16.522, Space Propulsion Prof. Manuel martinez-Sanchez Lecture 4: Re-positioning in Orbits Suppose we want to move a satellite in a circular orbit to a position A9 apart in the same orbit in a time At (assumed to be several orbital times at least). The general approach is to transfer to a lower(for positive A9)or higher(for 49<0)nearby orbit then drift in this faster (or slower) orbit for a certain time then return to the original orbit The analysis is similar for low and high thrust, because we have radius ratios very close to 1, so that in either case the satellite is nearly in orbit"even during thrusting periods, and Av's for orbit transfer amount (in magnitude) to the difference of the beginning and ending orbital speeds. In detail of course, if done at high thrust the maneuver involves a two-impulse Hohmann transfer to the drift orbit and one other two-impulse Hohmann transfer back to the original orbit For the low-thrust case, continuo us thrusting is used during both legs, with some guidance required to remove the very slight radial component of v picked up during spiral flight (although ignored here) We will do the analysis for the low-thrust case only then adapt the result for high- et 89 be the advance angle relative to a hypothetical satellite remaining in the original orbit and left undisturbed the general shape of the maneuver is sketched Forward △e Thrust △6-6 coasting se Δt-t1 Backwards Thrust Since the orbital angular velocity is its variation with orbit radius is 16.522, Space Propulsion Lecture 4 Prof. Manuel martinez-Sanchez Page 1 of 10
16.522, Space Propulsion Prof. Manuel Martinez-Sanchez Lecture 4: Re-positioning in Orbits Suppose we want to move a satellite in a circular orbit to a position ∆ϑ apart in the same orbit, in a time ∆t (assumed to be several orbital times at least). The general approach is to transfer to a lower (for positive ∆ϑ ) or higher (for ∆ϑ < 0 ) nearby orbit, then drift in this faster (or slower) orbit for a certain time, then return to the original orbit. The analysis is similar for low and high thrust, because we have radius ratios very close to 1, so that, in either case the satellite is nearly “in orbit” even during thrusting periods, and ∆V's for orbit transfer amount (in magnitude) to the difference of the beginning and ending orbital speeds. In detail, of course, if done at high thrust the maneuver involves a two-impulse Hohmann transfer to the drift orbit and one other two-impulse Hohmann transfer back to the original orbit. For the low-thrust case, continuous thrusting is used during both legs, with some guidance required to JG remove the very slight radial component of v picked up during spiral flight (although ignored here). We will do the analysis for the low-thrust case only, then adapt the result for highthrust. Let δϑ be the advance angle relative to a hypothetical satellite remaining in the original orbit and left undisturbed. The general shape of the maneuver is sketched below: Since the orbital angular velocity is Ω = µ 3 , r its variation with orbit radius is 16.522, Space Propulsion Lecture 4 Prof. Manuel Martinez-Sanchez Page 1 of 10
During thrusting, Sr is varying according to d v≡a F 日= d 1dr 2ar1/2 (3) and so d(a2) d(89)3 2ar1/2 3 (4) For integration, we will regard r=roas a constant(small variations) d(89) 3a =--t+ constant Starting from t=0, 69=00(89)20 e obtain t69=3at(t<t,) After(t=ti), we continue to drift at a constant rate dt and, since we start from (t,)2To the 59 during the coasting phase is 3 a t2-t1(t-t1)=- t At the end of coasting(t=At-t), we have then 16.522, Space P rtinez- sanchez Lecture 4 Prof. Manuel martinez
3 Ω δ r d (δϑ) δΩ = - = (1) 2 r dt During thrusting, δ r is varying according to d ⎛ µ ⎞ F r ⎝ a= M ⎠ ⎟ (2) dt ⎜ ⎝ - 2r ⎟ ⎠ ≅ M v ≅ a µ ⎛ F ⎞ ⎜ or µ dr ≅ a µ → 1dr = 2ar 12 (3) 2r2 dt r rdt µ and so d (δΩ) d2 (δϑ) 3 2ar12 3a = = - Ω = − (4) dt dt2 2 µ r For integration, we will regard r ≅ r0 as a constant (small variations): d (δϑ) 3a = - t constant + (5) dt r0 Starting from t= 0, = 0, δϑ d (δϑ) = 0, dt we obtain d (δϑ) = - 3a t ; δϑ = - 3 a t2 (t < t1 ) (6) dt r0 2 r0 After (t=t1), we continue to drift at a constant rate d (δϑ) 3a = - t1 dt r0 and, since we start from δϑ ( ) 3 a 2 t= - t1 , 1 2 r0 the δϑ during the coasting phase is t1 ⎞ δϑ = - 3 a t1 2 - 3a t1 (t - t1 ) = - 3at1 ⎜ ⎛ t - ⎟ (7) 2 r0 r r0 0 ⎝ 2 ⎠ At the end of coasting (t= ∆t t - ) , we have then 1 16.522, Space Propulsion Lecture 4 Prof. Manuel Martinez-Sanchez Page 2 of 10
Bat 6(△t+t)=-2 (8) and after the period t of reversed thrust we return to the initial orbit with O, and with 89 as in( 8), plus a further 89(t,) This is then the total 49 3at At -t (9) Clearly, the mission(given A9 and At )can be accomplished with different choices of thrusting time ti(but notice that t, At/ 2 in any case). The required thrust/mass ratio(a)and AV=2 at depend on this choice (10 3t1(△t-t1) 4v=2 3 4『o△6 2 We can see here that low thrust ends up as a penalty on Av, so that the thrusting time should be selected as short as possible within the available on-board power. 16.522, Space P pessan Lecture 4 Prof. Manuel martinez
3 ⎞ δϑ (∆tt ) = - 3at1 ⎜ ⎛ ∆t - t1 ⎟ - (8) 1 r0 ⎝ 2 ⎠ and, after the period t1 of reversed thrust, we return to the initial orbit with 0 d (δϑ ) =, and with δϑ as in (8), plus a further δϑ (t1 ) . dt This is then the total ∆ϑ : 3at1 ⎜ ⎛ ∆t - 3 ⎞ 3 a 2 ∆ϑ = - r0 ⎝ 2 t1 ⎠ ⎟ - 2 r0 t1 ∆ϑ = - 3at1 (∆t - t1 ) (9) r0 Clearly, the mission (given ∆ϑ and ∆t ) can be accomplished with different choices of thrusting time t1 (but notice that t1 < ∆t 2 in any case). The required thrust/mass ratio (a) and V=2 ∆ t a 1depend on this choice: 1 r0∆ϑ -a = 3 t1 (∆t - t1 ) (10) 2 r0∆ϑ ∆V= 3 (∆t t1 ) (11) - We can see here that low thrust ends up as a penalty on ∆V , so that the thrusting time should be selected as short as possible within the available on-board power. 16.522, Space Propulsion Lecture 4 Prof. Manuel Martinez-Sanchez Page 3 of 10
In the limit of impulsive thrust we realize that t, cannot really be any less than the Hohmann transfer time(1/2 orbit, or t,=-). A more detailed analysis of this case confirms that, for the high thrust case, Equations(10)and(11)are indeed valid witht,=r/Q2 The power per unit mass required is: P Fcac M 2n M P 1 040C M6nt1(△t-t1) It is useful to express results in terms of the coasting time At t=At-t 3 At +t t1(△t-t1) We then have △V 4。40 3△t+t P2c△6 M3n△t2-t coasting reduces Av, but increases P/M (not much if te /At is small) Sun-Moon Effects on a Geosynchronous Orbit- The Ns Drift Geosynchronous orbits are in the Equatorial plane, which is inclined 23.44 to the plane of the Ecliptic (path of Earth around the Sun). Because of this the tidal or gravity-gradient forces exerted by the Sun on a geosynchronous spacecraft will sometimes exert a" torque"on the orbit. This happens primarily at Solstice ( January and July) when the orbit dips the most out of the Ecliptic 16.522, Space P pessan Lecture 4 Prof. Manuel martinez
In the limit of impulsive thrust, we realize that t1 cannot really be any less than the π Hohmann transfer time (1 2 orbit, or t = ). A more detailed analysis of this case 1 Ω confirms that, for the high thrust case, Equations (10) and (11) are indeed valid with t1 = π Ω . The power per unit mass required is: P 1 F c a c = = M 2η M 2η P = 1 r0∆θ c M 6η t - 1 (∆t t 1 ) It is useful to express results in terms of the coasting time, t = ∆t - 2t c 1 ∆t t - c ∆t t + c t - 1 = 2 ⇒ ∆t t 1 = 2 , 2 c t - 1 (∆t t ) = ∆t2 + t , 1 4 We then have 4 r0∆θ ∆V = 3 ∆t t + c P = 2c r0∆θ 2 - 2 M 3η ∆t t c coasting reduces ∆V , but increases P M (not much if tc ∆t is small). Sun-Moon Effects on a Geosynchronous Orbit – The N/S Drift Geosynchronous orbits are in the Equatorial plane, which is inclined 23.44D to the plane of the Ecliptic (path of Earth around the Sun). Because of this, the tidal or gravity-gradient forces exerted by the Sun on a geosynchronous spacecraft will sometimes exert a “torque” on the orbit. This happens primarily at Solstice (January and July) when the orbit dips the most out of the Ecliptic: 16.522, Space Propulsion Lecture 4 Prof. Manuel Martinez-Sanchez Page 4 of 10
SUMMER GEOGRAPHIC SIDEREA SOLSTICE 23° S/C IN GEO s TIDAL ong vernal Equinox FORCES Ecliptic, solsticial line z: North sidereal EQUIN If r is the relative position vector of the spacecraft and s is the unit vector from the Sun, the Tidal Force( Gravity gradient) per unit mass is R (1) (This the imbalance between solar attraction and centrifugal force, which cancels on the earth, r=0). Of this, the r part gives no torque about the Earth's center. To compute the effect on the orbit, we smear"the torque r x f about the geosynchronous orbit, and calculate an orbit-average torque(per unit mass) (3) We project onto a set of axis(x, y, z)as defined in the sketch( fixed axis) r=y(=raising cosy here RG is the geosynchronous orbit radius(42, 200 Km)and y=23. 44. is the angle between the equatorial plane and the ecliptic 16.522, Space Propulsion Lecture 4 Prof. Manuel martinez-Sanchez Page 5 of 10
θ r r r Z f R ψ S Fi l x y o ~ G.N. TIDAL TORQUE, SUMMER SOLSTICE VERNAL EQUINOX TIDAL FORCES ES GEOGRAPHICAL NORTH SIDEREAL NORTH EARTH xed Axes: x: A ong Verna y: In Ec z: North s SUN 23 /C IN GEO s l Equinox liptic, solsticial line idereal G G If r is the relative position vector of the spacecraft, and s is the unit vector from the Sun, the Tidal Force (Gravity gradient) per unit mass is JG ( GG G G f = µs ⎡3 r.s s − r⎤ ) (1) 3 R ⎣ Es ⎦ (This the imbalance between solar attraction and centrifugal force, which cancels on G G the Earth, r = 0 ). Of this, the r part gives no torque about the Earth’s center. To G G compute the effect on the orbit, we “smear” the torque r f × about the geosynchronous orbit, and calculate an orbit-average torque (per unit mass): JG 1 2π G G q = × 2π ∫0 r f d θ (2) JG 3µ 1 2π G G G G q = 3 s (r.s ) (r × s) dθ (3) REs 2π ∫0 We project onto a set of axis (x, y, z) as defined in the sketch (fixed axis): ⎧ ⎫ x ⎧cos θ ⎫ G ⎪⎪ ⎪ ⎪ r= ⎨ ⎬y = RG ⎨sin θ cos γ⎬ (4) ⎪ ⎪ ⎪ ⎩ ⎭ ⎩sin θ sin γ ⎪ z ⎭ where RG is the geosynchronous orbit radius (42, 200 Km) and γ = 23.44 D is the angle between the equatorial plane and the ecliptic. 16.522, Space Propulsion Lecture 4 Prof. Manuel Martinez-Sanchez Page 5 of 10
Also S=sIny 0 So that rs=Rg(cos Cose siny Cosy sine) r×s=R。 cos y siny sine sin y Cos0-CoS y Cosy sine The earth does not advance much in its orbit in one day. Also, of course, Y In performing the averaging of Equation (3), we keep y(approximately) const constant. When(6)and(7)are multiplied together we obtain sine and cose quadratic combinations, and we use 1 sine cose de=0 1 sin e de case de 2兀 and therefore the mean torque is RG siny sin v cosy CoS y sIn y COS This torque will precess the geosynchronous orbit by changing its specific angular momentum (=RGOG nG(ng=unit vector along the Geographical North direction, OG 86400ads) le are interested in the long-term(secular)effects. For times much longer than year, we can also average the torque over the angle y. Noticing that, as before (sin y COs v)=0, -RG siny COS 0 16.522, Space P pessan Lecture 4 Prof. Manuel martinez Page 6 of 10
Also ⎧cos ψ⎫ G ⎪ s= ⎨sin ψ ⎪ ⎬ (5) ⎪ ⎩0 ⎪ ⎭ So that G G r.s = RG (cos ψ cos θ + sin ψ cos sin γ θ) (6) ⎧-sin ψ sin γ sin θ ⎫ G G ⎪ ⎪ r s=R × G ⎨cos ψ sin γ sin θ ⎬ (7) ⎪ ⎩sin ψ cos θ - cos ψ cos sin θ⎪ γ ⎭ In performing the averaging of Equation (3), we keep ψ (approximately) constant. The Earth does not advance much in its orbit in one day. Also, of course, γ is constant. When (6) and (7) are multiplied together we obtainsin θ and cos θ quadratic combinations, and we use 1 2π 1 2π 2π 1 d d d 2π ∫0 sinθ cosθ θ=0 ; 2π ∫0 sin2 θ θ= 2 1 π ∫0 cos2 θ θ= (8) 2 and therefore the mean torque is JG 3 µ ⎧- cos γ sin ψ⎫ q = 3 S 2 ⎪ ⎪ RG sin γ sin ψ ⎨cos γ cos ψ ⎬ (9) ⎪ 2 RES ⎩sin γ cos ψ ⎪ ⎭ This torque will precess the geosynchronous orbit by changing its specific angular G G G momentum 2 A = RG ω n (nG =unit vector along the Geographical North G G direction, ωG = 2π rad s ). 86400 We are interested in the long-term (secular) effects. For times much longer than 1 2year, we can also average the torque over the angle ψ . Noticing that, as before, sin ψ cos ψ = 0 , sin2 ψ = 1, 2 ⎧ ⎫ 1 JG 3 µs 2 ⎪ ⎪ q = − R sin γ cos γ ⎨ ⎬ 0 (10) sec ular 4 R3 G ES ⎪ ⎪ ⎩ ⎭ 0 16.522, Space Propulsion Lecture 4 Prof. Manuel Martinez-Sanchez Page 6 of 10
This shows a non-cancelling effect consisting on a mean torque which acts along the (negative) vernal equinox direction(as shown in the sketch) PRECESSION VE The effect of this torque along(-x)is to add a small to (, along(x), and hence dt to rotate(precess)the orbit along the axis perpendicular to the vernal direction, as In litude d i dt dt and equating to a)from(10), 3μssi dt 4R the angular frequency of the Earth's heliocentric motion is -SIn y COSY (13 4 16.522, Space P pessan Lecture 4 Prof. Manuel martinez
This shows a non-cancelling effect consisting on a mean torque which acts along the (negative) vernal equinox direction (as shown in the sketch): JJG The effect of this torque along (-x) is to add a small d dt A to G A , along (-x), and hence to rotate (precess) the orbit along the axis perpendicular to the vernal direction, as shown. In magnitude JJG d G A di 2 = A = RGω di (11) dt dt G dt JG and equating to q from (10), ⎛ di ⎞ 3 µs sin γ cos γ ⎜ 3 (12) ⎝ dt ⎟ ⎠sun = − 4 RES ωG Or, since the angular frequency of the Earth’s heliocentric motion is ω = , ΕS R µ 3 s ES 2 ⎛ di ⎞ 3 ωΕS ⎜ sin γ cos γ ⎝ dt ⎟ ⎠sun = − 4 ωG (13) 16.522, Space Propulsion Lecture 4 Prof. Manuel Martinez-Sanchez Page 7 of 10
Numerically, expressing Ogs and @G in( deg/yr ) 3(360 dt)an4(360×365sin2344c052344°=-0.27/yr The Moon effect is similar, except that the moon's orbital axis which is inclined at 5.15 with respect to sidereal North precesses slowly about that direction(once every 18.6 yr ) This means the angle between the Equator and the Moons orbit varies between y=2344°+5.15°=28.59°andw=2344-5.15°=18.29° The expression for/ di is similar to(12) 3 HM- sin lm COS (14) an HM M where M 1 is the Moon/Earth mass ratio, and Ome is the angular velocity of M81.3 the Moon about Earth(about 2/28 days) We then have 365 d i 31 481.3360×365 sin2344cos2344°=-0.56°/y (16) dt This is for the average ym in the 18.6 yr. lunar precession cycle At the peak of YN -0.65 and at the minimum d i Adding these values to the solar rotation(both act in the same direction) yields 16.522, Space P pessan Lecture 4 Prof. Manuel martinez
Numerically, expressing ωΕS and ωG in ( deg yr ), 2 ⎡ ⎤ ⎛ di ⎞ 3 (360 ) ⎢⎜ D ⎢⎝ dt ⎟ = − 4 360 × 365) sin23.44 cos 23.44 D = −0.27 D yr⎥ ⎣ ⎠sun ( ⎥ ⎦ The Moon effect is similar, except that the Moon’s orbital axis which is inclined at 5.15D with respect to sidereal North, precesses slowly about that direction (once every 18.6 yr.). This means the angle between the Equator and the Moon’s orbit D D D varies between γM = 23 .44 + 5 .15 D = 28 .59 D and γM = 23 .44 - 5 .15 D = 18 .29 . ⎛ di ⎞ The expression for ⎝ dt ⎟ ⎠moon is similar to (12); ⎜ ⎛ di ⎞ = − 3 µM sin γ cos γ (14) ⎜ ⎝ dt ⎠ ⎟ moon 4 R3 m m EM and ⎛M R µ µM µE = ⎜ m ⎞ 2 (15) 3 M = µ R3 ⎟ωΜΕ EM E EM ⎝ ME ⎠ Mm 1 where = is the Moon/Earth mass ratio, and ωΜΕ is the angular velocity of M 81.3 E the Moon about Earth (about 2 28 π days). We then have ⎡ ⎛ 365 ⎞ 2 ⎤ ⎢ ⎜360 28 ⎠ ⎟ ⎥ ⎢⎛ di ⎞ 3 1 ⎝ sin23.44 cos 23.44 D = −0.56 D yr ⎥ (16) ⎥ ⎢ ⎥ ⎢ ⎥⎦ ⎢⎝ ⎜ dt ⎟ ⎠moon = − 4 81.3 360 × 365 D ⎣ This is for the average γ in the 18.6 yr. lunar precession cycle. M At the peak of γM , ⎛ di ⎞ ⎝ dt ⎟ ⎠moon = −0.65 D yr , ⎜ and at the minimum, ⎛ di ⎞ ⎝ dt ⎟ ⎠moon = −0.46 D yr . ⎜ Adding these values to the solar rotation (both act in the same direction) yields 16.522, Space Propulsion Lecture 4 Prof. Manuel Martinez-Sanchez Page 8 of 10
d =-0.83/T (Maximum=-092, Minimum=-073) (17) This can be easily translated into an equivalent Av =3070m/s △V=44m/s/y (Maximum=49, Minimum=39) What is the effect of i if left uncompensated? Using some spherical trigonometry sini sine ④- sini cos2e (19) cosi) sine cos e sin s √1- sini cos0 (20) and for small rotations λ= i sine SOLSTICIAL e=-sine cos e So, during one day(0<< 360), the orbit describes as seen from the ground a figure 8 in the sky about its nominal position: 16.522, Space P pessan Lecture 4 Prof. Manuel martinez
⎛ di ⎞ ⎝ dt ⎟ ⎠Total = −0.83 D yr (Maximum=-0.92, Minimum=-0.73) (17) ⎜ This can be easily translated into an equivalent ∆V: ∆V=vG∆i; vG = µE = 3070 m/s RG ∆V=44 m/s yr (Maximum=49, Minimum=39) (18) What is the effect of i if left uncompensated? i � � ROTATED ORBIT � EQUATORIAL ORBIT Using some spherical trigonometry, sini sin θ sin λ = (19) 2 1 - sin i cos2 θ (1- cos i sinθ cos θ (20) ) sin ε = 2 1 - sin i cos2 θ and for small rotations i, λ i sinθ (21) SOLSTICIAL AXIS ε i 2 sin θ cos θ (22) 2 So, during one day (0< <360 D θ ), the orbit describes, as seen from the ground, a figure 8 in the sky, about its nominal position: 16.522, Space Propulsion Lecture 4 Prof. Manuel Martinez-Sanchez Page 9 of 10
6=0° The main deviation is (N/s direction), and this is called a N/s drift". Typically, communications geosynchronous satellites can only tolerate 0.05-0.1 of such drift before correction is needed 16.522, Space P pessan Lecture 4 Prof. Manuel martinez Page 10 of 10
The main deviation is λ (N/S direction), and this is called a “N/S drift”. Typically, D communications geosynchronous satellites can only tolerate 0.05 D − 0.1 of such drift before correction is needed. 16.522, Space Propulsion Lecture 4 Prof. Manuel Martinez-Sanchez Page 10 of 10
