Lecture 34: Performance to GEO

16.512, Rocket Propulsion Prof. Manuel Martinez-Sanchez Lecture 34: Performance to GEO ∆V Calculations for Launch to Geostationary Orbit (GEO) Idealized Direct GTO Injection (GTO = Geosynchronous Transfer Orbit) Assumptions: - Ignore drag and "gravity" losses Assume impulsive burns (instantaneous impulse delivery) Assume all elevations α>0 at launch are acceptable Launch is from a latitude L, directed due East for maximum use of Earth's rotation. The Eastward added velocity due to rotation is then vR = ΩE RE cosL = 463 cosL (m/s) (1) If the launch elevation is α, and the desired velocity after the first burn is V1, the rocket must supply a velocity increment (2) ∆V1 = V12 + vR2 − 2 V1 vR cos α vR v1 ∆V1 α The trajectory will then lie in a plane LOI through the Earth's center which contains the local E-W line. In order to be able to perform the plane change to the equatorial plane at GEO, we select the elevation α such as to place the apogee of the 1/3 2 ⎛ ⎞ transfer orbit (GTO) at the GEO radius R GEO = ⎜ µ T 2 ⎟ (T = 24 hr, µ = 3.986 × 1014 m3/s2) ⎝ 4π ⎠ = 42, 200 km North ΩE L V1 O α EQ UA TO R R GE O GTO I EQUATORIAL GEO ORBIT Fig. 1 Since OL is perpendicular to OI, the view in the plane of the orbit is: V1 α L r RE θ P I o RGEO Fig. 2 The polar equation of the trajectory is r = In our case p = R E (corresponding to θ = p ,>0 1 + e cos θ π ). The elevation is given by 2 ⎛ ⎛ dr ⎞ e sin θ tan α = ⎜ = ⎜ ⎟ ⎜ (1 + e cos θ )2 ⎝ r dθ ⎠ θ=π / 2 ⎝ ⎞ ⎟ = e ⎟ ⎠θ=π / 2 and, in turn, the eccentricity follows from (at θ = π ) R GEO = and so tan α = 1 − RE 1−e e =1− RE = 0.849 R GEO ; RE RGEO 0 α = 40.3 (3) The angular momentum (per unit mass) is h = µp = µR E . Equating this to R E V1 cosα V1 cos α = , µ RE (4) (i.e., the horizontal projection of the launch velocity is the local orbital speed, for any apogee radius, RGEO in this case) Combining (3) and (4), V1 = 2 ⎛ RE ⎞ ⎤ µ ⎡ ⎢1 + ⎜1 − ⎟ ⎥ RE ⎢ R GEO ⎠ ⎥ ⎝ ⎣ ⎦ (5) and this can now be substituted in (2): ∆V1 = 2 ⎛ R ⎞ ⎤ µ ⎡ µ ⎢1 + ⎜1 − E ⎟ ⎥ + vR 2 − 2vR RE ⎢ R R GEO ⎠ ⎥ E ⎝ ⎣ ⎦ ∆V1 = ⎛ µ ⎞ RE ⎞ µ ⎛ − vR ⎟ + ⎜⎜ ⎜1 − ⎟ ⎟ RE ⎝ R GEO ⎠ ⎝ RE ⎠ 2 2 (6) L = i Δ Upon arrival at I, there will have to be a second burn that will simultaneous µ accelerate the rocket to vGEO = , and rotate the plane to equatorial ( R GEO ). vGEO ∆i ∆Va va,GTO Fig. 3 The apogee velocity is va,GTO , given by R GEO va,GTO = ( V1 cosα) R E = µR E and so ∆Va = (7) 2 2 vGEO + va,GTO − 2vGEO va,GTO cos ∆i ∆Va = µ R GEO 1+ RE R - 2 E cos L R GEO R GEO (8) This second burn is probably provided by the spacecraft itself, or else by the launcher's upper stage. IDEALIZED TWO - BURN GTO INJECTION One difficulty with the direct injection scheme is the fact that GEO insertion at I must occur on the first pass, because the GTO perigee is actually below the Earth's surface (see Fig. 2). Most operators prefer a temporary parking of the spacecraft in a GTO orbit which has a perigee above the ground, so as to make functional tests and adjustments prior to the final apogee burn (over a period of 2-4 weeks). A modification of the launch sequence to accommodate this is: (1) Fire Eastwards with α selected for a low apogee ( ∼ 200 km above ground) at the equatorial crossing. (2) Fire again at equatorial crossing to raise the apogee to RGEO (no plane change) (3) At one of the apogee passes, perform the final (circularization + plane change burn). The formulation is very similar to the previous case. The elevation α is now given by tan α = 1 − ( Rp = perigee radius RE Rp (9) RE + 200 km ). This gives a very shallow trajectory, which is unrealistic; but it is a fair approximation to a real high-elevation launch, followed by a rapid rotation during the rocket firing. For RP - R∈ = 200 km , α = 1.740 . z ACTUAL IDEALIZED α x Fig. 4 Eqs. (5) and (6) still hold, with the quality R GEO replaced by Rp , and so 2 2 ⎛ µ ⎞ RE ⎞ µ ⎛ - vR ⎟ + ∆V1 = ⎜ ⎜1 ⎟ ⎜ R ⎟ RE ⎝ RP ⎠ E ⎝ ⎠ (10) which is now smaller, since we are going to a much lower apogee (at rp ). At this apogee (at the equatorial crossing), we have, as in Eq. (7), va = µR E Rp (11) and we next need to effect a second rocket firing that will increase velocity to that for the GTO perigee: vPGTO = µ 2R GEO R p R p + R GEO (12) No plane change is involved yet, so µ Rp ∆V2 = ⎡ 2R GEO RE ⎤ − ⎢ ⎥ R P ⎥⎦ ⎢⎣ R p + R GEO (13) This places the spacecraft on an elliptical GTO orbit, still in the original plane, with apogee at R GEO . The speed at this apogee is: va,GTO = µ R GEO 2R P RP + R GEO (14) and so, ∆Va = ∆Va = ∆Va = µ R GEO + vGEO2 + v2a,GTO − 2vGEO va,GTO cos L µ R GEO 2R P µ −2 R P + R GEO R GEO 2R P cos L R P + R GEO 2R P 2R P µ 1+ −2 cos L R GEO R P + R GEO R P + R GEO (15) vGEO L ∆Va va,GTO Fig. 5 Some numerical comparisons We will illustrate these ∆V 's by considering launches to GEO from two different locations: (1) Near the Equator, on at the French kouron complex, and (2) From mid-latitude, as from Café Canoveral ( L = 28.50 ). (1) Equatorial Launch Option (a): Ground to LEO (300 km), plus LEO-GEO Hohman transfer. No plane changes. Launch to the East. ∆V = ∆V1 + ∆V2 − VR + ∆V3 + ∆V4 GEO circularization GTOinjection To LEO, α= 0 ∆V = (8084 – 463) + (10,151 – 7725) + (3071 – 1573) = 7,621 + 2,426 + 1,498 = 11,545 m/s Notice this is more than to Escape from mean Earth ( ∆V Option (b): Direct injection into GTO from ground ∆V = ∆V1 + α= 0 launch to R = 42,200km ( −463 m / s for rotation) 11,200 m / s ) ∆V2 GEO circularization = (10,420 – 463) + (3071 – 1573) = 9,957 + 1,498 = 11,455 m/s (2) Launch from L = 28.5o. Launch to East, υR = 407 m / s Option (a): Direct injection to GTO, circularization + plane change at GEO. 2 firings, ∆V = ∆V1 + Launch with α= 40.30 ∆V2 GEO circularization andplane change = 10,070 + 2,102 = 12,172 m/s Note the two penalizations for latitude: the elevated launch increased ∆V1 , and the plane change at GEO increases ∆V2 . Option (b) Direct injection with 3 firings (LEO at 300km) ∆V = ∆V1 Launch to a300 km apogee + ∆V2 Firing to raise apogee to GEO = 7,512 + 2,605 + 1,830 = 11,947 m/s + ∆V3 Circularization + Plane change Is it true that plane change should be all done at end of GTO? Actually, a small turning combined with initial ∆V1 (say, from LEO) costs very little ∆V loss, even though V is then large. Try splitting into a ∆i1 and ∆i2 = ∆i − ∆i1 v2c1 + vp2GTO − 2vc1 vpGTO cos ∆i1 ∆V1 = ∆V = ∆V1 + ∆V2 ∆V2 = v2c2 + vp2GTO − 2vc2 vaGTO cos ( ∆i − ∆i1 ) +2vc1 VP sin ∆i1 +2 vc2 Va sin ( ∆i − ∆i1 ) d∆V = − =0 2 2 2 d∆i1 2 vc1 + VP − 2vc1 VP cos ∆i1 2 vc2 + Va2 − 2vc2 Va cos ( ∆i − ∆i1 ) vc1 = Call ρ = µ , R1 1+ 1+ρ µ , R2 2R 2 µ , R1 R1 + R 2 vp = va = 2R1 µ R 2 R1 + R 2 R2 R1 2 2ρ vc2 = sin2 ∆i ρ= ρ sin ∆i1 1+ρ 2ρ 2ρ −2 cos ∆i1 1+ρ 1+ρ 1 = ρ 1 2 sin ( ∆i − ∆i1 ) ρ1+ρ 1 1 2 2 + − ρ ρ1+ρ ρ 1 2 cos ( ∆i − ∆i1 ) ρ1+ ρ ⎤ ⎡ ⎤ 1 ⎡ 2 2 1 2 2ρ 2ρ cos ( ∆i − ∆i1 ) ⎥ = 2 sin2 ( ∆i − ∆i1 ) ⎢1 + cos∆i⎥ −2 −2 ⎢1 + 1+ρ 1+ρ 1+ρ 1+ρ ρ ⎢⎣ ⎥⎦ ρ 1 + ρ ⎢⎣ ⎥⎦ 42200 = 6.14265 6370 + 500 1.31148 Sin ∆i1 1 + 1.71999 − 2 × 1.31148 Cos ∆i1 2ρ = 1.31148 1+ρ = 1 6.14265 0.52916 Sin (28.5 − ∆i1 ) 6.14265 1 + 0.28001 − 2 × 0.52916 Cos (28.5 − ∆i1 ) Sin ∆i1 2.71999 − 2.62296 Cos ∆i1 = ∆i1 = 2.260 optimum 0.16280 Sin (28.5 − ∆i1 ) 1.28001 − 1.05832 Cos ( 28.5 − ∆i1 ) ∆i2 = 26.240 ⎛ ∆V ⎞ 2ρ 2ρ 1 1 2 2 −2 cos∆i1 + + − ⎜ ⎟ = 1+ ⎜ vc ⎟ 1+ρ 1+ρ ρ ρ1+ρ ρ ⎝ 1 ⎠op ⎛ ∆V ⎞ ⎜ ⎟ = 2.71199 − 2.62296 cos ∆i1 + ⎜ vc ⎟ ⎝ 1 ⎠op 1 6.14265 2 cos ∆i2 ρ (1 + ρ ) 1.21001 − 1.05832 cos ∆i2 = 0.30178 + 0.23227 = 0.53405 - small improvement Compare to same with ∆i1 = 0 ⎛ ∆V ⎞ ⎜ ⎟ = 0.29838 + 0.23868 = 0.53706 - small improvement ⎜ vc ⎟ ⎝ 1 ⎠ref Example: Effects of doing a small plane change ∆i2 simultaneous with the second (apogee-raising) firing in a 3-impulse direct GTO injection. 1.58 1.57 1.56 (R2-RE)/RE= 0.25 dVTot/vcE 1.55 0.2 0.15 1.54 0.1 1.53 1.52 (R2-RE)/RE=0.05 1.51 1.5 1 2 3 4 5 6 7 8 9 10 Di2(deg) Total dV for three-impulse launch from L=28.5 deg to GEO. Here vcE =sqrt(mu/RE) 0.975 (R2-RE)/RE= 0.25 0.97 dV1/vcE 0.965 0.96 0.955 0.95 (R2-RE)/RE= 0.05 0.945 1 2 3 4 5 6 Di2(deg) 7 8 9 10 dV1 for three-impulse launch from L=28.5 deg to GEO. Here vcE=sqrt(mu/RE) 0.4 0.39 0.38 (R2-RE)/RE=0.25 dV2/vcE 0.37 0.36 (R2-RE)/RE=0.05 0.35 0.34 0.33 1 2 3 4 5 6 7 8 9 10 Di2(deg) dV1 for three-impulse launch from L=28.5 deg to GEO. Here vcE=sqrt(mu/RE) 0.23 0.225 0.22 dV3/vcE (R2-RE)/RE=0.05 0.215 0.21 (R2-RE)/RE=0.25 0.205 0.2 0.195 0.19 1 2 3 4 5 6 7 8 9 10 Di2(deg) dV3 for three-impulse launch from L=28.5 deg.to GEO. Here, vcE=sqrt(mu/RE)