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Chapter 6

Q. 6.12

Suppose in the previous example that part of the plane change, Δi, takes place at B,the perigee of the Hohmann transfer ellipse, and the remainder, 28^{º} − Δi, occurs at the apogee C. What is the value of Δi which results in the minimum Δv_{total}?


Verified Solution

We found in Example 6.11 that if Δi = 0, then Δv_{total}= 5.3803   km/s,whereas Δi = 28^{◦} made Δv_{total}= 7.6307   km/s. Here we are to determine if there is a value of Δi between 0^{◦}   and   28^{◦} that yields a Δv_{total} which is smaller than either of those two.
In this case a plane change occurs at both B and C. Recall that the most efficient strategy is to combine the plane change with the speed change, so that the delta-vs at those points are (Equation 6.21)

\Delta v=\sqrt{v^{2}_{1}+v^{2}_{2}-2v_{1}v_{2}\cosδ}                        (6.21)

\Delta v_{B}=\sqrt{v^{2}_{B_{1}} +v^{2}_{B_{2}} -2v_{B_{1}}v_{B_{2}}cos\Delta i} =\sqrt{7.7258^{2} + 10.152^{2} − 2 · 7.7258 · 10.152 · \cos\Delta i} =\sqrt{162.74 − 156.86 \cos\Delta i}


\Delta v_{C}=\sqrt{v^{2}_{C_{2}} +v^{2}_{C_{3}} -2v_{C_{2}}v_{C_{3}}cos (28^{◦}-\Delta i)}=\sqrt{1.6078^{2} + 3.0747^{2} − 2 · 1.6078 · 3.0747 · \cos(28^{◦}-\Delta i)}

=\sqrt{12.039 − 9.8871\cos(28^{◦}-\Delta i)}


\Delta v_{total }=\Delta v_{B}+\Delta v_{C}

=\sqrt{162.74 − 156.86\cos \Delta i}+\sqrt{12.039 − 9.8871\cos (28^{◦}-\Delta i)}                    (a)

To determine if there is a Δi which minimizes Δv_{total} , we take its derivative with respect to Δi and set it equal to zero:

\frac{d\Delta v_{total }}{d\Delta i}=\frac{78.43\sin \Delta i}{\sqrt{162.74-156.86\cos \Delta i} }-\frac{4.9435\sin (28^{◦}-\Delta i)}{\sqrt{12.039-9.8871\cos (28^{◦}-\Delta i)} }=0

This is a transcendental equation which must be solved iteratively. The solution, as the reader may verify, is

\Delta i= 2.1751^{◦}                   (b)

That is, an inclination change of 2.1751^{◦} should occur in low-earth orbit, while the rest of the plane change, 25.825^{◦}, is done at GEO. Substituting (b) into (a) yields

\Delta v_{vtotal }= 4.2207   km/s

This is 21 percent less than the smallest Δv_{total} computed in Example 6.11.