An Error Analysis of Elliptical Orbit Transfer Between Two Coplanar Circular Orbits

Transcription

1 An Error Analysis of Elliptical Orbit Transfer Between Two Coplanar Circular Orbits Marc A. Murison U.S. Naval Observatory, Washington, DC 9 November, 006 Abstract We consider transfer orbits between two coplanar, confocal circular orbits. We calculate the semimajor axis eccentricity of the elliptical transfer orbit, as well as the energy velocity magnitude changes required to accomplish the transfer. We assume the energy changes i.e., thruster firings) occur over a short time interval compared to the relevant orbital periods. We then consider the consequences of small errors in the thruster firings. Subject headings: celestial mechanics orbit transfer The XML version of this document is available on the web at The PDF version of this document is available on the web at 1

2 1 Motivation Orbital Elements of the Transfer Orbit 3 Energy Considerations 3 4 Energy Changes to Accomplish the Transfer 4 5 Changes in Relative Velocity Magnitude to Accomplish the Transfer Using Physics Using Algebra Expansions 6 7 Effects of Errors in the Velocity Changes Impulse Error at Pericenter Perturbed Transfer Orbit Elements Circularize the Outer Orbit at the Perturbed Radius Eliminating the First-Order Term An Eccentric Perturbed Outer Orbit Impulse Error at Apocenter Motivation Consider two coplanar, circular, confocal orbits of semimajor axes >. Suppose we wish to transfer an artificial satellite from the inner to the outer orbit do it in such a way that the transfer orbit is elliptical as opposed to hyperbolic or parabolic) confocal to the circular orbits. This type of orbit was first considered by W. Hohmann in 195. At some point during the inner circular orbit, a thruster firing perpendicular to the radius vector occurs, at which point the satellite s motion switches to the pericenter of the elliptical transfer orbit. The ellipse is such that its apocenter distance coincides with the outer circular orbit distance. After coasting from the pericenter out to the apocenter, another thruster firing again in the orbital plane perpendicular to the instantaneous radius vector) places the satellite on the outer circular orbit. This process is symmetric, so one can just as easily transfer from an outer orbit to an inner one, requiring only a few changes in sign in what follows. Hence, what are the semimajor axis a eccentricity e of the transfer orbit, what are the changes in energy velocity magnitude required to perform the transfer? Orbital Elements of the Transfer Orbit Now, the pericenter of the transfer ellipse call it P) coincides with the radius of the smaller circle, a e) =, the apocenter call it A) coincides with the radius of the larger

3 circle, a1 + e) =. Thus, we may combine these two relations to find the semimajor axis eccentricity of the transfer orbit: a = 1 + ) 1) e = + ) Thus, we have a tidy result. The semimajor axis of the transfer ellipse is the average of the radii of the circular orbits, the eccentricity is the fractional difference of those radii. The amount of time spent in the transfer orbit is, by definition, half the orbital period of the transfer orbit. Kepler s third law can be written = n a 3 3) where = Gm 1 + m ), n =π/t is the mean motion, T is the orbital period. Using 1), we therefore find T = π a 3 = π + ) 3 4) 8 3 Energy Considerations The specific energy E of a two-body orbit is E = 1 v 5) r where v is the magnitude of the relative velocity vector, r is the magnitude of the relative position vector. The specific energy is the energy of the two-body system, E tot, divided by the reduced mass, m 1 m E = 1 m 1 m v Gm 1m = E tot 6) m 1 + m m 1 + m r One can also show that v = r 1 ) 7) a which is a statement of conservation of energy known as the vis viva integral. Combining 5) 7) lets us write the specific energy as E = 8) a Thus, the difference in energy henceforth we shall assume energy means specific energy ) between the two circular orbits is ΔE = 1 1 ) 9) 3

4 4 Energy Changes to Accomplish the Transfer From 8) 1), we can write the energy of the transfer orbit, E = 10) + Hence, we discover that the changes in energy that must occur at P at A are, from 10) 8), ΔE P = E E P = + = ee P 11) ΔE A = E A E = = ee A 1) + where E P = E A = are the energies of the respective circular orbits. It is quite delightful that the changes in energy are just the respective orbital energies scaled only by the transfer orbit eccentricity. One can easily show that 11) 1) add up to 9): ΔE P +ΔE A = ee P + E A )=ΔE 13) 5 Changes in Relative Velocity Magnitude to Accomplish the Transfer Let us assume that the changes in orbital energy are accomplished by thruster firings which are very short compared to the orbital periods involved. Then, since the changes occur at pericenter apocenter of the elliptical transfer orbit, our assumptions let us put ṙ = 0, the energy changes involve only the kinetic energies hence the relative velocity magnitudes. 5.1 Using Physics Differentiating 5), we have de dt = v dv dt + dr r dt = v dv 14) dt since during the thruster firings ṙ = 0. Integrating through the thruster firing from say) times t 0 to t 0 +Δt, we have E 0 ˆ+ΔE E 0 de = v 0 ˆ+Δv v 0 vdv 15) ΔE = 1 v 0 +Δv) 1 v 0 = 1 Δv + v 0 Δv 16) 4

5 where v 0 = vt 0 ) is the magnitude of the velocity the instant before the thruster firing, Δv is the velocity magnitude impulse arising from a change in energy ΔE. Thus, the changes in relative velocity magnitude are Δv = v 0 ± ΔE + v0 17) at the respective orbital locations. At P, v 0 P )= 18) which we obtained from eq. 7) for the interior circular orbit, while at A we similarly find e v 0 A) = +e = + 19) for the elliptical orbit. Plugging 18) 19), 11) 1), into 17), we find that ) ) a Δv P = 1 ± 1+e = 1 ± + 0) ) a1 Δv A = ± 1 + 1) We must choose the signs in 0) 1) to match physical circumstances. The first thruster firing is in the same direction as the direction of motion around the inner circular orbit, so the change in velocity magnitude is positive. Likewise, the second thruster firing is also along the direction of motion, so that change, too, is positive. Therefore, since we are considering the case >, the changes in velocity magnitude are Δv P = ) ) a 1+e 1 = 1 + Δv A = a1 + In each case the change in velocity magnitude is a fraction of the respective circular velocity v c = /a, the fractions being the expressions in parentheses in eqs. ) 3). 5. Using Algebra The previous derivation uses conservation of energy as the starting point. It is therefore satisfying in that is makes use of a fundamental physical principle. If we are willing to forgo ) ) 3) 5

6 physics, then here is a short algebraic derivation. From 7) the circular velocities at P at A are v c P )= v c A) = 4) The velocities on the transfer orbit at those same points are again using 7)) 1+e v P = e = 5) + e v A = +e = Thus, the changes in velocity magnitude are just which yield eqs. ) 3). 6 Expansions + 6) Δv P = v P v c P ) 7) Δv A = v c A) v A 8) Suppose. Then we can exp on ε = /, ) 3) become ) 1 3 Δv P = ε 8 ε3/ + 5 ) 16 ε5/ which we can also write as ) 1 Δv P = ε 3 8 ε + 5 ) 16 ε3 ε 1 Δv A = + ε3/ 3 ) 8 ε5/ which, similarly, we can write as Δv A = ε + 1 ε 3 ) 8 ε3 From 9) 31) we see that the thruster firings consist of a large kick of order the circular velocity at the corresponding radius) followed by higher order terms. 9) 30) 31) 3) 6

7 The ratio of velocity kicks is Δv A = 1+ε ε ε 33) Δv P 1+ε A series expansion of 33) yields [ Δv A Δv P = u ) 1 1 u ) ) u u 3) 7 4 ) u 1) 4 u 5) ) u 1) 7 u 8) ) 4 u 5 u 6) ) 1) 6 u 9 u 10) ] 34) where we have set u = ε = /. We carry out the expansion to an impractical number of terms in order to show the interesting pattern of the expansion coefficients. 7 Effects of Errors in the Velocity Changes 7.1 Impulse Error at Pericenter Perturbed Transfer Orbit Elements Suppose, as is always the case in the real world, an error occurs the change in velocity magnitude imparted by the first thruster firing is in error by some small amount, say Δv P =Δv 0 P + δv P,whereΔv 0 P is the desired change in velocity given by ) δv P is the error. What are the effects on the transfer orbit semimajor axis eccentricity, on the outer orbit? Using ), we find the variation in Δv P is δδv P = δv P = Δv P δ = 1 + ) 3 δ 35) Thus, we can write the resulting error in the radius of the resulting circular orbit assuming, ideally, that a compensating adjustment of the thrust at A circularizes it), or 1 + ε) 3 δ = [ ] 1/ + ) 3 δv P 36) 1+ 3 ε ε ε 1 ) δvp 16 ε3 v c A) δ δv P = ε v c A) = where v c A) = / is the unperturbed outer orbit circular velocity ε = /. 37) 7

8 From eq. ) for e, we have the error in the transfer orbit eccentricity, δe = + ) δ = ε 1 + ε) δ = ε ε +3ε 3 4ε 4 )δ 38) while the corresponding error in the transfer orbit semimajor axis is δa = 1 δ 39) 7.1. Circularize the Outer Orbit at the Perturbed Radius What is the impulse required to circularize the outer orbit? That would be the difference in velocities between the circular orbit of perturbed radius + δ the apocenter of the perturbed transfer ellipse of semimajor axis a + δa. Thus, from 4) 6), we may write Δv A = 40) + δ + + δ + δ which, to first order, becomes or Δv A =Δv 0 A δδv A = [ Δv A = [ a1 + ε ε 1+ε 1+ε a ε 1+ε ) ] δ ) ] δa 41) 4) where Δv 0 A is the unperturbed circularizing impulse, eq. 3). Hence, the adjustment impulse the small change in velocity that we must subtract from the unperturbed velocity change given by 3) is, to first order, δδv A = 1 ) ε +ε δ 43) 1+ε 1+ε Substituting eq. 36) for δ, we can express the corrections in terms of the original velocity error at pericenter. Then eq. 4) becomes ) [ ] Δv A ε 1+ε v c A) = δv P 1 + ε) + ε) 44) 1+ε ε v c A) Another way to achieve the same results is to consider instead the variation of Δv A when varies. From 4), 6), 8), δδv A = Δv A δa [ )] = δa a1 + 45) ) = a1 + δa a + + 8

9 which we see is identical to 43). Yet a third way is to consider eq. 6) plug in the variations directly. We have v A = e +e = a + δa e δe 1+e + δe for the perturbed transfer orbit aphelion velocity. Use 1) ) for a e, then38) 39) for δe δa. One finds 46) + ) δ v A = = + + δ + )+ δ + The circular velocity at the perturbed radius is v c + δ )= = δa ) + δ The difference is then the circularization impulse: a ) 1 + δ + 47) 48) Upon substituting 47) 48), one finds again eq. 41) Eliminating the First-Order Term Δv A = v c + δ ) v A 49) We see from 43) 44) that for some value of the ratio / the first-order correction will be zero. Solving the expression in brackets in 44) is equivalent to solving 1+ε) 3 = ε 4 + ε) 50) Eq. 50) has three real solutions with two of them being negative hence unphysical. Thus, the adjustment δδv A is a positive quantity when < 1 10cos 3 3 arctan 3 111) ) 5 = ) 3 Therefore, oddly enough, if the ratio of unperturbed circular orbits is the value given by equality in eq. 51), then the adjustment impulse is second order in the perturbation δ. Numerically, the expansion 41) becomes [ ) ) 3 δa δa Δv A = v c A) ] 5) eq. 44) becomes [ ) ) 3 δvp δvp Δv A = v c A) ] v c A) v c A) 53) where v c A) = / is the unperturbed outer orbit circular velocity. 9

10 7.1.4 An Eccentric Perturbed Outer Orbit The foregoing consider the consequences that result from an error δv at the pericenter point P of the transfer orbit, assuming the Δv A impulse is adjusted to compensate so that we end at A with a circular orbit now of radius + δ. If the velocity error at P is not detected or for some other reason the impulse at A is not adjusted to compensate, then application of the unperturbed Δv A from eq. 3) will result in an eccentric orbit. Now, the velocity at the apocenter of the perturbed transfer orbit is given to first order by 47). Application of the planned thrust Δv 0 A at the apocenter, eq. 3), results in the new velocity magnitude We find v = v A +Δv 0 A 54) v = a δ ) ) ε +εδ = v c A) 1+ε 1+ε 55) This velocity is now the pericenter or apocenter of the final orbit, which is an ellipse. We would like to know the orbital elements of this ellipse. distance is, respectively, r = a1 ± e). Thus, we can write The apocenter or pericenter a1 ± e) = + δ 56) From the vis viva integral 7), we have v = a1 ± e) 1 ) = 1 e 57) a ± e Now we may solve 56) 57) to get ±e =1 + δ v 58) + δ a = v a 59) + δ ) Using 55) for v, we find the first-order results ) ε +ε ±e = 1+ε 1+ε 1 δa = ) δa ε 1+ 60) ) a ε +ε δa =1+ =1+ ) δa ε 61) 1+ε 1+ε If the right h side of 60) is positive/negative, the thruster firing point is the final ellipse apocenter/pericenter. 10

11 7. Impulse Error at Apocenter Suppose the impulse at P occurs as planned, but the thrust at the transfer orbit apocenter has an error. An impulse error at A, δv A, will cause the outer orbit to be eccentric rather than circular. The velocity magnitude after the erroneous thruster firing is then v =Δv A + δv A 6) where the unperturbed circularizing impulse Δv A is given by 3). Similar to eq. 56), we have while eq. 57) still holds. Solving 57) 63), we have a1 ± e) = 63) ±e =1 a ) v v =1 64) v c A) a = 1 v = 1 v Substituting 6) for v 3) for Δv A, we find ±e =1 exact) v ca) ) ΔvA + δv ) ) A ε ε =1 v c A) 1+ε 1+ε a 1 = ΔvA + Δv Avc A) δv v vca) c A) Δv A) A 1+ε = +1+ε) 1 ε1+ε)+ε 1+ ε1+ε) ε ) 65) δv A 1+ ε1+ε) ε δv A v c A) ) δva 66) v c A) v ca) 67) where v c A) = / 67) is to first order in δv A.Iftheratioε = / is small, then ±e = ε ε ) ) ) δva ε v c A) δva 68) v c A) a = ) ε +10ε + 5 ) δva ε +36ε 69) v c A) 11