Elements of the small circular orbit = 6.828. m, ![v[c1] := sqrt(mu/r1) = sqrt(3.986005088*10^14/(6.828*10^6))](http://www.maplesoft.com/view.aspx?SI=151351/0df1985d1f40860b93ea7f100323c342.gif)  Elements of the large circular orbit = . m, Elements of the elliptic orbit The semi-major axis: =  m. Eccentricity: ![e:=((r[max]-r[mini]))/((r[max]+r[mini]))=((6.828*10^(6)- 7.578*10^(6)))/((6.828*10^(6)+7.578*10^(6)))=0.05206164098. V[perigee]:=sqrt((2 mu)/()*(1/(r[mini])-1/(2 a)))=sqrt((2 mu)/()*(1/(r[mini])-1/(r[max]+r[mini])))=sqrt((2 mu)/()*((r[max])/(r[mini] (r[max]+r[mini]))))=sqrt((2 mu)/()*((r[max])/(r[mini] 2 a))).](http://www.maplesoft.com/view.aspx?SI=151351/aa6d1d4b856a3ef337da20ebe2261283.gif) m/s. m/s. = . seconds. = kg. . ![L := h^2/(G*M[Earth])](http://www.maplesoft.com/view.aspx?SI=151351/e7c47e237aebad9a110b2de2f9417d93.gif) =  Problem #1 in details Sally Maneuver To inject the satellite into elliptic orbit from point B (Figure-1 (a)) on the large circular orbit r2 = ( apogee of the elliptic orbit) Sally needs to decrease her velocity from to : so that the decrement of the velocity is: m/s. To inject the satellite into small circular orbit at point A she has to reduce her velocity by: m/s. When Sally reaches point A, Igor would be at point (Figure-1 (a)). The difference in degrees between Igor at and Sally at A is angle : , + 4.5 = degrees. While on the smaller circular orbit Sally has to make a few revolutions waiting for Igor to get into the right relative position symmetrical with position ( degrees) before firing again her spaceship to inject it into the transfer ellipse. The gap between Sally at A after a few revolutions on the smaller circle and Igor at is the small angle degrees. The gap between them is in fact the large angle = 360 -  degrees. A 360 degrees revolution on the small circular orbit corresponds to a fraction of a revolution on the large circular orbit for Igor. The difference in degrees is : degrees. The number of revolutions that Sally has to go through till Igor gets to point is: revolutions. When the time of these revolutions has elapsed: = hours, Sally should start firing at this moment in order to go on elliptic orbit again at A. For this to happen she needs to increase her velocity from to : hence the increment of the velocity is: m/s. She will coast along this elliptic orbit while Igor is moving along the large circular orbit. When their paths are parallel ( hence having the same potential energy) which occurs at B, she has to fire her rocket to increase her forward speed by: m/s, and she would sail alongside Igor with no lagging between them.
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sp; What was wrong with problem #1? The problem that we just solved doesn't give the correct answer for Sally to be at point A when Igor is at point on his orbit. If the number of revolutions were an integral number then Sally would've been back into A when Igor got at point on his orbit. However the number of revolutions was not an integral number it was found to be = 6 revolutions + of a revolution = 6 revolutions + 114.1259144 degrees. Hence when Igor was at the right position at , Sally was not at A . She has already passed this point by 114 degrees. Firing her forward rocket won't get her at point B: her new point of contact with the large circular orbit will be ahead of Igor by 114 degrees! This error is the result of the constraint that comes from setting the radius of the smaller circular orbit to a fixed value ( 450 km) in advance. We should leave this radius as unknown, say (x), then solve our problem till we get to the expression that gives the number of revolutions as function of (x). We then set this expression as equal to an integral number, say 6, then with the help of Maple to 'fsolve' it for (x). Doing so we get meters. Now if restating the problem by specifying that meters, then the number of revolutions that Sally has to go through on the smaller circular orbit till Igor gets to point on his orbit is: 5.9999999 ≈ 6 revolutions. Now Sally can be sure that rendezvous with Igor will occur at point B. Finding the time Δt during which the engine should be fired Knowing 1- the velocity change each time Sally has to pass form one orbit to another in m/s, 2- the thrust that her rocket (or retrorocket) can deliver in Newtons , 3- the mass of the spaceship in kg, Sally will use the impulse - linear momentum change to find the necessary time during which she has to fire her rocket. From this relation : , we get : . Problem #2 Here we use two-stage rocket to raise the payload to the apogee (point B in Figure(1)(a)) of an initial elliptic orbit which is supposed to be tangent at B to the circular orbit of the station. Then at this point the second stage is ignited to increase its velocity to that of the large circular orbit to put the payload on it. The timing for both stages is very important as we found above in order that the rendezvous with the space station occurs.
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![with(plottools); with(plots); h1 := 4.12463*10^5; r[mini] := h1+R[Earth]; r[2] := r[maxi]; r2 := r[maxi]; r[c2] := r[maxi]; r1 := r[mini]; r[1] := r[mini]; r[c1] := r[mini]](http://www.maplesoft.com/view.aspx?SI=151351/fa43aa8fb0eec2ce926ed07611b8b358.gif)
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![a[E] := (r[maxi]+r[mini])*(1/2); e := (r[maxi]-r[mini])/(r[maxi]+r[mini]); V[perigee] := sqrt(mu*(1+e)/(a[E]*(1-e))); V[apogee] := sqrt(mu*(1-e)/(a[E]*(1+e))); h := r[mini]*V[perigee]; L := h^2/(G*M[Earth]); v[c2] := sqrt(mu/r[maxi]); TC2 := 2*pi*r[maxi]/v[c2]; v[c1] := sqrt(mu/r[mini]); TC1 := 2*pi*r[mini]/v[c1]; tau := 2*pi/sqrt(mu/a[E]^3)](http://www.maplesoft.com/view.aspx?SI=151351/f63fd935648cacbcff09eddd1e1b6907.gif)
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Kepler equation to get: E = eccentric anomaly, then we get θ from the relation : where θ = true anomaly.
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![Kepler := E-e*sin(E)-t*sqrt(G*M[Earth]/a[E]^3) = 0](http://www.maplesoft.com/view.aspx?SI=151351/4ec9d33ac1fb45296c55f29a91c7490c.gif)
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If in Kepler's equation we replace ( t ) by a small interval Δt = period/100: we get E then angle θ and finally r (the position of the satellite) after this interval of time. We then get these data for , , ,..., ,...etc till one complete period where n = 100. Plotting these positions gives the elliptic orbit. With this substitution the equation becomes: . This is a transcendental equation that is solved using Newton's iterative method.
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Plot of the three orbits: the elliptic, the small circular orbit & the large circular orbit
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![plot([(1/1000)*r[mini]*cos(`ϕ`), (1/1000)*r[mini]*sin(`ϕ`), `ϕ` = 0 .. 2*pi], color = red); C1 := %; plot([(1/1000)*r[maxi]*cos(`ϕ`), (1/1000)*r[maxi]*sin(`ϕ`), `ϕ` = 0 .. 2*pi], color = gold); C2 := %; plot([0.1e-2*L*cos(`ϕ`)/(1+e*cos(`ϕ`+(1/2)*pi)), 0.1e-2*L*sin(`ϕ`)/(1+e*cos(`ϕ`+(1/2)*pi)), `ϕ` = 0 .. 2*pi], color = blue); EL := %; display(C1, C2, EL)](http://www.maplesoft.com/view.aspx?SI=151351/ce4e9400c1b6661484a4883f794ca001.gif)
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This is the procedure for the animation. This procedure has so many points to plot that the total size of the file went from 300 kilobytes to 20 megabytes! For this reason I left this procedure non executed in order to reduce the size of the file to a mere 300kB for easy uploading. To see the animation just remove (#) in the last two commands then execute them. Be patient since the loading of the data for plotting takes at least 20 to 30 seconds on my computer. The animation starts when Sally, being at B and Igor lagging behind her by 4.5 degrees, fires her spaceship retrorocket to move into elliptic orbit apogee. At A she moves into smaller orbit by reducing her speed ( using retrorocket again) to that of the smaller circular orbit (red). She makes exactly 6 revolutions (count them) on the small circular orbit to be at A when Igor is at the right point on his larger circular orbit (gold). At A she uses her forward rocket to put her spaceship into elliptic orbit (blue) again going to point B which she reaches at the same time as Igor. At B she uses again her forward rocket to get on the larger circular orbit alongside Igor. I use different colors for the spaceships ( blue for Sally & red for Igor) as well as for the 3 different orbits (blue for Sally on elliptic then red on the smaller circuar orbit & gold for Igor orbit). Enjoy it.
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![sat := proc (n) local i, E1, E2, r, j, k, theta, GEL, GC1, GC2; global pt1, D2, D1, D3; j := n; k := n-650; E1 := 0.6283185308e-1*j; if 50 <= n and n <= 650 then j := 50 end if; if 650 < n then j := 50+k end if; for i to 50 do E2 := E1+(-1)*0.5481010740e-1*sin(E1)+(-1)*0.6283185308e-1*j; if abs(E2) < 1/1000000 then break end if; E1 := E1-E2/(1-e*cos(E1)) end do; theta := arccos((e-cos(E1))/(e*cos(E1)-1)); if pi < E1 and E1 < 2*pi then theta := 2*pi-theta end if; r := L/(1+e*cos(theta+pi)); pt1 := [r*cos(theta), r*sin(theta)]; if 50 <= n then GEL := disk([-(1/1000)*pt1[2], (1/1000)*pt1[1]], 1, color = blue) else GEL := disk([-(1/1000)*pt1[2], (1/1000)*pt1[1]], 150, color = blue) end if; if 650 < n then GEL := disk([-(1/1000)*pt1[2], (1/1000)*pt1[1]], 150, color = blue) end if; if 50 <= n then GC1 := disk([(1/1000)*r1*cos((1/50)*Pi*n+(1/2)*pi), (1/1000)*r1*sin((1/50)*Pi*n+(1/2)*pi)], 150, color = blue) else GC1 := disk([0, -(1/1000)*r1], 1, color = red) end if; if 650 < n then GC1 := disk([0, -(1/1000)*r1], 1, color = red) end if; if n <= 50 then GC2 := disk([(1/1000)*r2*cos((1/50)*Pi*n*tau/TC2+(1/2)*pi-0.7853981635e-1), (1/1000)*r2*sin((1/50)*Pi*n*tau/TC2+(1/2)*pi-0.7853981635e-1)], 150, color = red) else GC2 := disk([(1/1000)*r2*cos((1/50)*Pi*(n-50)*TC1/TC2+(3/2)*pi+(-1)*18.34589276*pi/180), (1/1000)*r2*sin((1/50)*Pi*(n-50)*TC1/TC2+(3/2)*pi+(-1)*18.34589276*pi/180)], 150, color = red) end if; if 650 < n then GC2 := disk([(1/1000)*r2*cos((1/50)*Pi*(n-650)*tau/TC2+(3/2)*pi+13.84589277*pi/180), (1/1000)*r2*sin((1/50)*Pi*n*tau/TC2+(3/2)*pi+13.84589277*pi/180)], 150, color = red) end if; display(GEL, GC1, GC2, C1, C2, EL) end proc](http://www.maplesoft.com/view.aspx?SI=151351/e1c02fbdf4ba88e56854f0eaab0475f6.gif)
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