered mass launcher running on a flexible track, creativity in physics teaching and learning,” Phys. Teach. 55, The final speed when the impulse is fired instead at A or C would also serve as a demonstration of these principles. Stu- 25–28 (Jan. 2017). can be found from Eq. (10) by setting dB = dA. dents should appreciate that Newtonian conservation laws 16. See the online supplements at TPT Online, http://dx.doi. apply to rocket propulsion for all inertial observers, as long org/10.1119/1.5126818 under the Supplemental tab. Rocket fuel carries chemical and mechan- as mechanical energy changes in the rocket and exhaust are ical energy taken into account. Philip Blanco credits his interest in spaceflight to his early childhood Equations (7) and (10) provide a remarkable result. If we Additional PowerPoint slides for presenting this problem memories of watching live TV coverage of the Apollo missions. [email protected] increase the valley depth dB, the rocket’s final mechanical en- to students, and an STK animation of Fig. 3, are provided in ergy, and hence its final speed, can be as large as one wants! If an online supplement.16 Thanks to Devin Potratz for assis- Carl Mungan enjoys teaching across the entire undergraduate physics dB is deep enough, it is even possible for a rocket to start at A tance in producing Fig. 1. curriculum, and thinks using clicker questions such as the one in Fig. 1 and escape on the right with vR,f > v, eventually passing an is beneficial for undergraduates of all levels, and not just introductory identical rocket that starts from rest on the valley’s right side References students. and fires its burst there. 1. E. Ginsburg, “Simple examples of the interpretation of changes [email protected] in kinetic and potential energy under Galilean transforma-

THE PHYSICS TEACHER ◆ Vol. 57, October 2019 441 Supplement to “Rocket propulsion, classical relativity, and the Oberth Effect”, published in The Physics Teacher, 2019 Authors: Philip Blanco, Grossmont College Carl Mungan, United States Naval Academy Contents: Figure 1 (without annotations) Alternate Figure 1 Additional figures Animations with Systems Tool kit (separate PPT File): Impulse fired at periapsis (cf. Figure 3) Impulse fired at apoapsis for comparison Figure 1 (annotations removed) This rocket-powered skateboard is about to roll downhill. To escape on the right with the maximum speed, where should it fire a brief impulse? (Or does it not matter?)

A C

B

Thanks to student Devin Potratz for assistance with the graphics Alternate Figure 1

Thanks to student Devin Potratz for assistance with the graphics

This rocket-powered skateboard is about to roll downhill. To escape on the right with the maximum speed, where should it fire a brief impulse? (Or does it not matter?)

A C

B You get moreΔK for a fixed Δv if you are already moving fast! 100 K = ½ Explanation: mv2 80 Impulse F δt = mΔv is fixed, but Work = F δx =F vδt ΔK increases with speed. 60 Δv

40

But where does this “extra” 20 kinetic energy come from? ΔK Kinetic Energy per kg (J/kg) Δv 2 4 6 8 10 Speed (m/s) Where does this “extra” kinetic energy boost come from?

v Δv ex M mex R v v For the rocket: “extra” 2 ΔK = 1 M v + Δv − 1 M v2 = 1 M Δv2 + vM Δv R 2 R ( ) 2 R 2 R R For the exhaust: 2 ΔK = 1 m v − v − 1 m v2 = 1 m v2 − vm v ex 2 ex ( ex ) 2 ex 2 ex ex ex ex Since mexvex = MRΔv, when we add up the total ΔK, ⇒ ΔK = ΔK + ΔK = 1 M Δv2 + 1 m v2 = ΔE total R ex 2 R 2 ex ex chem - invariant! Classical Relativity è invariance of impulse and total energy change (Rocket converts same chemical è kinetic energy independent of speed.)

For a rocket already moving fast, in our reference frame, • more ΔKtotal goes to the rocket…and less to the exhaust, so • rocket “steals” kinetic energy from exhaust! Orbital Oberth effect (Figure 3)

2GM Planet surface escape speed v = P SE R P Set rocket mass MR ≡16 mex 2 →ΔEchem = 17/32 mex vex Set exhaust velocity vex ≡ 1.46286 vSE 10 →Δv = vex/16 = 0.09143vSE Animation at →ΔE = 2.274 GMm /R . chem ex P https://vimeo.com/286053315 Initial rocket+fuel orbit (ellipse):

Set semi-major axis a ≡ 7.50 RP, eccentricity e ≡ 0.8.

→periapsis distance = 1.5 RP, →periapsis speed = 0.77460 vSE 5 17 → Etotal,i = - /15GMmex/RP = -0.4984 ΔEchem

Resulting rocket orbit (hyperbola):

Semi-major axis a = -6.00 RP, eccentricity e = 1.250. 4 → Etotal,f = + /3GMmex/RP →ΔER = +2.4 GMmex/RP =+1.0555 ΔEchem .

Resulting exhaust orbit-10 (ellipse): -5 5 10

Semi-major axis a = 2.591 RP, eccentricity e = 0.4211. →ΔEex = -0.1263GMmex/RP = -0.0555 ΔEchem. B

-5

-10