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## transfer from the LEO orbit to the GEO orbit

May 23, 2003   Compiled on May 24, 2020 at 7:01am

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 $$\beta$$ 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 $$\alpha )$$ at the perigee of the Hohmann orbit, then apply the remaining inclination correction(angle $$\beta -\alpha$$) at the apogee. This is the case we will solve for here to ﬁnal the optimal amount of angle $$\alpha .$$ For $$\alpha =0$$, this case becomes the same as case (2) above.

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

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

$$\Delta V_{1}=\sqrt{V_{c1}^{2}+V_{p}^{2}-2V_{c1}V_{p}\cos \left ( \alpha \right ) }$$

$$\Delta V_{2}=\sqrt{V_{c2}^{2}+V_{a}^{2}-2V_{c2}V_{a}\cos \left ( \beta -\alpha \right ) }$$

$$\Delta V_{total}=\Delta V_{1}+\Delta V_{2}$$

To ﬁnd the minimum $$\Delta V_{total}$$ for a given $$\alpha$$, take $$\alpha$$ as the independent variable, and minimize $$\Delta V_{total}$$ as a function of $$\alpha .$$

Hence solve $$\frac{\partial \Delta V_{total}}{\partial \alpha }=0$$

Let $$f\left ( \alpha \right ) =\Delta V_{total}=\sqrt{V_{c1}^{2}+V_{p}^{2}-2V_{c1}V_{p}\cos \left ( \alpha \right ) }+\sqrt{V_{c2}^{2}+V_{a}^{2}-2V_{c2}V_{a}\cos \left ( \beta -\alpha \right ) }$$

$$\frac{\partial f}{\partial \alpha }=\frac{2V_{c1}V_{p}\sin \left ( \alpha \right ) }{\sqrt{V_{c1}^{2}+V_{p}^{2}-2V_{c1}V_{p}\cos \left ( \alpha \right ) }}-\frac{2V_{c2}V_{a}\sin \left ( \beta -\alpha \right ) }{\sqrt{V_{c2}^{2}+V_{a}^{2}-2V_{c2}V_{a}\cos \left ( \beta -\alpha \right ) }}=0$$ for minimum

so

$$\frac{2V_{c1}V_{p}\sin \left ( \alpha \right ) \sqrt{V_{c2}^{2}+V_{a}^{2}-2V_{c2}V_{a}\cos \left ( \beta -\alpha \right ) }-2V_{c2}V_{a}\sin \left ( \beta -\alpha \right ) \sqrt{V_{c1}^{2}+V_{p}^{2}-2V_{c1}V_{p}\cos \left ( \alpha \right ) }}{\sqrt{V_{c1}^{2}+V_{p}^{2}-2V_{c1}V_{p}\cos \left ( \alpha \right ) }\sqrt{V_{c2}^{2}+V_{a}^{2}-2V_{c2}V_{a}\cos \left ( \beta -\alpha \right ) }}=0$$

$$V_{c1}V_{p}\sin \left ( \alpha \right ) \sqrt{V_{c2}^{2}+V_{a}^{2}-2V_{c2}V_{a}\cos \left ( \beta -\alpha \right ) }-V_{c2}V_{a}\sin \left ( \beta -\alpha \right ) \sqrt{V_{c1}^{2}+V_{p}^{2}-2V_{c1}V_{p}\cos \left ( \alpha \right ) }=0$$ (1)

let

$$F\left ( \alpha \right ) =V_{c1}V_{p}\sin \left ( \alpha \right ) \sqrt{V_{c2}^{2}+V_{a}^{2}-2V_{c2}V_{a}\cos \left ( \beta -\alpha \right ) }-V_{c2}V_{a}\sin \left ( \beta -\alpha \right ) \sqrt{V_{c1}^{2}+V_{p}^{2}-2V_{c1}V_{p}\cos \left ( \alpha \right ) }$$

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

$$F^{^{\prime }}\left ( \alpha \right ) =V_{c1}V_{p}\cos \left ( \alpha \right ) \sqrt{V_{c2}^{2}+V_{a}^{2}-2V_{c2}V_{a}\cos \left ( \beta -\alpha \right ) }+V_{c1}V_{p}\sin \left ( \alpha \right ) \left ( \frac{-V_{c2}V_{a}\sin \left ( \beta -\alpha \right ) }{\sqrt{V_{c2}^{2}+V_{a}^{2}-2V_{c2}V_{a}\cos \left ( \beta -\alpha \right ) }}\right ) -$$

$$\left [ V_{c2}V_{a}\cos \left ( \beta -\alpha \right ) \sqrt{V_{c1}^{2}+V_{p}^{2}-2V_{c1}V_{p}\cos \left ( \alpha \right ) }+V_{c2}V_{a}\sin \left ( \beta -\alpha \right ) \frac{V_{c1}V_{p}\sin \left ( \alpha \right ) }{\sqrt{V_{c1}^{2}+V_{p}^{2}-2V_{c1}V_{p}\cos \left ( \alpha \right ) }}\right ]$$

To solve for $$\alpha ,$$ I wrote a small MATLAB function that uses Newton root ﬁnding method to ﬁnd the root of $$F\left ( \alpha \right )$$ for a given $$r_{1},r_{2},\beta$$. where $$r_{1}\$$and $$r_{2}$$ are the radius of the smaller and the larger circular orbits respectively, and $$\beta$$ is the angle (in degrees) of the inclination of the ﬁrst circular orbit relative to the second circular orbit, measured positive clockwise. The function returns back the angle $$\alpha$$ for which $$\Delta V_{total}$$ is minimum. For an initial guess for $$\alpha$$, and since $$\alpha$$ is expected to be small compared to $$\beta$$, I selected $$\alpha _{0}=0.1\ \beta$$.

Apply Newton iterative root ﬁnding:

$$\alpha _{i+1}=\alpha _{i}-\frac{F\left ( \alpha _{i}\right ) }{F^{^{\prime }}\left ( \alpha _{i}\right ) }$$

The tricky part in this problem was ﬁnding a good initial guess for the root. When I selected $$\alpha _{0}$$ to be $$50\%$$ of $$\beta$$, Newton method did not converge. Selecting $$\alpha _{0}$$ to be close to where one expects it to be (which is very small compared to $$\beta$$) worked and a root was found.

For the design project, we are given $$\beta =15^{0}.$$ Using this matlab function I found that $$\alpha =1.28891^{0}$$ is the solution, for the given geometry and parameters. Hence, this is the angle I will use.