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transfer from the LEO orbit to the GEO orbit

Nasser M. Abbasi

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

The problem: We are given 2 non co-planer circular orbits of different radius. For example, a LEO orbit that is inclined at an angle β  to the plane of another, larger, circular orbit (for example, a GEO orbit).

We wish to transfer from the LEO orbit to the 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 Hohmman transfer from LEO to GEO. In other words, all of the correction is made at the perigee of the Hohmann ellipse. This is the most fuel costly maneuver.
2.
Perform a Hohmann transfer to move from the LEO orbit to a GEO orbit. Then apply all of the plane inclination correction at the apogee of the Hohmann elliptical orbit. This is less costly than above procedure, and is a common maneuver.
3.
Apply a small and partial plane inclination correction (say angle α)  at the perigee of the Hohmann orbit, then apply the remaining inclination correction(angle β − α  ) at the apogee. This is the case we will solve for here to final the optimal amount of angle α.  For α = 0  , this case becomes the same as case (2) above.

See the following diagram to illustrate the geometry of the 3rd case above.

PIC

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

       ∘ -------------------------
ΔV1 =    Vc21 + V2p − 2Vc1Vp cos(α)

       ∘ --2----2---------------------
ΔV2 =    Vc2 + Va − 2Vc2Va cos(β − α )

ΔVtotal = ΔV1 + ΔV2

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∂tαotal=  0

Let                  ∘ -------------------------
                     2     2                   ∘ --2----2---------------------
f (α ) = ΔVtotal = V c1 + Vp −  2Vc1Vpcos (α ) +   Vc2 + V a − 2Vc2Va cos(β − α )

∂f = ∘----2Vc1Vpsin(α)----−  ∘---2Vc2Va-sin(β−-α)-----= 0
∂α     V2c1+Vp2−2Vc1Vpcos(α)     V2c2+Va2− 2Vc2Vacos(β−α)  for minimum

so

         ∘ --------------------            ∘ -----------------
2Vc1Vpsin(α)-V∘2c2+V-2a−2Vc2Vacos(β−α-)−∘2Vc2Vasin(β−-α)-V2c1+V-2p−2Vc1Vpcos(α) = 0
             V2c1+Vp2−2Vc1Vpcos(α) Vc22+V2a−2Vc2Va cos(β− α)

            ∘ --2--------------------------                   ∘ --2----------------------
Vc1Vpsin(α )  Vc2 + Va2− 2Vc2Va cos(β − α ) − Vc2Va sin(β − α ) Vc1 + Vp2− 2Vc1Vp cos(α ) = 0  (1)

let

                    ∘ --2-----2--------------------                   ∘  -2-----2-----------------
F (α) = Vc1Vpsin (α )  V c2 + Va −  2Vc2Vacos (β − α) − Vc2Vasin(β −  α)   Vc1 + Vp − 2Vc1Vpcos (α)

This is a non-linear equation in α  . Solve for α  using Newton root finding method.

                                                                   (                      )
 ′                   ∘ -----------------------------                     −Vc2Va sin(β− α)
F (α ) = Vc1Vp cos(α )  V 2c2 + Va2− 2Vc2Va cos(β −  α) + Vc1Vpsin(α )  ∘--2--2--------------  −
                                                                      Vc2+Va−2Vc2Va cos(β− α)

[                 ∘ -------------------------                                      ]
 V  V  cos(β − α )  V 2 + V 2− 2V  V  cos(α ) + V V  sin (β − α) ∘----Vc1Vpsin(α)-----
  c2 a               c1    p      c1  p           c2 a              Vc21+V2p− 2Vc1Vpcos(α)

To solve for α,  I wrote a small MATLAB function that uses Newton root finding method to find the root of F (α)  for a given r ,r ,β
 1  2  . where r
 1   and r
 2   are the radius of the smaller and the larger circular orbits respectively, and β  is the angle (in degrees) of the inclination of the first circular orbit relative to the second circular orbit, measured positive clockwise. 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  .

Apply Newton iterative root finding:

            F(αi)
αi+1 = αi − F′(αi)

The tricky part in this problem was finding a good initial guess for the root. When I selected α0   to be 50%  of β  , Newton method did not converge. Selecting α0   to be close to where one expects it to be (which is very small compared to β  ) worked and a root was found.

For the design project, we are given        0
β =  15 .  Using this matlab function I found that             0
α =  1.28891   is the solution, for the given geometry and parameters. Hence, this is the angle I will use.