Meccanica (2007) 42:465–476DOI 10.1007/s11012-007-9065-7

A complete closed form vectorial solution to the Keplerproblem

Daniel Condurache · Vladimir Martinusi

Received: 4 May 2006 / Accepted: 12 March 2007 / Published online: 6 July 2007© Springer Science+Business Media B.V. 2007

Abstract The paper gives an exact vectorial solu-tion to the Kepler problem. A vectorial regularizationthat linearizes the Kepler problem is given using aSundman transformation. Closed form expressionsdescribing the Keplerian motion are deduced. A uni-fied approach to the classic Kepler problem is offered,by studying both rectilinear and non-rectilinear Keple-rian motions with the same instrument. The approachis an elementary one and only simple vectorial compu-tations are involved.

Keywords Kepler problem · Vectorialregularization · Sundman transformation · ClassicalMechanics

1 Introduction

The present paper focuses on the famous Kepler prob-lem described by the initial value problem:

r + µ

r3 r = 0, r (t0) = r0, r (t0) = v0. (1)

We denote by: r = r (t) the position vector of thebody related to the attraction center; v = r the velocity

D. Condurache (B) · V. MartinusiDepartment of Theoretical Mechanics, TechnicalUniversity “Gheorghe Asachi”, Iasi, Romaniae-mail: danielcondurache@rdslink.ro

V. Martinusie-mail: vladmartinus@gmail.com

of the body; r = r (t) the magnitude of the positionvector r ; µ > 0 the gravitational parameter, µ = k M,

with k the universal attraction constant and M the massof the attraction center; t0 ≥ 0 the initial moment oftime.

The prime integrals of the initial value problem (1)are (see [6,9,13]):

r × v = r0 × v0def= � (2)

1

2v2 − µ

r= 1

2v2

0 − µ

r0

def= h (3)

v × (r × v)

µ− r

r= v0 × (r0 × v0)

µ− r0

r0

def= e (4)

(“def=” = “defined to be”). Eq. 2 represents the spe-

cific angular momentum conservation, Eq. 3 the specificenergy conservation and Eq. 4 the eccentricity vectorprime integral (see [1,16,19]).

Depending on the specific angular momentum �,the Keplerian trajectory is a conic section or a sectionof a straight line.

In classical approaches, the prime integrals (2) and(3) are used to study the trajectories and the motion onthe trajectories (see [6,10,13]). The eccentricity vectorprime integral (4) is related to the Laplace-Runge-Lenz prime integral via a scalar term (see [5]).It has an interesting geometrical interpretation: vector ehas the direction of the symmetry axis of the trajectory,its sense indicates the pericenter and its magnitude rep-resents the eccentricity of the conic (i.e., the Keplerian

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466 Meccanica (2007) 42:465–476

trajectory), which is fundamental for the solution pre-sented in this paper.

The singularity that occurs in Eq. 1 for r = 0 waseliminated through several regularizations. Sundman(1912) (see [15]) and Levi-Civita (1920) (see [12])introduce a transformation of the independent variable:

dt = r dτ (5)

in an attempt to solve the restricted three body problem.The planar case of the Kepler problem is regularizedby Levi-Civita with the help of complex numbers.

Kustaanheimo and Stiefel (1965) ([7,8]) use theSundman transformation (5) and some coordinatetransformation and regularize the spatial Kepler prob-lem by using spinors. Burdet (1967) [2] uses Carte-sian coordinates and a Sundman-like transformationsimilar to Eq. 5 and regularizes the Kepler problem(1). He transforms the non-linear initial value problem(1) into a linear one, with constant coefficients. Theelliptic Kepler problem becomes a three-dimensionaloscillator problem. The key to Burdet approach is theLaplace-Runge-Lenz vector, expressed also in Carte-sian coordinates. Vivarelli [16–19] uses an ingeniousgeometrical construction together with the transforma-tion (5) in the study of the Kepler problem and obtainsa vectorial regularization of the Kepler problem. Otherapproaches use quaternions (see [17,21,22]).

More recent papers also take into consideration therectilinear case (i.e., the specific angular momentumprime integral is zero), see [1] for example. The col-lision destroys the physical system and here the studystops, as the velocity magnitude increases to infinity.

The present paper uses the time transformation (5)and the three prime integrals (2)–(4) in order to pro-vide a unified way of obtaining vectorial closed formsolutions to the Kepler problem (1). The methods thatare used are purely vectorial and they cover two differ-ent types of situations that may occur: one situationcomprises both non-degenerated and rectilinear trajec-tories, and the other situation refers to the manner theinitial value of the independent variable τ is chosen. Inthe latter case, Kepler’s equations and Kepler’s gener-alized equations respectively are obtained via a simplecomputation (see also [14]).

The aim of this paper is to present a unified Kep-ler problem approach, using only elementary vectorialand differential computations. The results arepresented in a vectorial form, that is independent onthe choice of some specific coordinate system. The

Sundman-like vectorial regularization plays only acatalyst role, the final results being expressed as func-tions of time. Moreover, the way Kepler’s equationsand Kepler generalized equations are deduced is uni-fied and the eccentric anomaly is revealed without anygeometrical considerations, but intimately related to theindependent variable τ .

The approach is structured as follows:Section 2 succinctly presents the Keplerian trajecto-

ries and the velocity hodograph starting from the primeintegrals (2)–(4).

In Sect. 3, the Kepler problem (1) is transformed intoa linear initial value problem with constant coefficients.

The time-transformation and the linearization of theinitial value problem describing the Keplerian motionlead to explicit vectorial expressions for the state ofthe body (i.e., the position vector and the velocity).These expressions are explicit in the new independentvariable, which is called fictitious time. They are inti-mately related to the specific energy of the motion andthe eccentric anomaly (in the non-rectilinear case). Onemay remark that no geometrical approaches are neededto introduce the eccentric anomaly of the Keplerianmotion. Moreover, in the rectilinear case, where theeccentric anomaly loses its geometrical sense, this newindependent variable allows exact expressions for posi-tion and velocity in all possible cases. A comprehensivestudy of the Keplerian motion is performed in Sect. 4,including a complete study of the collision situation.Phase plane portraits in both variables (the “regular”time t and the fictitious time τ ) are presented.

The solution presented in this paper is elementary.It involves only simple vectorial computations.

2 The Kepler problem: the trajectories

The approach is structured with respect to vector �.

2.1 Non-Zero specific angular momentum � �= 0

From Eq. 2 the trajectory results to be a plane curve.The vectorial equation of plane � where the motiontakes place is:

r · � = 0, F ∈ �, (6)

where F denotes the attraction center. Eq. 4 may berewritten:

v × (r × v) = µ (e + er) , (7)

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Meccanica (2007) 42:465–476 467

Fig. 1 The generic non-rectilinear Keplerian trajectory is a conicsection

where erdef= r/r is the unit vector associated to vector

r. By dot-multiplying in (7) by r and taking eq (2) intoaccount, it results:

r· (e + er)def= p, (8)

which is the vectorial equation of a conic in plane �

(see Fig. 1). Here p = �2/µ represents the semilatusrectum (or the parameter) of the conic. Vector e repre-sents the vectorial eccentricity of the conic: its direc-tion is identical with the main axis of the conic, itssense indicates the pericenter and its magnitude givesthe genre of the conic: e < 1 : ellipse, e = 1: parabola,e > 1: hyperbola (see [6]). The particle moves on thistrajectory according to the energy conservation law (3).

Vector e is situated in plane � and its magnitude is(from Eq. 4):

e =√

1 + 2h�2

µ2 . (9)

From Eq. 9 it results: e < 1 ⇔ h < 0; e = 1 ⇔h = 0; e > 1 ⇔ h > 0.

The position vector of the pericenter relative to theattraction center and the velocity in this point have thevectorial expressions:⎧⎪⎪⎪⎨⎪⎪⎪⎩

rP = �2

µ (1 + e)

ee;

vP = µ (1 + e)

�2e� × e.

(10)

when e �= 0. If e = 0, the trajectory is circular; wemay consider the pericenter to be situated at the initialposition of the body. It results:

{rP = r0;vP = v0.

(11)

By cross-multiplying Eq. 7 by � and taking Eq. 2 intoaccount, it results:

v = ααα× (e + er) . (12)

where αααdef= (

µ/�2)�. From Eq. 12 it results that the

velocity hodograph is a circular section. If e < 1, theorigin of the velocity space is situated inside the circle,if e = 1 the origin of the velocity space is situated onthe circle and if e > 1 the origin of the velocity spaceis situated outside this circle.

2.2 Zero specific angular momentum: � = 0

In this situation the trajectory is rectilinear, as it followsfrom Eq. 4 which becomes:

− rr

= −r0

r0= e. (13)

Vector e loses its geometrical sense of eccentricityvector. Nevertheless, it still remains an important primeintegral: in the rectilinear situation, it becomes a unitvector that has the same direction of the straight linewhere the motion takes place. The vectorial equationof the straight line (d) where the motion takes place is:

r × r0 = 0, F ∈ (d) . (14)

The body may or may not reach the attraction cen-ter, depending on the initial conditions. The study ofthe collision case is to be done in Sect. 4.

3 The Kepler problem: a vectorial regularization

We introduce a new Sundman-like variable τ = τ (t)that will be called as it follows fictitious time. For anydifferentiable vectorial map u, the following resultholds:

d

dτu = r

d

dtu, (15)

where r is the magnitude of the solution to the initialvalue problem (1). From Eq. 15 it results that τ is infact subject to the standard Sundman transformation(5). But Eq. 5 is not sufficient by itself to uniquelydetermine a functional relation between t and τ. Twoimportant situations may occur:

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468 Meccanica (2007) 42:465–476

1. If we consider τ = 0 at the moment tP when thebody passes through the pericenter of the trajectory,then:

t − tP =∫ τ

0r (ξ) dξ (16)

(we extend the notion of “pericenter” to “the pointon the trajectory closest to the attraction center”:in case of collision with the attraction center, tP

denotes the moment of time when the particle issituated in the attraction center; the moment of timetP may be smaller or greater than t0, depending onthe initial conditions—this will be studied furtherin this paper).

2. If we consider the fictitious time variable τ = 0 atthe initial moment of time t0, when the Keplerianmotion study begins, then:

t − t0 =∫ τ

0r (ξ) dξ. (17)

From Eq. 15 it follows that a vectorial map u is con-stant with respect to the regular time t iff it is constantwith respect to the fictitious time τ .

By denoting with ( )′ the derivative with respect tothe fictitious time τ , Eq. (15) may be rewritten:

u′ = r u. (18)

After computing:

r′ = r r =rv; (19)

r′′ = (r r)′ = (r · v) v−µ

rr =2hr−µe;

and by taking into account Eqs. 1–4, 10, 11, the initialvalue problem (1) becomes:

1. When τ = 0 at the moment of time t = tP :r′′ − 2hr = −µe,⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

r|τ=0 =rP =

⎧⎪⎪⎨⎪⎪⎩

�2

µ (1 + e)

ee, if e �=0

r0, if e=0

;

r′∣∣τ=0 =rP vP =

⎧⎨⎩

1

e� × e, if e �=0

r0v0, if e=0.

(20)

2. When τ = 0 at the moment of time t = t0 :

r′′ − 2hr = −µe,

⎧⎨⎩

r|τ=0 = r0

r′∣∣τ=0 = r0v0.

(21)

The initial value problems (20) and (21) regularizethe initial value problem (1) by eliminating the singu-larity for r = 0 . The non-linear vectorial differentialequation that describes the Keplerian motion is trans-formed into a linear one, with constant coefficients. Thesolutions to the initial value problems (20) and (21)have explicit expressions with respect to the fictitioustime τ .

4 The Kepler problem: the motionon the trajectories

A vectorial closed form solution to the Kepler prob-lem is about to be given with the essential help ofthe regularization introduced in Sect. 3. All possiblecases (h < 0, h = 0, h > 0) will be studied and ex-act vectorial solutions will be obtained, with sugges-tive interpretations. Expressions for the law of motionand the velocity will be deduced with respect to theregular time variable. Implicit relations (Kepler’s equa-tions) between the fictitious time τ and the regular time-variable t will be deduced in each case. In eachsituation, the same algorithm leads to the results:

1. The linear initial value problem (20) or (21) issolved. The exact solution is determined in eachsituation.

2. The magnitude of the position vector is computed.3. The velocity vectorial map is determined using:

v = d

dtr =1

r

d

dτr. (22)

4. By replacing the magnitude of the position vectorcomputed at step (2) of the present algorithm inEq. 16 or 17, Kepler’s equations (or Kepler’s gen-eralized equations) are deduced in a unified way ineach case.

5. Starting from the expressions for the law of motionand velocity computed at steps (1) and (3), the ini-tial value with respect to time τ (t0) is determinedfor the fictitious time variable τ .

6. The moment of time tP is computed depending onthe initial conditions:

tP = tP (r0, v0) . (23)

7. An implicit functional equation that has as solutionthe map τ = τ (t) is deduced.

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Meccanica (2007) 42:465–476 469

When we choose τ (tP ) = 0, in the situation h �= 0,

all results are expressed with respect to one scalar andtwo vectorial prime integrals based on the prime inte-grals (2)–(4):

n = (2 |h|) 32

µ; a =

{ µ

2e |h|e, e �= 0

r0, e = 0;

b =

⎧⎪⎪⎨⎪⎪⎩

1

e√

2 |h|� × e, e �= 0,

1

nv0, e = 0.

(24)

Here n is also known as the mean motion (in theelliptic case) and it is involved in Kepler’s equation.An interesting relation between these new prime inte-grals and those from Eqs. 2–4 is:

� = n (a × b) . (25)

Vectors a and b are related to the spatial orbital ele-ments of the trajectory and they have an amazing geo-metrical interpretation: in case h �= 0, they representthe vectorial semiaxes of the conical trajectories. When� �= 0 they determine plane � where the trajectory issituated. If � = 0, vector b is zero and a gives thedirection of the straight line where the motion takesplace.

All computations in the rectilinear situation (� = 0)are realized by taking b = 0. The results are quite sim-ilar.

The same algorithm is used in order to provide vec-torial closed form expressions for the Keplerian motionin the case when the fictitious time map τ = τ (t) satis-fies τ (t0) = 0. The results are quite similar, includingthe generalized Kepler equation, obtained by the samemethod from step 4 of the previous algorithm. The com-mon issue of the results obtained in these both situationsis that in the case of non-zero specific energy (ellipticor hyperbolic Keplerian motion), the final results areexpressed depending on the conic conjugate semi-diameters; in case τ (tP ) = 0, these semi-diametersrepresent the semimajor and semiminor axes of theconic, a and b, and in case τ (t0) = 0, they are given bythe initial position and velocity vectors: r0−[µ/ (2h)] eand r0v0/

√2 |h|). Of course, both situations unify

when the body is launched from the conic pericenter(tP = t0).

In the special case when the conic is degenerated(vectors r0 and v0 are collinear), all results obtained

from the non-degenerated situation transform bytaking e = −r0/r0 and taking into consideration thatv0 = [

(r0 · v0) /r20

]r0.

Respecting the steps of the algorithm, the followingresults are deduced:

4.1 Non-zero specific angular momentum (� �= 0) :planar Keplerian motion

4.1.1 The elliptic trajectory (e < 1 ⇔ h < 0)

This is the most important case from the point of viewof Orbital Mechanics, because it applies in satellitemotion. The solution to the perturbed Kepler prob-lem (non-spherical attraction source) is also based onthis elementary solution. Similar results (obtained viaa quaternionic approach) in the elliptic case are givenin [21].

� The case τ (tP ) = 0The solution to the initial value problem (20) is:

r (t) = a [cos E (t) − e] + b sin E (t) , (26)

where a and b are given in Eqs. 24 and

E (t) = √2 |h|τ (t) ; t ∈ [t0,+∞) (27)

represents the eccentric anomaly of the ellipticKeplerian motion. The Keplerian trajectory and thevelocity hodograph in case h < 0 are presented inFig. 2.

The magnitude of the radius vector (26) is:

r (t) = a [1 − e cos E (t)] , t ∈ [t0,+∞). (28)

The velocity vector is equal to:

v (t) = n

1 − e cos E (t)[−a sin E (t) + b cos E (t)] ,

t ∈ [t0,+∞) (29)

where n is the mean motion defined in Eq. 24.By replacing the expression of r from Eq. 28 into

Eq. 16, Kepler’s equation for the elliptic case isdeduced:

n (t − tP ) = E (t) − e sin E (t) , t ∈ [t0,+∞). (30)

By making t = t0 in Eqs. 28 and 29 and denoting

E0not= E (τ (t0)), it results:

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470 Meccanica (2007) 42:465–476

Fig. 2 The elliptic Keplerian motion and the velocity hodograph

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

r0 =√

2 |h|n

(1 − e cos E0) ;

v0 =√

2 |h|r0

[−a sin E0 + b cos E0] .

(31)

By dot-multiplying by a the second relation inEq. 31 and making all possible computations, it resultsthat E0 ∈ [0, 2π) is uniquely defined by:⎧⎪⎪⎪⎨⎪⎪⎪⎩

cos E0 = 1

e

(1 − nr0√

2 |h|)

;

sin E0 = n

2 |h| e(r0 · v0) .

(32)

The moment of time tP is computed from Eq. 30taking into account the expression of E0 deduced fromEqs (32):

tP = t0 − 1

n(E0 − e sin E0) . (33)

The implicit functional equation defining the mapτ (t) = (

1/√

2 |h|) E (t) is deduced from Eq. 30 tak-ing Eqs. 32 and 33 into account:

E (t) − e sin E (t)=n (t − t0)+(E0 − e sin E0) . (34)

� By taking τ (t0) = 0, it results:

r (t) = − µ

2 |h|e +(

r0 + µ

2 |h|e)

cos H (t)

+ r0v0√2 |h| sin H (t) (35)

r (t) = µ

2 |h| +(

r0 − µ

2 |h|)

cos H (t)

+ r0 · v0√2 |h| sin H (t) (36)

v (t) = [r (t)]−1 · [r0v0 cos H (t)

+2hr0 − µe√2 |h| sin H (t)

](37)

where

H (t) = √2 |h|τ (t) (38)

plays the role of a generalized eccentric anomaly. Byintegrating in Eq. 36, it results the generalized Keplerequation for elliptic trajectories ([14]):

t − t0 = 1

nH (t) + r0 · v0

2 |h| [1 − cos H (t)]

+(

r0 − µ

2 |h|)

sin H (t)√2 |h| . (39)

The circular trajectory: e = 0All computations from Eqs. 26, 28, 29 modify by

taking:

a = r0; b =1

nv0; E (t) = v0

r0(t − t0) . (40)

The trajectory is circular iff e = 0, which is equiv-alent to:⎧⎪⎨⎪⎩

r0 · v0 = 0;v0 =

õ

r0.

(41)

The vectorial expressions that describe the circularKeplerian motion are:

r (t) = cos [n (t − t0)] r0 + sin [n (t − t0)]v0

n(42)

r (t) = r0 (43)

v (t) = −n sin [n (t − t0)] r0 + cos [n (t − t0)] v0.

(44)

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Meccanica (2007) 42:465–476 471

4.1.2 The parabolic trajectory (e = 1 ⇔ h = 0)

� When τ (tP ) = 0, the solution to the initial valueproblem (20) is:

r (t) = 1

2

(p − µτ 2 (t)

)e+τ (t) (� × e) . (45)

Eq. 45 represents the vectorial equation of a parab-ola with the semilatus rectum p = �2/µ. The direc-tion of the symmetry axis of the parabola is givenby unit vector e. The Keplerian parabolic trajectoryand its associated velocity hodograph are presented inFig. 3.

The magnitude of the radius vector (45) is:

r (t) = 1

2

(p + µτ 2 (t)

), t ∈ [t0,+∞). (46)

The velocity vector is:

v (t) = 2

p + µτ 2 (t)[−µτ (t) e + (� × e)] ,

t ∈ [t0,+∞). (47)

By replacing the expression of r from Eq. 46 intoEq. 16 and making all possible computations, Kepler’sequation in the parabolic case is deduced:

t − tP = 1

2

[pτ (t) + µ

τ 3 (t)

3

], t ∈ [t0,+∞). (48)

In order to compute the initial value τ0def= τ (t0) ,

we make t = t0 in Eqs. 46 and 47. It results:

v0 = 1

r0[−µτ0e + (� × e)] . (49)

By dot-multiplying by e in Eq. 49 it results:

τ0 = r0 · v0

µ. (50)

The moment of time tP is now computed usingEqs. 48 and 50:

tP = t0 − r0 · v0

2µ

[p + µ

3

(r0 · v0

µ

)2]

. (51)

If r0·v0 > 0, it results tP < t0. The body is launched“after” the pericenter moment. In case r0 · v0 = 0, thebody is situated in the pericenter at the initial moment

of time and in case r0 · v0 < 0 it results tP > t0, so thebody passes through the pericenter after the launch. Inthis particular situation, the period of time between thelaunch and the pericenter passing is equal to tP − t0.

Map τ = τ (t) is solution to the functional Eq. 48.Its time-explicit expression is:

τ (t) = 13√

µ

3

√3 (tP −t)+

√9 (tP −t)2+ p3/µ

+ 13√

µ

3

√3 (tP − t) −

√9 (tP − t)2 + p3/µ.

(52)

� By choosing τ (t0) = 0, it results:

r (t) = r0 + τ (t) r0v0 − τ 2 (t)

2µe; (53)

r (t) = r0 + τ (t) (r0 · v0) + µτ 2 (t)

2; (54)

v (t) = [r (t)]−1 [r0v0 − τ (t) µe] , (55)

and the generalized Kepler equation for parabolic tra-jectories:

t − t0 = r0τ (t) + τ 2 (t)

2(r0 · v0) + µ

τ 3 (t)

3. (56)

4.1.3 The hyperbolic trajectory (e > 1 ⇔ h > 0)

� When τ (tP ) = 0, the solution to the initial valueproblem (20) is:

r (t) = a [e − cosh E (t)] + b sinh E (t) , (57)

where:

E (t) = √2hτ (t) ; t ∈ [t0,+∞) (58)

and vectors a and b are given in Eq. 24. Eq. 57 rep-resents the equation of a hyperbolic oscillator. MapE = E (t) represents the eccentric anomaly for theKeplerian hyperbolic motion. The Keplerian hyperbolictrajectory and its associated velocity hodograph arepresented in Fig. 4.

The magnitude of the position vector is given by:

r (t) = a [e cosh E (t) − 1] , t ∈ [t0,+∞). (59)

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472 Meccanica (2007) 42:465–476

Fig. 3 The parabolic Keplerian motion and the velocity hodograph

Fig. 4 The hyperbolic Keplerian motion and the velocity hodograph

The velocity vector is:

v (t) = n

e cosh E (t) − 1[− sinh E (t) a

+ cosh E (t) b] , t ∈ [t0,+∞). (60)

By replacing r from Eq (59) into Eq (16) and makingall computations, Kepler’s equation for the hyperbolictrajectory reveals:

n (t − tP ) = e sinh E (t) − E (t) , t ∈ [t0,+∞). (61)

In order to compute the initial value E0 of map E,

we consider t = t0 in Eq. 60 and dot-multiply by a. Itresults:

E0 = sinh−1 n (r0 · v0)

2eh. (62)

The moment of time tP may now be computed fromEq. 61 by considering t = t0 and taking Eq. 62 intoaccount. It results:

tP = t0 − 1

n

[n (r0 · v0)

2h− sinh−1 n (r0 · v0)

2eh

]. (63)

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Meccanica (2007) 42:465–476 473

In case r0 · v0 > 0, it results tP < t0; the body islaunched “after” the pericenter. In case r0 · v0 < 0,

then tP > t0, so the body will be in the pericenter atthe moment of time tP . In case r0 · v0 = 0, the body islaunched exactly from the pericenter of the hyperbola.

The eccentric anomaly E = E (t) is given by theimplicit functional equation:

e sinh E (t) − E (t) = n (t − t0) + e sinh E0 − E0,

t ∈ [t0,+∞). (64)

� By choosing τ (t0) = 0, it results:

r (t) = µ

2he +

(r0 − µ

2he)

cosh H (t)

+ r0v0√2h

sinh H (t) (65)

r (t) = − µ

2h+ r0 · v0√

2hsinh H (t)

+(

r0 + µ

2h

)cosh H (t) (66)

v (t) = [r (t)]−1 [r0v0 cosh H (t)

+2hr0 − µe√2h

sinh H (t)

](67)

where H (t) = √2hτ (t) is the generalized eccentric

anomaly. The generalized Kepler equation is:

t − t0 = −1

nH (t) + r0 · v0

2h[cosh H (t) − 1]

+(

r0 + µ

2h

) sinh H (t)√2h

. (68)

4.2 Zero specific angular momentum (� = 0):the rectilinear Keplerian motion

In this case, the motion takes place on a section of astraight line that has the vectorial Eq. 14. All compu-tations from paragraphs 4.1.1, 4.1.2, 4.1.3 modify bytaking e = −r0/r0, e = 1, and b = 0 in case h �= 0,� = 0 in case h = 0. The results are presented below.

4.2.1 Negative specific energy h < 0

� When τ (tP ) = 0, the solution to the initial valueproblem (20) is:

r (t) = a [1 − cos E (t)]r0

r0, t ∈ [t0, tP + 2π

n) (69)

where a = µ/ (2 |h|) . The magnitude of the positionvector (69) is:

r (t) = a [1 − cos E (t)] , t ∈ [t0, tP + 2π

n). (70)

The velocity vector is:

v (t) = na sin E (t)

1 − cos E (t)

r0

r0, t ∈ [t0, tP + 2π

n). (71)

Kepler’s equation is deduced in the same way:

n (t − tP )= E (t)−sin E (t) , t ∈ [t0, tP + 2π

n). (72)

E0 ∈ [0, 2π) is uniquely defined by:⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

cos E0 = 1 − nr0√2 |h| ;

sin E0 = n

2 |h| (r0 · v0) .

(73)

The moment of time tP is computed now from:

tP = t0 − 1

n(E0 − sin E0) . (74)

As the trajectory is bounded and rectilinear, the bodywill “return” to the attraction center. The “next”moment of time the body reaches the attraction center istc, that represents the collision moment. It correspondsto E = 2π . From Eqs. 72 and 74, it results:

tc = t0 − 1

n(E0 − sin E0) + 2π

n. (75)

� By choosing τ (t0) = 0, it results:

r (τ ) =[

µ

2 |h| +(

r0 − µ

2 |h|)

cos H (t)

+ r0 · v0√2 |h| sin H (t)

]r0

r0(76)

r (t) = µ

2 |h| +(

r0 − µ

2 |h|)

cos H (t)

+ r0 · v0√2 |h| sin H (t) (77)

v (t) = 1

r (t)[−√

2 |h|(

r0 − µ

2 |h|)

sin H (t) +

+ (r0 · v0) cos H (t)]r0

r0. (78)

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474 Meccanica (2007) 42:465–476

The generalized Kepler equation is:

t − t0 = 1

nH (t) + r0 · v0

2 |h| [1 − cos H (t)]

+(

r0 − µ

2 |h|)

sin H (t)√2 |h| . (79)

4.2.2 Zero specific energy h = 0

� When τ (tP ) = 0, the solution to the initial valueproblem (20) is:

r (t) = 1

2µτ 2 (t)

r0

r0. (80)

The magnitude of the position vector (80) is:

r (t) = 1

2µτ 2 (t) . (81)

The velocity has the expression:

v (t) = 2

τ (t)

r0

r0. (82)

Kepler’s equation is deduced in the same way:

t − tP = µτ 3 (t)

6. (83)

The initial value τ0 = τ (t0) is computed fromEq. 82 by dot-multiplication with v0. It results:

τ0 = r0 · v0

µ. (84)

The moment of time tP may now be computed fromEqs. 83 and 84:

tP = t0 − µ

6

(r0 · v0

µ

)3

. (85)

Map τ (t) has the explicit expression (from Eq. 83):

τ (t) =[

6 (t − tP )

µ

] 13

. (86)

Two major situations occur:

1. The case r0 · v0 > 0 : The body is launched in theopposite direction relative to the attraction center.The motion takes place for t ∈ [t0,+∞) (in thiscase tP < t0).

2. The case r0 ·v0 < 0 : The body is launched towardsthe attraction center. The motion takes place for

t ∈ [t0, tP ) (in this case tP > t0). The collisionmoment is given in Eq. 85.

� By choosing τ (t0) = 0, it results:

r (t) =[

r0 + (r0 · v0) τ (t) + µτ 2 (t)

2

]r0

r0(87)

r (t) = r0 + (r0 · v0) τ (t) + µτ 2 (t)

2(88)

v (t) = [r (t)]−1 [(r0 · v0) + µτ (t)]r0

r0. (89)

The generalized Kepler equation is:

t − t0 = r0τ (t) + (r0 · v0)τ 2 (t)

2+ µ

τ 3 (t)

3. (90)

4.2.3 Positive specific energy h > 0

� When τ (tP ) = 0, the solution to the initial valueproblem (20) is:

r (t) = a [cosh E (t) − 1]r0

r0. (91)

where a = µ2h . The magnitude of the position vector

is:

r (t) = a [cosh E (t) − 1] (92)

and the velocity is computed using Eq. 22:

v (t) = na sinh E (t)

cosh E (t) − 1

r0

r0. (93)

Kepler’s equation becomes:

n (t − tP ) = sinh E (t) − E (t) . (94)

The initial value E0 of map E is computed from:

E0 = sinh−1 n (r0 · v0)

2h. (95)

The moment of time tP is computed using Eqs. 94and 95:

tP = t0 − 1

n

[n (r0 · v0)

2h− sinh−1 n (r0 · v0)

2h

]. (96)

Two major situations occur:

1. The case r0 · v0 > 0 : The body is launched in theopposite direction relative to the attraction center.The motion takes place for t ∈ [t0,+∞) (in thiscase tP < t0).

2. The case r0 ·v0 < 0 : The body is launched towardsthe attraction center. The motion takes place for

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Meccanica (2007) 42:465–476 475

t ∈ [t0, tP ) (in this case tP > t0). The collisionmoment is given in Eq. 96.

Map E = E (t) is the solution to the implicitfunctional equation:

sinh E (t) − E (t) = n (t − t0) + sinh E0 − E0. (97)

� By choosing τ (t0) = 0, it results:

r (t) = [− µ

2h+

(r0 + µ

2h

)cosh H (t) (98)

+r0 · v0√2h

sinh H (t)]r0

r0

r (t) = − µ

2h+

(r0 + µ

2h

)cosh H (t)

+r0 · v0√2h

sinh H (t) (99)

v (t) = 1

r (t)[(r0 · v0) cosh H (t)

+2hr0 + µ√2h

sinh H (t)

]r0

r0. (100)

The generalized Kepler equation is:

t − t0 = −1

nH (t) + r0 · v0

2h[cosh H (t) − 1]

+(

r0 + µ

2h

) sinh H (t)√2h

. (101)

4.2.4 The phase plane portrait for the rectilinearKeplerian motion

By denoting

r = xr0

r0, x : [t0,+∞) → R, (102)

the initial value problem that describes the motion inthis particular case becomes:

x + µ

|x |3 x = 0, x (t0) = r0, x (t0) = r0 · v0

r0. (103)

The prime integrals of the initial value problem (103)yield from the prime integrals (2), (3) and (4). Theyhave the particular form:

1

2x2 − µ

|x | = 1

2v2

0 − µ

r0= h (104)

Fig. 5 The phase plane portrait in the rectilinear Keplerianmotion with respect to the time variable t : for h = 0, the delim-itating orbit is obtained; it separates the motions with vanishingvelocity (h < 0) from the motions without annulation of velocity(h > 0)

(the energy conservation);

rr

= r0

r0(105)

(the motion is rectilinear).From (104), it results:

x2 = 2h + 2µ

|x | . (106)

The phase plane portrait with respect to the time-variable t is given in Fig. 5. It is obtained from Eq. 106.

Taking into consideration the regularized initial valueproblem that describes the Keplerian motion (20) andthe notation in Eq. 102, the phase plane portrait withrespect to the fictitious time variable τ introduced byEq. 16 is made by using:

(x ′)2 = 2 |x | (µ + h |x |) . (107)

The phase plane portrait with respect to the fictitioustime τ is presented in Fig. 6. It is worth noticing thatthe orbits in the phase plane are: pairs of symmetricalellipses for h < 0, pairs of symmetrical parabolas forh = 0 and pairs of symmetrical hyperbolas for h > 0.

Looking to the rectilinear Keplerian motion withrespect to the time variable t , the attraction point isreached with infinite magnitude of the velocity. Themathematical model stops functioning at this point. Incase of collision, the time until the body reaches theattraction center is finite. The collision moment of time

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476 Meccanica (2007) 42:465–476

Fig. 6 The phase plane portrait in the rectilinear Keplerianmotion with respect to the fictitious time variable τ : for h = 0,

the delimitating orbit is obtained; it separates the motions withvanishing velocity (h < 0) from the motions without vanishingvelocity (h > 0).

is given by Eqs. 74 or 75 in case h < 0, by Eq. 85 whenh = 0 and by Eq. 96 when h > 0. To our knowledge,these results have not been pointed out until now.

5 Conclusions

The paper presented a closed form unified solution tothe classic Kepler problem. The key to this comprehen-sive study is the vectorial regularization that leads to anew initial value problem with an exact solution.

All cases are studied, including the rectilinear one.Also, the study is made with respect to the manner thevalue zero is chosen for the fictitious time: in the peri-center of the trajectory or at the initial moment of time.Closed form vectorial expressions for the law of motionand the velocity are obtained. The collision momentof time (in the occurring situations) is explicitly com-puted.

Without geometrical considerations, Kepler’s equa-tions and Kepler’s generalized equations are unitarilydeduced in each case. Phase plane portraits (in the rec-tilinear case) are presented with respect to both vari-ables: the time t and the fictitious time τ introduced bythe standard Sundman transformation.

All considerations in this paper start only from thethree prime integrals of the Kepler problem. The solu-tion is elementary and it involves only simple vectorialcomputations.

A study of the Kepler problem in non-inertial ref-erence frames using a similar approach might be thesubject for a forthcoming paper.

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