Hohmann and beyond. by Luigi Amedeo Bianchi. Abstract

Transcription

1 Hohmann and beyond by Luigi Amedeo Bianchi Abstact This ae just some peliminay notes fo an intoduction to Astodynamics. They efe often to Kebal Space Pogam. Theoem 1. (Hohmann) Given two cicula obits with the same cente and lying in the same plane (concentic and complana) of adii 1 and f > 1 espectively, the two-impulse tansfe manoeuve fom the st to the second obit that minimises the total velocity incement v TOT is the tansfe along the elliptical obit of peiapsis 1 and apoapsis f, the same obital plane as the two given obits and the same diection as the stating obit. Figue 1. Hohmann tansfe Remak. Possible situations launch fom the suface of a planet, tansfe between two cicula obits aound Kebin. q Poof. Set = M G. We have that the velocity of the st cicula obit is, with =, q 1 while the nal velocity is the one of the cicula obit of adius f, i.e. v cf, with v cf =. Othe vaiables of inteest v~ 1, velocity of the st impulse, v~ f, one of the nal impulse, ' is the initial elevation, i.e. the angle between v~ 1 and, while is the nal one, the angle between v~ f and v cf. f 1

2 We ae aiming to minimise v TOT = jv~ 1 j + jv cf v~ f j = v 1 + v f We can simplify the poblem at hand by using the peseved quantities E = v 1 = v f 1 =) v f = v 1 ( f 1 ) f 1 f (enegy, fom vis-viva equation) and (impulse / angula momentum) c = 1 v 1 cos ' = f v f cos =) cos = 1 f v 1 v cos ' Now we can take advantage of those to expess both v 1 and v f in tems of v 1 q v 1 = jv~ 1 j = v 1 + v c1 v 1 cos ' q v f = jv~ f v cf j = v f + v cf v f v cf cos = v cf + v 1 ( f 1 ) v cf v 1 1 cos ' f 1 f p We can escale eveything in tems of the speed in the intenal obit, = / 1 by setting x = v 1/ vc1. We get then s p = x x cos ' x v TOT 1 f 3 / cos ' 1 ( f 1 ) f = D(x ') and we want to minimise D(x '), with the additional constaint of the obit having apoapsis geate o equal than f (because we want to each that obit). If we conside the patial deivative with espect to ' = v 1 sin ' + v 1 v 1 cf / f sin ' v 1 v f we have that the minimum is attained in ' = 0, independently of the speed, hence of x. We can then conside only the case D(x) = D(x 0) and dieentiate in x dd dx = v x 1 x ( 1 / f ) 3/ c 1 + v 1 v f We can obseve that x>1, othewise we'd have apoapsis at 1, and then it follows that the deivative is always positive and the solution to ou poblem is the minimum value of x (i.e. the minimum value of v 1 ) such that we can actually each the nal obit. Fo obvious geometical easons it coesponds to the value v 1 such that the ellipse is tangent in the apoapsis too, i.e. v f v 1 = 1 + f / 1

3 This coves the case whee the tansfe obit lies on the same plane of the initial and nal obit. Leaving the plane is in this case not such a geat idea, as we'd just add moe tems to v TOT. Remak 3. The time needed fo the Hohmann tansfe is the time equied fo tavelling half the ellipse, that is s 1 + f 3 T H = We can also expess vtot in tems of f, which will tun out to be useful to compae Hohmann 1 tansfe to othe manoeuves. We have v TOT = v 1 = s + v f v cf f / f / 1 = v 1 + v c f v f f / 1 f / 1 (1 + / f 1 ) An easy computation povides us with the inteesting insight that the maximum of this atio is attained fo f / 1 = 15.58, in which case v TOT = Remak 4. If the obits ae elliptical due to the Obeth eect it is usually optimal to leave the intenal obit at the peiapsis. We still need to check both cases, as the conguation of the two obits might cause it to be bette to leave the intenal obit at the apoapsis. Rocket equation and specic impulse Why do we want to educe v anyway? Let's conside a ocket ying with velocity v~ and subject to some extenal foces of sum F ~ (e.g. gavitational, atmospheic fiction...). Let v~ 1 e be the ejection velocity of the popellant (with espect to the ocket) in a (innitesimal) time dt. We want to get the module v of the coesponding velocity vaiation of the ocket. We set the mass of the ocket (including engine mass and fuel mass and payload mass) to m. In a time dt the ocket is subject to the following vaiation of momentum [(m dm) (v~ + dv~) + dm (v~ v~ e )] m v~ = F ~ dt whee in the st tem we have the dieence between the momentum at time t 0 + dt and the momentum at time t 0. Since extenal foces ae not impulsive, we have that F ~ dt!0. If we conside a st ode appoximation and foget about the second ode tem dm dv~, we get which pojected along v~ gives m dv~ = v~ e dm hence dv v e dm m v = v e log m0 m f 1. Maybe a dieent notation is bette, as fo masses we have m_e denoting the mass of the engine... So maybe v_f (m_p is the mass of the payload) 3

4 with m 0 the mass of the ocket at time 0, and m f the mass at the end of the manoeuve. The fuel used fo the v vaiation is then m 0 m f. It makes sense to intoduce the following quantity, called specic impulse I sp = v e g with g being the gavitational acceleation at sea level (9.81 on Kebin o Eath). The specic impulse (which has seconds as its unit, at least in the most common fomulation, given above), is a measue of the eciency of the engine used (namely it depends on the engine and the fuel used). As it is clea fom the pevious equation fo the v, the highe the specic impulse is, the bette. Engine I sp (vac) mass thust TWR Rocketdyne F1 63 s 8400 kg 6.77 MN 8.7 R s 356 kg.8 MN Melin 1D 340 s 490 kg 0.8 MN 150 RD 171 M 337 s 9300 kg 7.9 MN 86.9 RD 107 A 310 s 1190 kg 1 MN 89.9 PPS kw 1650s 5.3 kg N Table 1. Eath engines Engine I sp (vac) mass thust TWR Rockomax Mainsail 310 s 6000 kg 1.5 MN 5.48 LV-T45 30 s 1500 kg 0. MN LV-N* 800 s 3000 kg 0.06 MN.04 PB-ION 400 s 50 kg kn Table. Kebal engines Might be woth noting that usually (ocket) engines have dieent values fo the specic impulse depending on being o not in the atmosphee (compaed to be in the vacuum). Also woth stessing is that this ocket equation (o Tsiolkovskij equation) holds fo engines that (appoximately) e in a single impulse, and does not hold fo engines such as electical ones (vey low thust, even if they have a huge Isp). Moeove it is not necessaily a good measue of compaison fo engines, not just fo poblems as the one above (thust vs isp, instant vs long time), but also because an engine with a vey high Isp might have a huge mass, thus using a lot of its powe to move itself, so it is bette to compae the wok povided by each engine. Reminde thust is given by T ~ = v~ dm dt and its unit is Newtons, while thust to weight atio (TWR) by and is a pue numbe. TWR = T m g Bi-elliptic tansfe and compaison with Hohmann. The Hohmann theoem above tells us that the Hohmann tansfe is optimal among those with impulses, but what if we elax this hypothesis and conside the case with 3 impulses? We get the bi-elliptic tansfe. 4

5 Figue. Bielliptic tansfe So we stat once again in I, on the cicula obit of adius 1, we have a st bun v 1 along the same diection as (pogade) and we stat the st elliptical tansfe obit, with semi-majo axis 1 + a 1 =, whee we stay until we each the apoapsis P (lying at a distance > f fom ou gavity cente). In this point we bun pogade again (v ), aising the peiapsis to height f and stating + f a second elliptical tansfe obit of semi-majo axis a =. When we each the peiapsis of this obit (at F), it is time fo the thid and last bun v f, which is etogade and ciculaises the obit. Using the same ideas as fo the Hohmann Theoem, we can compute v 1 (I) v 1 (P ) 1 + v (P ) + f v (F ) f + f And now, since v 1 = v 1 (I) v = v (P ) v 1 (P ) v f = v (F ) v cf we can compute the thee v (o bette, thei atio with ) and v 1 = / 1 + / v f = v = f / 1 ( / s / 1 f 1 + / f / 1 / 1 ( f / 1 + / 1 ) 1 ) 1 / f 1 / 1 (1 + / Now we can set x = f / 1 and y = / v 1 and we can wite B / vc1 in tem of these as v B = y 1 + y 1 + x y (x + y) y (1 + y) + y x (x + y) 1 x 1 ) 5

6 We ae inteested in when this manoeuve is bette than Hohmann, i.e. when its fuel consumption is less, i.e. when the v equied is less, so we study whee v B v H < 0, o equivalently when v B / v vc1 H / vc1 < 0. The answe is povided by this inequality (holding fo y >x, i.e. when > f ) y 1 + y 1 + x y (x + y) y (1 + y) + and is epesented in the following pictue y x (x + y) 1 x x x x + x (1 + x) < 0 Figue 3. Compaison between Hohmann and bielliptic So, when f / 1 > any choice of > f povides some gain, while fo f / 1 < Hohmann tansfe is always bette. In between if we take big enough we can gain with a bielliptic tansfe. The best case is fo = 1, fo which we'd have v B p q q 1. This case shows + 1 f us the poblem that we might have with a bielliptic tansfe we have to tavel two half ellipses, so the total time is s 1 + s 3 + f 3 T B = + Anyway, if we ae not in a huy, fo which kind of missions can we use such manoeuve to save fuel, i.e. when does it happen that the atio of the adii is such that it allows a bielliptic tansfe? In KSP we can use it to tavel fom LKO (o MKO) to an highe obit o to the Mun (o Minmus, but we have a poblem...). On the othe hand the bielliptic manoeuve is not suitable fo tansfes between planets, as the Kebol System is quite small even if we wee to send a pob fom Kebin to Eeloo when it is at his futhest it would be a atio of 8.69 between the adii, so that would not be fuel spaing. On the othe hand in ou good ol' Sola System we have not only the tansfes between obits aound the Eath, but also fom LEO (o MEO) to the Moon, and fom Eath to Uanus ( f / 1 19), to Neptune ( f / 1 30) o to Pluto ( f / 1 39) and beyond... Remak 5. It might also be woth noting that the highest possible saving of v with a bielliptic tansfe compaed to a Hohmann tansfe is 8%, so nothing spectacula, but still not quickly discaded, consideing that a big pat of the mass of a spacecaft consists of fuel. 6

7 Change of plane. Befoe, when I mentioned Minmus, I stated that we have a poblem with that, and the poblem is that Minmus' obit lies on a dieent plane than an equatoial obit aound Kebin. So moving to Minmus equies a change of plane. Let's see how we do it! We stat (and nish) with a cicula obit of adius, so thei speed is q = v c = v c = and if we ae changing the plane by an angle # we have v = v c sin # / v = sin v # / c This means that to change the plane of an angle # = 60, we need v = v c, while to change it of an angle of 180 (i.e. to evese obit) we need v = v c (vey easonable). Note that we e in a diection which is othogonal to both the tangent to the obit (we want the obit to stay the same, adius-wise) and the vecto to the cente of mass. This diection is called the nomal diection. Also note that the point whee we execute the bun is a point that lies on both the oiginal obit and the new one, i.e. is in the intesection of the two planes. One must not think of the point whee we bun as ising in space. Remak 6. The futhe away we ae fom the gavity cente, the cheape the manoeuve is. This emak tells us something maybe we can gain some fuel if we send the pobe fa enough so that the v equied fo the change of obit, change plane and come back is less than the fuel necessay to change plane staying in the same obit. So we aim fo a 3-buns manoeuve. The st pogade bun of v 1 takes us on an obit, on the same plane, of apoapsis f > 1, then we have a manoeuve that just changes the plane (while we ae at the apoapsis), by a bun of v = v sin # /, and nally we have a etogade bun at the peiapsis to lowe the apoapsis back to 1 and ciculaise the obit once again. The v fo this last bun is exactly the same needed fo v 1, so the total is v TOT = v 1 + v Now we use the same change of vaiables as befoe, i.e. x = f / 1 and we conside whence v =) v = v c f 1 + f x (1 + x) At the same time we have v = sin # v c x (1 + x) v 1 v c =! x x v and we can poceed to minimise TOT / vc with espect to x, fo xed #. We get (computations!!! Inset in futue, maybe) the following answe x ott = sin #/ 1 sin # / 7

8 which holds fo x ott > 1. We can check how, fo xed x, the function that we ae minimising is is behaving wt #. We go and see when it happens that x ott = 1, which is fo sin # / = 1 / 3, i.e. fo # So fo 0 < # < we have that a single bun manoeuve is moe ecient, fo # 6 60 we can have a moe ecient 3-buns manoeuve by choosing x = x ott. Actually, any 1 < x 6 x ott povides some impovement wt single bun. When we each 60 we hit x ott = 1, so any choice of x > 1 gives some benet, and the bigge the x the bette. (The eason is that the deivative is always negative, but inceasing to 0) Again fo pactical easons we will choose some x that gives us enough benets while it doesn't take too long (the time taken is the time fo an elliptic obit of semi-majo axis ). 1 + f Remak 7. Thee is an obvious genealisation to this manoeuve, that is to split the angle in 3 contibutes # 1 + # + # 3 = # and include each of them in one of the thee buns. This way we can pefom even bette. Hohmann tansfe with change of plane. If we need to change the plane we can split the contibutions in two, one of each bun. This is exactly the idea of the pevious emak. To choose the optimal to be taken cae of in the st bun, we need to solve numeically a minimisation poblem. Flybys. We can use gavity to save fuel. We get the so-called slingshot o gavity assist. We can use that to acceleate o deceleate, o just to modify ou tajectoy. Fo example the Cassini spacecaft was able to each Satun with km/s v instead of the 15.7 km/s equied by the Hohmann tansfe thanks to the ybys with Venus, one with the Eath and one with Jupite. Fom a time pespective it needed 6.7 yeas instead of the 6 equied by Hohmann. Also Ulysses (change of plane, out of the ecliptic). Poblem you can't foce the planets to be whee you want them! Thee ae situations whee this is not a poblem fo example in the Mun landing tutoial you can use a yby with the Mun (clockwise) to slow you caft down, saving fuel fo a soft landing (in contaposition with had landing, jagon fo cashing). We aleady mentioned in passing the Obeth eect it states that a pogade manoeuve is the most ecient if pefomed at the peiapsis. The Obeth eect is easy to pove fomally, but is tickie to undestand fom an intuitive point of view. We have that when we pefom the bun the whole caft + fuel system peseves the kinetic enegy (because the potential enegy is the same in the point), so we have 1 (M + m) v = 1 M (v + v) + 1 m (v v e) On the othe hand, if we conside only the spacecaft we have E = 1 M (v + v) 1 M v = 1 M v + M v v and so we see that the highe v is, the moe ecient the bun, as we gain moe kinetic enegy than just 1 / M v. We could say that we ae stealing it fom the fuel that we ae leaving behind. Of couse one can combine the two techniques pesented in a poweed yby, the only poblem being, in eal life, the pecision of such manoeuve and the timing issues due to communications lags (ty playing with RemoteTech...). What else? Something we can't use in KSP, due to the SOI and bodies appoximation (hence no Lagangian points and so on... Low enegy tansfes, ballistic captue and Inteplanetay tanspot netwok. Just a passing emak on an Eath-Moon LET the fuel savings can be up to 5%. Fee etun tajectoy. O how to bing touists aound the Mun, get some science and land back on Kebin with a single bun fom Kebin obit! Philipp would complain about my dawing, hee, so I will just mention it. 8

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