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LEO to GEO orbit design project report

Nasser M. Abbasi

May 23, 2003 page compiled on July 28, 2015 at 9:10pm

Contents

1 Problem
2 Assumptions
3 Method
4 Analysis
 4.1 Common calculations
 4.2 Decide how to correct the LEO plane inclination
  4.2.1 First scenario. All plane correction at perigee.
  4.2.2 Second scenario. All plane correction at apogee.
  4.2.3 Third Scenario. Partial plane correction at perigee. Rest at apogee.
  4.2.4 Summary of scenarios to correct plane inclination
 4.3 Calculate time to move from LEO to GEO
 4.4 Decide when to inject to GEO. Calculate lead angle β  and rendezvous with first satellite
  4.4.1 First option. Inject to GEO at t = 0
  4.4.2 second option: phase-wait in the LEO orbit until the correct lead angle β  reached at one end of lines of nodes.
  4.4.3 Third option: phase-wait in the LEO orbit until the correct lead angle β  reached at the other end of lines of nodes.
  4.4.4 Fourth option. phase-wait in the LEO orbit until smallest difference to β  reached first time.
  4.4.5 Summary of lead angle β  scenarios
 4.5 Rendezvous with the second satellite
 4.6 Stay locked in with the second satellite.
 4.7 reposition to final destination.
5 Summary and Results
6 Appendix

1 Problem

Inject spacecraft directly into circular parking orbit of altitude 100km. At burnout flight path angle is zero and velocity is Vbo  .

Even though launch at equator (line of nodes), Orbit has an undesired 150   inclination in the geocentric frame and longitude of ascending node Ω = 200   . Need to correct this inclination so that you may rendezvous with a satellite in the GEO (35,860 km) on the equatorial plane with zero inclination.

This target satellite was 400   behind you at the time you entered your parking orbit of 100 km.

After you rendezvous with the first satellite, you need to transfer to a second satellite in GEO which was   0
10   ahead of you at time you entered your parking orbit.

Finally after making one complete orbit with this second satellite you need to take your final position in GEO which is another 50   ahead of the second satellite.

Design the above sequence. You may opt to minimize the time to complete or the fuel needed (ΔV  )  . Specify your design criteria.

2 Assumptions

No drag while in LEO orbit. This allows the spacecraft to orbit as many times as needed to improve rendezvous conditions with the first target in GEO.

All impulses applied are assumed to have infinitely small time durations.

In addition, all assumptions used to derived Kepler equation apply as well.

Launch site latitude effect on ΔV  are ignored. In practice, ΔV  requirement need to be modified by small magnitude depending on the launch site location on the surface of the earth.

3 Method

The geometry of the problem is illustrated in the figure below, which depicts the state of the system at  t0   . The time the space vehicle enters its parking orbit at burn out Vbo  .

PIC

The analysis part will show an outline of the maneuvers to achieve the goal of the project.

The criteria for selecting the maneuvering sequence is:

The minimization of fuel

Which directly relates to the minimization of the ΔV.

4 Analysis

Now, we will show each phase of the maneuver, with different scenarios to achieve each phase. Before starting, there are common calculations that will be done now that will be shared by many scenarios, so that we do not have to re-calculate these each time.

4.1 Common calculations

For the Hohmann transfer from LEO to GEO:

Let Vp  be the speed on the Hohmann orbit at the perigee point.

Let Va  be the speed on the Hohmann orbit at the apogee point.

Let 2a  be the semi major axis for the Hohmann orbit.

Let Th  be the period of the Hohmann orbit.

For the LEO and GEO orbits:

Let Vleo  be the speed on the LEO orbit.

Let Vgeo  be the speed on the GEO orbit.

Let rleo  be the radius of the LEO orbit.

Let rgeo  be the radius of the GEO orbit.

Let Tleo  be the period of the LEO orbit

let Tgeo  be the period of the GEO orbit

Then

rleo  = 6378.145 + 100 =  6478.145  km

rgeo  = 6378.145 +  35,860 = 42,238.145  km

    r  +r
a = -leo2-geo = 6478.145+422238.145 = 24358.145  km

Vp =  ∘  ----------
     -2   1
   μ(rp − a)  but rp = rleo

so, Vp =  ∘ ----------------------------------
                5(---2---  ----1-- )
  3.986012 × 10   6478.145 − 24358.145 = 10.3294  km/sec = 37, 185.84  km/hr

V  =
 a  ∘  ----------
   μ(-2 − 1)
     ra   a  but r  = r
 a    geo

so, V  =
 a  ∘ -----------------------------------
  3.986012 × 105 (---2--- − ---1--- ) =
                  42238.145   24358.145  1.584  km/sec = 5,702.4  km/hr

      ∘ ----  ∘  -----------
Vleo =    μ--=    3.986012×105 = 7.844
         rleo        6478.145  km/sec =  28,238.82  km/hr

      ∘  ----  ∘ -----------
V   =    -μ- =    3.986012×105=  3.072
 geo      rgeo       42238.145  km/sec = 11, 059.089  km/hr

2a = rleo + rgeo = (6378.145 + 100 ) + (6378.145 + 35860 ) = 48,716.29  km. Hence a = 24,358.145  km.

         ∘ r3leo-    ∘  -6478.1453---
Tleo = 2π   -μ- = 2π   3.986012×105 =  5,189  sec = 1 hr 26 minutes and 28 seconds.

         ∘ r3geo     ∘ --42238.1453--
Tgeo = 2 π  -μ- =  2π   3.986012×105 =  86,391  sec = 23 hrs and 59 minutes and 50 seconds.

       ∘  a3-    ∘  -24358.1453--
Th = 2π   μ =  2π   3.986012×105 = 37,833  sec = 10  hr 30 min 22 seconds

4.2 Decide how to correct the LEO plane inclination

The problem is that we are given two non-coplanar circular orbits of different radices. A LEO orbit that is inclined at an angle to the plane of another, and larger circular GEO orbit.

We wish to transfer from the inclined (      0
i = 15 )  LEO orbit to the equatorial (     0
i = 0 )  GEO orbit.

We must correct the plane inclination to be able to transfer to the desired GEO orbit.

There are 3 possible ways to achieve this1 :

1.
Correct all of the plane inclination before performing a Hohmann transfer from LEO to GEO. In other words, all of the inclination correction is made at the perigee of the Hohmann ellipse where the Hohmann transfer speed is largest. This will turn out to be the most fuel costly maneuver.
2.
First perform a Hohmann transfer to transfer from the inclined LEO orbit to an inclined GEO orbit, and then apply all of the plane inclination correction at the apogee of the Hohmann elliptical orbit where the ellipse speed will be smallest. This is less costly in ΔV  than the above sequence, and is a common maneuver.
3.
Apply a small and partial plane inclination correction (say an angle α )  at the perigee of the Hohmann orbit, then apply the remaining inclination correction (angle β − α  ) at the apogee. Notice that for α =  0  , this case becomes the same as case (2) above.

Now, we will analyze each case above in details and find the ΔV  for each case and select the maneuver with the smallest ΔV.

4.2.1 First scenario. All plane correction at perigee.

Move from the initial circular parking orbit (which has    0
15   degrees inclination) to a new circular orbit of the same radius but on the equatorial ( 0
0   degrees inclination). This requires one impulse to adjust the inclination. This impulse applied at the point where the parking orbit intersects the equator (line of nodes).

Next, and immediately, apply a coplanar Hohmann transfer (2 impulses) to transfer from the LEO orbit to the outer GEO orbit. (We Can combine the inclination correction velocity impulse vector with the first Hohmann velocity impulse vector using vector additions.)

This is illustrated in figure below.

PIC

To find impulse 1:

                  0
Δ1V  = 2Vleosin 15-
                 2

hence, Δ1V   = 2 (7.84412 )sin 1520

Δ1V  =  2.048km ∕sec = 7,371.81km  ∕hr

To find impulse 2:

Δ2V  =  |Vleo − Vp| = |7.844 − 10.3298| = 2.4858km  ∕sec = 8,948.88km  ∕hr

(speed up).

To find impulse 3:

Δ3V   = |Vgeo − Va| = |3.072 − 1.584| = 1.488km ∕sec = 5,356.8km  ∕hr

(speed up).

Hence, total impulses is found by summing the above

ΔV  =  2.048 +  2.4858 +  1.488 =  6.0218km  ∕sec = 21, 678.48km ∕hr

4.2.2 Second scenario. All plane correction at apogee.

Transfer from the initial LEO orbit (which has    0
15   degrees inclination) to a GEO orbit at 35860  km altitude (which still has a 150   degrees inclination). This is achieved using a normal Hohmann transfer (2 impulses). Next, perform an orbit plane inclination correction (one impulse) to move into the equatorial (0o  inclination) GEO circular orbit on which the first target is currently orbiting.

This is illustrated in figure below.

PIC

To find impulse 1:

Δ1V   = Vleo − Vp = |7.844 − 10.3298| = 2.4858  km/sec =  8,948.88 km/hr

To find impulse 2:

Δ2V  = |Vgeo − Va| = |3.072 − 1.584| = 1.488 km/sec  = 5, 356.8 km/hr

(speed up). To find impulse 3

                  0
Δ3V  = 2Vgeosin 15--= 0.80195 km/sec  = 2,887.031  km/hr
                 2

Hence, total impulses is founding by summing the above total.

ΔV  = 2.4858 + 1.488 + 0.80195 = 4.77575  km/sec =  17,192.7 km/hr

4.2.3 Third Scenario. Partial plane correction at perigee. Rest at apogee.

In this scenario, we will apply a partial orbit plane correction at the perigee and the remaining correction at the apogee. See figure below.

PIC

At the perigee of the Hohmann transfer, apply the law of the cosines to obtain

pict

To find the minimum ΔVtotal  for a given α  , take α  as the independent variable, and minimize ΔVtotal  as a function of α  . Hence solve

∂ ΔV
-----total= 0
   ∂α

Let

                 ∘  --------------------------  ∘ -------------------------------
f (α) = ΔVtotal =   Vl2eo + Vp2− 2VleoVp cos(α) +   V 2geo + Va2− 2VgeoVa cos(β − α )

Hence for minimum

∂f- =  ∘------2VleoVp-sin-(α)--------− ∘-------2VgeoVasin(β-−-α-)--------= 0
∂ α        2     2                        2     2
         Vleo + Vp −  2VleoVp cos(α)      Vgeo + V a − 2VgeoVacos (β − α)

Hence

             ∘  -2------2----------------------                    ∘ --2-----2-----------------
2VleoVp sin (α)   Vgeo + Va − 2VgeoVacos (β −  α) − 2VgeoVasin(β − α )  V leo + Vp − 2VleoVpcos (α)
-----------------∘---------------------------∘--------------------------------------------------= 0
                   V 2leo + V2p − 2VleoVpcos (α )  Vg2eo + V2a − 2VgeoVa cos(β − α)

Then

            ∘  -------------------------------                   ∘ --------------------------
VleoVp sin (α)   V2geo + Va2− 2VgeoVacos (β −  α) − VgeoVa sin(β − α )  V 2leo + V 2p − 2VleoVpcos (α ) = 0

Let

                     ∘ -------------------------------                   ∘ --------------------------
                         2      2                                            2     2
F (α) = VleoVpsin(α )  V geo + Va − 2VgeoVa cos(β − α )− VgeoVa sin(β − α )  V leo + V p − 2VleoVp cos(α )

This is a non-linear equation in α  . we solved for α  using Newton root finding method. Hence we need to find   ′
F  (α )  as follows

                        ∘ -------------------------------
    ′                       2     2
  F  (α ) = VleoVp cos(α)   Vgeo + V a − 2VgeoVa cos (β − α)
                                       (                                  )
                                       (  -------− VgeoVa-sin-(β-−-α)------)
                         + VleoVp sin (α)   ∘ --2-----2----------------------
                                            Vgeo + Va − 2VgeoVa cos(β − α)
  ⌊                                                                                              ⌋
                     ∘  -2------2-----------------                   --------VleoVp-sin-(α)--------
− ⌈VgeoVa cos (β −  α)   Vleo + Vp − 2VleoVpcos (α) + VgeoVasin(β − α )∘  -2-----------------------⌉
                                                                        Vleo + Vp2− 2VleoVpcos (α)

To solve for α,  I wrote a MATLAB function that uses Newton root finding method to find the root of F (α )  for a given rleo,rgeo,β, μ  where β  is the total angle (in degrees) of the inclination of the first circular orbit relative to the second circular orbit, and μ  is the gravitational constant.

The function returns back the angle α  for which ΔVtotal  is minimum.

For an initial guess for α  , and since α  is expected to be small compared to β  , I selected α0 =  0.1 β  .

Applying Newton iterative root finding:

            F (α )
αi+1 = αi − --′--i-
            F  (αi)

The tricky part in this problem was finding a good initial guess for the root (common problem with using Newton roots finding method).

I had to try different values for an initial guess before the root was converged to. For example, when I selected α0   to be 50%  of β  , Newton method did not converge to the root. Selecting α0   to be close to where one expects it to be (which is a small value compared to β  ) did work and a root was found.

For this design project, we are given that        0
β =  15 .  Using this matlab function2 I found that

α = 1.288910

is the solution. Hence, this is the angle I will use for the correction to apply at the perigee.

So, at the perigee, Apply a correction of α  angle, and at the apogee, apply a correction of 15 − 1.28891 =  13.711  .

To find the impulse needed to correct inclination at the perigee and transfer from LEO to a Hohmann orbit with a correction of α

       ∘  --------------------------  ∘ -----------------------------------------------------
ΔV1  =    V2leo + Vp2− 2VleoVpcos (α) =   7.844 2 + 10.32942 − 2(7.844)(10.3294 )cos(1.288910)

so, ΔV1  = 2.4936  km/sec = 8,977.085  km/hr

To find impulse needed to correct inclination at the apogee and transfer from Hohmann to the GEO orbit for an angle (β − α) :

       ∘  -------------------------------  ∘ -------------------------------------------------
ΔV2  =    V 2 + V 2 − 2VgeoVacos (β − α) =   3.072 2 + 1.5842 − 2(3.072)(1.584) cos(13.711090)
           geo    a

so, ΔV2  = 1.578  km/sec = 5682.388  km/hr

So, for this scenario, total ΔV  is given by

ΔV1 + ΔV2  = 2.4936 + 1.578  = 4.0716 km/sec  = 14657.76km/hr

4.2.4 Summary of scenarios to correct plane inclination

Comparing the ΔV  from the above 3 scenarios, we see this:

scenario 1: 6.0218  km/sec

scenario 2: 4.77575  km/sec

scenario 3: 4.0716  km/sec

So, we can see that splitting the plane correction between the perigee and the apogee points leads to a more economical maneuver.

Hence Choose scenario 3 for the next sequence.

4.3 Calculate time to move from LEO to GEO

The vehicle was at the lines of node at t = 0  , hence the time to reach GEO orbit is half the period of the Hohmann transfer orbit.

t = Th-= 18, 916.77sec = 5.25hr
    2

The above time is the same regardless if we inject at one end of the lines of nodes or at the other end. Also, this time is independent of what inclination the Hohmann transfer orbit was at the time of injection.

4.4 Decide when to inject to GEO. Calculate lead angle β  and rendezvous with first satellite

In the previous step, the time it takes to move from LEO to GEO over a Hohmann orbit was found.

Now, find when to make this transfer. That is, we need to find the time to inject into the transfer orbit such that the overall ΔV  is minimized.

The injection must occur when the space vehicle is on the line of nodes. Since this line is where the LEO and the desired GEO plane intersects at.

Hence, there are only 2 points on the LEO orbit that we can use to launch to GEO. (Both ends of the lines of nodes, at both sides of earth).  Let me call one end of the lines of nodes, the top end, and the other end, the bottom end. Where the top end is that end which the spacecraft was at when it first reached LEO, i.e. at time=0.

In addition to the above restriction, if we want to eliminate the need to make any phasing loops when we arrive at the desired GEO orbit in order to rendezvous with the first target, then injection must occur only when the correct lead angle β   is encountered. This additional synchronization requirement will turn out to be costly in time to achieve. If LEO and GEO orbits had been coplanar, then we can inject from any point on the LEO orbit as long as the lead angle β  requirement is met. There will not be an additional requirement of the injection having to be from only two points in the LEO orbit.

We know that the synodic period between LEO and GEO is -TleoTgeo-   5189×86391-
Tgeo−Tleo =  86391−5189 =  5520  seconds, or 92 minutes. This means the LEO and the GEO objects will be aligned along a radial vector originating from the center of the earth once every 92 minutes.

But due to the restriction that this radial vector be only the lines of nodes of the space vehicle, using this synodic period is not too much help for me here.

So, what options do we have?

These are the options to investigate:

1.
Inject from the top end of the lines of nodes. Reach GEO orbit and then phase-wait in that orbit to rendezvous with the target.
2.
phase-wait in the LEO orbit until the correct lead angle β  with the target is reached. Calculate this for when the spacecraft is on the top end of the lines of nodes.
3.
The same as above, but for the case when the spacecraft is on the bottom end of the lines of nodes.
4.
phase-wait in the LEO orbit until the lead angle is closest to β  the first time this happens. (i.e. within the first 2π  )

At the end, select the option which gives the smallest ΔV  as long as the time cost is reasonable.

4.4.1 First option. Inject to GEO at t = 0

we know that at t = 0  we have this state as shown in figure below

PIC

From previous calculations, we found the time needed for space vehicle to reach GEO orbit is 18,916.77  sec. Hence, angle that the GEO satellite will travel in this time is found from

---------18916-----------  =  -x-
(0.9972696 )(24)(60)(60 )     2 π
                       x   =  1.3793765886   rad
                                    0
                           =  79.03

Hence When Vehicle reaches GEO, the first satellite will be (see diagram)

                              0
180 − (79.03 − 40) = 140.9675

behind the space vehicle.

Hence

ΔL  =  − 140.96750

The reason a minus sign is used, is because ΔL  is measured positive if the change in longitude desired is eastwards, and since our target is behind us (westwards relative to the spacecraft), this change is negative.

           0
− 140.9675                (-π-)
= − 140.9675  180 =  − 2.46  radians.

For n = 1  ,        −2.46
Pph = 1×-6.3 + 0.9972696 = 0.60676  days.

Hence  ∘   -ΔL--   --−2.46--
L =  n Pph = 1(0.60676) =  − 4.054  radians/day           (180)
=  − 4.054  π   =  − 232.2771  deg/day.

Hence ΔV   = 5.8(− 232.2771) =  −  1347.20718  m/sec = − 1.347  km/sec.

For n = 2  , Pph = -−2.46 + 0.9972696  = 0.802
      2× 6.3  days.

∘
L = 2−(20..84062) = − 1.5337  rad/day or −  87.875  deg/day

Hence ΔV   = 5.8(− 87.875) =  − 509.675  m/sec = − 0.5  km/sec.

Continuing the above process, I obtain this table.





n  ΔV  (Km/s) Period of phase orbit (in days) Total time in phasing period in days




1  − 1.347  0.6067934  0.6067934




2 − 0.509  0.80203  1.60406




3 − 0.314  0.86711  2.60133




4 − 0.2272  0.8996  3.5984




5 − 0.17788 0.91917436 4.5958718




6  − 0.1462  0.93219  5.59314




12  − 0.07062  0.96472  11.57664




24  − 0.0347  0.980999  23.543976




96  − 0.0085  0.993202  95.347392





Any one of the above choices for n  will achieve rendezvous with the first satellite.

Plotting ΔV  against the total time in the phasing orbit results in the plot shown in figure below

PIC

Figure above shows that most saving in fuel is made by staying in the phasing orbit for less than 20 days. For more than 20 days, the additional saving in fuel is not justified by the time wasted in the phasing orbit.

Zooming in the region of interest in the plot shown in above figure results in figure below

PIC

From the above, it is clear that the slope after 6 days in the phasing orbit is less steep than earlier. The two options I see is to choose n = 6  and save some fuel, or choose n = 1  (the smallest possible value) and save time.

For n = 6

time  = 5.59314 days   ΔV  = − 0.1462 km/sec.

For n = 1

time = 0.6067934  days  ΔV   = − 1.347 km/sec.

Since the fuel saved is so small compared to the initial fuel needed to send the vehicle into GEO orbit, I decided to use the smaller time here.

4.4.2 second option: phase-wait in the LEO orbit until the correct lead angle β  reached at one end of lines of nodes.

PIC

Looking at above figure, the problem can be seen as the following: we need to find the time it takes for the spacecraft to be at the top end of the lines of nodes when the target is at the correct lead angle β  . Because in this case, the spacecraft can injects into the Hohmann orbit and will meet the target at the apogee. This would result in the spacecraft not having to do any phase-waiting loops in the GEO orbit. Hence saving ΔV  . A trade is made between time and ΔV.

How do we find this time value?

First, find β  . To do this, equate travel time for target and spacecraft.

Travel time for spacecraft is half the Hohmann orbit period = 37,833  sec (see common calculations section for derivation)

Travel time x  for target is found from

distance travelled in radians         x
----------------------------  =  -------
            2 π                  Period
                          x   =  P--(𝜃-−-β-)
                                     2π
Where period
          ∘ ----
            r3
P  =   2π   -geo
             μ
          ∘ ------------3--
   =   2π   --42,238.145---
            3.986012 × 105
   =   86390 sec
Hence, and since 𝜃 = π  , when equating times of travel, we get this relation 86390(π−-β)=  37833
   2π

Solve for β

37833 (2π ) =   86390 (π − β)
37833 (2π)
----------- =   π − β
  86390
                    37833-(2π-)
         β  =   π −    86390
            =   0.389980 rad
                      0
            =   22.344

PIC

So, now that we know β  , we need to find when will the spacecraft be at the top end of the lines of nodes when the target has this β  with it.

The angle that the target will move by for each one full period that the spacecraft makes in the LEO orbit is Tleoϖ  , where Tleo  is the LEO period found in the common calculations section to be 5,189  seconds, and ϖ  is the average angular speed of the GEO target in radians per second which is 2π  each 24  hrs (since GEO).

Exactly, ϖ  = -------2π-------=  0.00007292115609102
     0.9972696∗24∗60∗60  rad per second.

Hence, target will travel 5,189 × 0.00007292115609102  =  0.378388  rad        0
= 21.68   for each one full LEO period.

Now that we know the angle the target will travel for each one full LEO period, we need to find how many times we have to do this so that target will end at the correct β  location starting from t=0.

This becomes a simple counting problem. If we imagine a straight line, starting at t=0, and then we move a stick from its left end to its right end at an equal increments of 21.68 units, we just need to find when this stick will land at the correct point (or close enough) on the line where the point β  is located.

When the stick reaches the right end of the line, we carry the reminder back to the start of the line and continue the process.

Figure below shows how to do this counting. Notice that the point I am interested in finding, which is the angle β,  needs to be compensated for by adding the initial 400   to it (since the counting is starting from the epoch). In other words, this is the degrees the spacecraft was ahead of the target when counting starts.

PIC

So, the point on the line we are looking for is 40 + β = 40 + 22.344 =  60.3440   , measured from t=0.

Figure below shows how the counting is actually done.

PIC

I wrote a small MATLAB3  function called nma findPointOnLine.m to do the counting. This function accepts as input the step size, the length of the line, and the distance we are looking for (60.344 in the above example), and how close to the target we want to be (the tolerance).

It returns the number of steps needed to achieve the synchronization.

This is an example call

>> stepSize=21.68; lineLength=360; target=60.344; tol=0.1;  
>> nSteps=nma_findPointOnLine(stepSize,lineLength,target,tol)  
 
nSteps =  
 
        3274

This is the result of the MATLAB function





tolerance (in degrees) number of LEO periods needed time to achieve (seconds) time in hrs




1 36 5189 ×  36 =  186804 51.89




0.1 3274 5189 ×  3274 = 16988786  4719




It is clear that to achieve synchronization to 0.1 degree is too costly in time.

For the case of 1 degree tolerance, it will take 36 LEO loops to achieve the optimal situation with the target at the correct β.

This means, if we spend this time in LEO, we can inject and will meet the target at the same time when we reach GEO. Hence no additional ΔV  would be needed in GEO to phase-wait. We have traded time for fuel.

4.4.3 Third option: phase-wait in the LEO orbit until the correct lead angle β  reached at the other end of lines of nodes.

This case is the same as above, except now we want the spacecraft to be at the other end of the lines of nodes at injection. The only difference is that now t=0 have been shifted to become time after making one half LEO period. We can find this shift since we know the angle the target will travel in one half LEO period. We found from above that target will travel 21.680   for one full LEO period, hence it will travel 10.840   per half that period.

So, the only thing we need to do is determine where the point is located that we need to synchronize with, as illustrated by the figure below.

PIC

So, use the same MATLAB function to find the number of full LEO rotations needed.

>> stepSize=21.68; lineLength=360; target= 231.544; tol=0.1;  
>> nSteps=nma_findPointOnLine(stepSize,lineLength,target,tol)  
 
nSteps =  
 
        3614  
>> nSteps=nma_findPointOnLine(stepSize,lineLength,target,1)  
 
nSteps =  
 
   293

Hence, we see that for a tolerance of 1 degree, we have to wait 293 full LEO loops.

Compare this with the earlier case where we looked for the other end of the lines of nodes, which achieved the same synchronization for only 36 LEO loops. Hence this maneuver will not be accepted.

4.4.4 Fourth option. phase-wait in the LEO orbit until smallest difference to β  reached first time.

To solve this problem, I will calculate the lead angle β  with the first target for a number of time increments of 0.5P  each, where P  is the LEO period, and for each such time increment, will calculate where the first target will be at the end of a Hohmann transfer. Then will calculate the ΔV  needed to phase-wait in GEO to close this final angle gap.

At t = 0.5Tleo

Figure below illustrates this case.

PIC

Total time from epoch to reach GEO =  0.5T   +  0.5T   = 0.5(5189 + 37833 ) =
       leo       h  21, 511sec

Hence, angle that the GEO satellite will travel in this time is found from

         21511                 x
------------------------- =   ---
(0.9972696 )(24) (60)(60)      2π
                       x  =   1.568606 rad
                          =   89.874 0.

Hence When Vehicle reaches GEO, the first satellite will be

89.874 − 40  = 49.8740

ahead.

At t = Tleo

Total time from epoch to reach GEO =  Tleo + 0.5Th = 5189 + 0.5 (37833 ) =  24,105.5sec

Hence, angle that the GEO satellite will travel in this time is found from

         24105.5               x
-------------------------  =   ---
(0.9972696 )(24)(60) (60 )      2π
                        x  =  1.7578 rad
                           =  100.710

Hence When Vehicle reaches GEO, the first satellite will be

                             0
180 − (100.71 − 40 ) = 119.29

behind

At t = 1.5Tleo

Total time from epoch to reach GEO =  1.5Tleo + 0.5Th  = 1.5(5189 ) + 0.5(37833 ) =  26, 700sec

Hence, angle that the GEO satellite will travel in this time is found from

--------26700----------     -x-
0.9972696  ∗ 24 ∗ 60 ∗ 60 =  2π
                     x  =   1.946995 rad
                                     0
                        =   111.5546

Hence When Vehicle reaches GEO, the first satellite will be

                      0
111.555 − 40 =  71.555

ahead.

Continue this process. The result is illustrated in this table






time total time to reach GEO (sec) ϕ  target travelled angle β  lead angle  β  sense relative to spacecraft





0 18,916  79.030   140.96750   behind





0.5Tleo  21,511  89.874  0   49.8740   ahead





Tleo  24,105  100.710   119.290   behind





1.5Tleo  26,700  111.5546 0   71.5550   ahead





2Tleo  29,294  122.39460   97.6050   behind





2.5Tleo  31,889  133.23460   93.2350   ahead





3Tleo  34,483  144.07460   75.92540   behind





3.5Tleo  37,078  154.91460   114.9150   ahead





4Tleo  39,672  165.75470   54.24530   behind





4.5Tleo  42,267  176.59470   136.5950   ahead





5Tleo  44,861  187.43470   32.56530   behind





5.5Tleo  47,456  198.27470   158.2750   ahead





6Tleo  50,050  209.11470   10.88530   behind





6.5Tleo  52,645  219.95470   179.9570   ahead





7Tleo  55,239  230.79470   − 10.79470   behind





From the above table, we see that the closest the target gets to the lines of nodes (within the first 2π  ) at the same time as the vehicle is 10.88530   and is achieved after 6  periods in the LEO orbit.

What is left to do is to determine is the ΔV  needed in the GEO orbit to wait-phase so as to close this final remaining ΔL  = − 10.88530

Now, for the − 10.88530   case calculate the cost ΔV  for phase-waiting in the GEO orbit:

Let P
 ph  be the period of the phasing orbit while in the GEO orbit.

Let ωE  be the angular rate of axial rotation of the earth, which is 360.985647  deg/day = 360.985647  1π80 = 6.3  radians/day.

Let Po  be the period of the geosynchronous orbit, which is 1436.068  min or 0.9972696  days.

Let ΔL  be the change in longitude desired. Which we found it to be     0
− 10          (-π-)
= − 10  180 =  − 0.174533  radians.

Let n  be the number of revolutions spent in the phasing orbit.

Let ∘
L  be the drift rate, positive eastwards.

First, find the period of the phasing orbit Pph  using the equation

P   = -ΔL-- + P
 ph   n  ωE     o

for a specific n.  Then solve for ∘
L  from the equation

      ∘
ΔL  = Ln  Pph

Then find ΔV  corresponding to this  ∘
L  from figure 7.14, Orbital Mechanics book or by using the relation

          ∘
ΔV   = 5.8L.

Where 5.8 is the slope of the line relating ∘
L  to ΔV  . (The above slope is not exact, but it is close enough).

Try the above for a number of different values for n  .

For n = 1  ,       −-0.174533
Pph =   1× 6.3  + 0.9972696 =  0.96957  days.

Hence  ∘
L =  -ΔL--=  −0.174533-=
     n Pph   1(0.96957)  − 0.18  radians/day = − 0.18(180) =
          π  − 10.31385  deg/day.

Hence

ΔV  = 5.8 (− 10.31385 ) = − 59.82033 m/sec = − 0.0598 km/sec.

I do not need to look for n=2 and higher for this case, since the saving in ΔV  is clearly not worth spending an extra day for each additional increment in n  . We see that the ΔV  is very small already (0.0598 km/sec), and for larger n, it will only get smaller and smaller.

4.4.5 Summary of lead angle β  scenarios

For first option, move to GEO without phase-waiting in LEO, and instead phase-wait in GEO, results in ΔV   = − 1.347  with time spend (0.6067934 )(24)(60)(60) = 52426  seconds, or 14  hrs and 33  minutes.

For the second and third options, the time spend is all in LEO, and zero ΔV  was needed to phase-wait in GEO.

For 4th option, part of time spend is in LEO, and some part of time spend is in GEO. ΔV  is not zero, but smaller than first option.

Notice that in this table, the ΔV  cost refer only to the cost of phase wait in GEO to rendezvous with the first target.





scenario ΔV  cost LEO loop total time cost to rendezvous




first. Inject at t=0 1.347  km/sec 0 52,426 (sec)  = 14 hrs 33 minutes




second. optimal β   at top end 0  km/sec 36 186,804 (sec) = 51.89 hrs = 2 days 3 hrs 53 minutes




third. optimal β   at bottom end 0  km/sec 293 1,520,377 (sec)




fourth. smallest β   at apogee 0.0598  km/sec 6 133,820 (sec) = 1 day, 13 hrs, 10 minutes, 20 seconds.




Clearly option 3 is bad. With option 2, we get the same saving for much less time.

Between options 1 and 4, I prefer option 4, since for the cost of about 1.5 days, we reduced ΔV  from 1.347  to 0.0598  km/sec.

So, the final choice is between option 4 and option 2.

With option 2, we have zero ΔV  but we have to spend about 15 more hours in LEO to save 0.0598  km/sec. Is this good or not?

Compared to the ΔV  needed to inject from LEO to GEO which is 4.0716  km/sec, this amount is 1.5% of that. It takes about 5 hrs to go from LEO to GEO. So, should I spend about 3 times as many hours to save 1.5% as many in fuel? (This is interesting that spending more time flying ends up saving fuel! only in space this is possible).

Will not consider option 2 as the time needed is not worth the saving in fuel.

Option 4 is selected

4.5 Rendezvous with the second satellite

Simply perform an in-orbit repositioning using a phasing orbit to rendezvous with the second satellite. The second satellite is 500   ahead of the first satellite (this is given), hence ΔL  = 500 =  0.87267  rad. Apply the same process of in-orbit repositioning to decide on the procedure to select. This table was generated:





n  ΔV  (Km/s) Period of phase orbit (in days) Total time in phasing period in days




1  0.255  1.13578  1.13578




2 0.13596  1.066529  2.133058




3 0.093  1.04344  3.13032




4 0.0703  1.0318993  4.1275972




5 0.0566 1.0249734 5.124867




6  0.04746  1.020356  6.122136




12  0.02396  1.008812  12.105744




24  0.012  1.00304122  24.0729




96  0.003  0.998712  95.87635





Plotting ΔV  against the total time in the orbit results in

PIC

Similarly, at n =  6  , the fuel saving is best for the time spend in the phasing orbit. This gives 6.122136  days in the phasing orbit, and

ΔV  =  0.04746  km/sec.

However, since the fuel saved as a percentage of initial fuel is small, it does not seem that spending extra 6 days in orbit is worth it. I will use n = 1

Time =  1.3578 days  ΔV  = 0.255 km/sec.

Also, since from the specifications, it seems that it says that the second satellite needs to be reached quickly, so I choose n = 1  , the smallest possible value.

4.6 Stay locked in with the second satellite.

Stay in orbit for one complete orbit revolution. This adds one day to the total time in flight. No additional ΔV  needed.

4.7 reposition to final destination.

Perform an in-orbit repositioning using a phasing orbit to reposition ΔL  =  50 = 0.087266  rad ahead.

Apply the same process of in-orbit repositioning to decide on the procedure to select. This table was generated:





n  ΔV  (Km/s) Period of phase orbit (in days) Total time in phasing period in days




1  0.02868  1.0111213  1.0111213




2 0.0144  1.0041954  2.0083908




3 0.00965  1.00188684  3.00566052




4 0.007  1.00073253  4.00293012




5 0.0058 1.00003994 5.0001997




6  0.00484  0.9995782  5.9974692




12  0.00242  0.9984239  11.9810868




24  0.001211  0.99784675  23.948322




96  0.0003  0.9974138  95.7517248





As before, choose n =  1  with

Time  = 1.0111213  day  ΔV  = 0.02868 km/sec.

This adds little over one day to the total flight time.

This completes the required sequence.

5 Summary and Results

See figure below for the final decision tree.

PIC

This is a summary, in table format, of the entire orbiting sequences using the most ΔV  optimal maneuvers selected out of the different scenarios shown above. This table below is the final result of selecting the best option out of each phase. This table only shows the ΔV  needed starting from LEO orbit. It does not include the ΔV  needed to reach LEO which I will add next.

sequence: Inject to LEO. wait in LEO for 6 full orbits. Perform Hohmann transfer with partial plane correction at each end. phase-wait in GEO to rendezvous with first target. Phase-wait in GEO to rendezvous with second target. Lock into second target. Reposition to final location in GEO.




sequence ΔV  (km/sec) duration



wait in LEO 0 50,050 (sec)



Hohmann transfer 4.0716 18,916 (sec)



Rendezvous with 1st target 0.0598 83,770 (sec)



Rendezvous with 2nd target 0.255 98,131 (sec)



lock with 2nd target 0 86,400 (sec)



position to final destination 0.02868 87,360 (sec)



TOTAL 4.41508 424,627 (sec)=117.95 hrs = 4 days 21hrs 57 min



To reach LEO, we have found that Vleo = 7.844  km/sec. To be more realistic, we need to account for the gravitational loss and drag. Typical time to reach LEO is about 2 minutes or 120 seconds. Hence additional ΔV  to account for gravity loss is gΔt  = 9.8(120 ) = 1176  m/sec = 1.176  km/sec.

For the drag effect, it of course depends on the rocket cross sectional area, the drag coefficient and air density. A typical value I have seen in the literature for rockets is to use 5%  of the LEO velocity to account for drag. Hence an additional ΔV  for drag will be 5%  of 7.844  km/sec or 0.392  km/sec.

Hence total ΔV  to reach LEO = 7.844 + 1.176 + 0.392 =  9.412  km/sec

Hence the total ΔV  to achieve final position of spacecraft is

4.41508 + 9.412 = 13.827 km/sec

see figure below for a graphical display of the deltaV used at each stage.

PIC

6 Appendix

This is the MATLAB function that solves for α  to find what partial correction in inclination angle can be done at the perigee for the Hohmann transfer.

Caller script:

  
  function nma_testfindAlphaForMinDeltaV
  
  r0=6378.145;
  r1=100+r0;
  r2=35860+r0;
  beta=15;
  mu=3.986012*10^5;
  
  alpha=nma_findAlphaForMinDeltaV(r1,r2,beta,mu)

This is the function that solves for alpha.

  
  function alpha=nma_findAlphaForMinDeltaV(r1,r2,beta,mu)
  %function alpha=nma_findAlphaForMinDeltaV(r1,r2,beta,mu)
  %
  % Finds the minimum alpha (initial inclination correction)
  % for an orbit relative to a larger circular orbit.
  % see design note for more details.
  %
  %INPUT:
  %  r1: The radius of the smaller circular orbit
  %  r2: the radius of the larger circular orbit
  %  beta: the angle, in degrees, in which the two
  %        circular orbits are non co-planers to
  %        each others.
  % mu: gravitational constant
  %
  % OUTPUT:
  %  alpha: The angle in degrees to use for initial
  %         correction such that minimum delta V is
  %         obtained to move from the smaller circular
  %         orbit to the larger circular orbit
  %
  % Author: Nasser Abbasi
  %         May 19,2003
  
  a  = (r1+r2)/2;
  rp = r1;
  ra = r2;
  
  Va=sqrt( mu*(2/ra - 1/a ));
  Vp=sqrt( mu*(2/rp - 1/a ));
  
  Vc1=sqrt( mu/r1 );
  Vc2=sqrt( mu/r2 );
  
  beta=beta*pi/180;

  
  root(1)=0.1*beta;
  keepLooking = true;
  i=0;
  
  while(keepLooking)
      i=i+1;
      root(i+1)=root(i)- ( F(Vc1,Vc2,Vp,Va,root(i),beta)/dF(Vc1,Vc2,Vp,Va,root(i),beta) );
      root(i+1)
      if( abs ( (root(i+1) - root(i)) / root(i+1) ) * 100  < 0.001 )
          keepLooking=false;
      else
  
          if( ( root(i+1) * root(i) )<0.0)
              error('jumped out of root');
          end
      end
      if(i>100)
          error('Failed to converge');
      end
  end
  
  alpha=root(end);
  alpha=alpha*180/pi;
  
  %%%%%%%%%%%%%%%
  %
  %
  %%%%%%%%%%%%%%%%
  function v=F(Vc1,Vc2,Vp,Va,alpha,beta)
  
  v=Vc1*Vp*sin(alpha)*sqrt(Vc2^2+Va^2-2*Vc2*Va*cos(beta-alpha)) ...
      - Vc2*Va*sin(beta-alpha)*sqrt(Vc1^2+Vp^2-2*Vc1*Vp*cos(alpha));
  
  %%%%%%%%%%%%%%%
  %
  %
  %%%%%%%%%%%%%%%%
  function v=dF(Vc1,Vc2,Vp,Va,alpha,beta)
  
  v=Vc1*Vp*cos(alpha)*sqrt(Vc2^2+Va^2-2*Vc2*Va*cos(beta-alpha)) ...
      +Vc1*Vp*sin(alpha)* ...
      ( -Vc2*Va*sin(beta-alpha)/sqrt(Vc2^2+Va^2-2*Vc2*Va*cos(beta-alpha))) ...
      - (Vc2*Va*cos(beta-alpha)*sqrt(Vc1^2+Vp^2-2*Vc1*Vp*cos(alpha)) ...
      + Vc2*Va*sin(beta-alpha)*...
      ( Vc1*Vp*sin(alpha) /  sqrt(Vc1^2+Vp^2-2*Vc1*Vp*cos(alpha))));
  

  function nSteps=nma_findPointOnLine(stepSize,lineLength,target,tol)
  %function nSteps=nma_findPointOnLine(stepSize,lineLength,target,tol)
  %
  % Function to find how many steps needed to reach withing tolearance
  % close to a point on a line by taking fixed number of steps. Line wrapes
  % around.
  %
  %INPUT:
  %   stepSize: the step size
  %   lineLength: The line length
  %   target: The distance from leftend of line that we want to reach
  %   tol: tolerance in abs. value
  %
  %OUTPUT:
  %   nSteps: number of steps needed
  %
  %Author Nasser Abbasi. May 22, 2003
  %
  
  currentDist = 0;
  nSteps      = 0;
  
  while true
      currentDist = currentDist+stepSize;
      nSteps = 1;
      while(currentDist <= lineLength)
          if( abs (currentDist-target) < tol )
              return;
          end
          rem = lineLength-currentDist;
          if(rem < stepSize)
              currentDist = stepSize-rem;
          else
              currentDist = currentDist+stepSize;
          end
          nSteps = nSteps+1;
      end
  end