Paraphrase of Basics of Space Flight
The link Basics of Space Flight gives a thorough set of equations in terms of the burnout speed, v_{1}, the range, r_{1}, from the center of a large mass, GM, and the flight angle _{γ}_{1} of the vector velocity with respect to the vector range. From these we need to find the value of the initial true anomaly angle _{ν}_{1}, the ellipse semi-major axis a, the ellipse eccentricity e, and the ellipse semi-minor axis b.
There are two conservation laws that pertain to this problem: Energy and Angular Momentum.
The energy law is
$$\frac{m{v}^{2}}{2}-\frac{GMm}{r}=\frac{m{v}_{1}^{2}}{2}-\frac{GMm}{{r}_{1}^{}}$$ |
(1.1) |
And the angular momentum law is:
$$rv\mathrm{sin}\gamma ={r}_{1}{v}_{1}\mathrm{sin}{\gamma}_{1}$$ |
(1.2) |
At both the periapsis, R_{p}, and the apapsis, R_{a}, $\gamma =\pi /2$, so equation (1.2) becomes:
$${R}_{p}{v}_{p}={R}_{a}{v}_{a}$$ |
(1.3) |
Combining equations (1.3) and (1.1) we obtain the following equations for the speeds at periapsis and apapsis
$$\begin{array}{l}{v}_{p}=\sqrt{\frac{2GM{R}_{a}}{{R}_{P}({R}_{a}+{R}_{p})}}\\ {v}_{a}=\sqrt{\frac{2GM{R}_{p}}{{R}_{a}({R}_{a}+{R}_{p})}}\end{array}$$ |
(1.4) |
We can also use these expressions to get R_{p} and R_{a}
_{}_{} |
_{$$\begin{array}{l}{R}_{a}=\frac{{R}_{p}}{\frac{2GM}{{R}_{p}{v}_{p}^{2}}-1}\\ {R}_{p}=\frac{{R}_{a}}{\frac{2GM}{{R}_{a}{v}_{a}^{2}}-1}\end{array}$$}_{ } |
_{(}_{1}_{.}_{5}_{)} |
Again using equation (1.2) for v_{p} and R_{p} we can write
$${v}_{p}=\frac{{r}_{1}{v}_{1}\mathrm{sin}{\gamma}_{1}}{{R}_{p}}$$ |
(1.6) |
And again using equation (1.1) we can write:
$$\frac{{r}_{1}^{2}{v}_{1}^{2}{\mathrm{sin}}^{2}{\gamma}_{1}}{{R}_{p,a}^{2}}-{v}_{1}^{2}=2GM\left(\frac{1}{{R}_{p,a}}-\frac{1}{{r}_{1}}\right)$$ |
(1.7) |
Which can be solved for R_{p} by multiplying first by $-{R}_{p}^{2}/{({r}_{1}{v}_{1})}^{2}$ which results is the equation
$$\begin{array}{l}{\left(\frac{{R}_{p,a}}{{r}_{1}}\right)}^{2}(1-C)+\left(\frac{{R}_{p,a}}{{r}_{1}}\right)C-{\mathrm{sin}}^{2}{\gamma}_{1}=0\\ where\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}C=\frac{2GM}{{r}_{1}{v}_{1}^{2}}\end{array}$$ |
(1.8) |
Which results in
$$\frac{{R}_{p,a}}{{r}_{1}}=\frac{-C\pm \sqrt{{C}^{2}+4(1-C){\mathrm{sin}}^{2}{\gamma}_{1}}}{2(1-C)}$$ |
(1.9) |
By definition of an ellipse with semi-major axis a and eccentricity e we have
$$\begin{array}{l}{R}_{a}=a(1+e)\\ {R}_{p}=a(1-e)\end{array}$$ |
(1.10) |
These equations can be solved for e by eliminating a
$$e=\frac{\frac{{R}_{a}}{{R}_{p}}-1}{\frac{{R}_{a}}{{R}_{p}}+1}$$ |
(1.11) |
And we already have expressions for R_{p} and R_{a} in equation (1.9).
Also from equations (1.10) we have a result for the semi-major axis
$$a=\frac{{R}_{p}+{R}_{a}}{2}=\frac{-C}{2(1-C)}{r}_{1}$$ |
(1.12) |
Which means that a is independent of initial flight angle _{γ}_{1.}
A much more convenient expression for e is
$$e=\sqrt{{\left(\frac{{r}_{1}{v}_{1}^{2}}{GM}-1\right)}^{2}{\mathrm{sin}}^{2}{\gamma}_{1}+{\mathrm{cos}}^{2}{\gamma}_{1}}$$ |
(1.13) |
We also need the “true anomaly” angle ${\nu}_{1}$ from the periapsis point to the initial position vector end, r_{1}, (see Figure) and this is obtained from the equation
$$\mathrm{tan}{\nu}_{1}=\frac{\frac{{r}_{1}{v}_{1}^{2}}{GM}\mathrm{sin}{\gamma}_{1}\mathrm{cos}{\gamma}_{1}}{\frac{{r}_{1}{v}_{1}^{2}}{GM}{\mathrm{sin}}^{2}{\gamma}_{1}-1}$$ |
(1.14) |
Another easier expression for a is
$$a=\frac{1}{\frac{2}{{r}_{1}}-\frac{{v}_{1}^{2}}{GM}}$$ |
(1.15) |
Also, since we have equation (1.13) for e, and equation (1.15) for a, we can obtain R_{p} and R_{a} from equations(1.10) without the need for the quadratic equation solution (1.9).
Parameters e and a specify the shape of the ellipse while R_{p}_{ }defines the distance of closest approach to the large mass, GM, and ${\nu}_{1}$ expresses the starting angle.
Note, for any ellipse, that the semi-minor axis, b, is
$$b=a\sqrt{1-{e}^{2}}$$ |
(1.16) |
We need one more expression: the rate of increase of the true anomaly angle $\nu $ .
Three equations are involved here. First one is called the eccentric anomaly,E, and is
$$E={\mathrm{cos}}^{-1}\left(\frac{e+\mathrm{cos}\nu}{1+e\mathrm{cos}\nu}\right)$$ |
(1.17) |
The second one is for the mean anomaly angle, M
$$M=E-e\mathrm{sin}E$$ |
(1.18) |
The third equation relates the change of M to the change of time:
$$M-{M}_{0}=\sqrt{\frac{GM}{{a}^{3}}}(t-{t}_{0})$$ |
(1.19) |
So we solve equation (1.17) for E in terms of $\nu $ , then M is related to E and the progression of M with respect to time is
$$\frac{dM}{dt}=\sqrt{\frac{GM}{{a}^{3}}}$$ |
(1.20) |
And using the chain rule we can write
$$\frac{dM}{dt}=\frac{dE}{dt}(1-e\mathrm{cos}E)=\sqrt{\frac{GM}{{a}^{3}}}$$ |
(1.21) |
And then use the chain rule on equation (1.17) to obtain and expression for $\dot{\nu}$ in terms of $\nu $ .
Referring to mean and eccentric anomaly as we did above is more complicated than necessary. Instead we refer to Kepler Laws of Planetary Motion from which we obtain the following equations for the period of the elliptical motion, T, distance from the large mass, r, and the rate of angular motion $\dot{\nu}$
$$T=\frac{2\pi {a}^{\frac{3}{2}}}{\sqrt{GM}}$$ |
(1.22) |
$$\dot{\nu}=\frac{2\pi ab}{T{r}^{2}}$$ |
(1.23) |
$$r=a\frac{1-{e}^{2}}{1+e\mathrm{cos}\nu}$$ |
(1.24) |
Combining equations (1.16), (1.22), and (1.24) in equation (1.23) we get the following expression for the rate of change $\dot{\nu}$
$$\dot{\nu}=\frac{\sqrt{GM}}{{[a(1-{e}^{2})]}^{\frac{3}{2}}}{\left(1+e\mathrm{cos}\nu \right)}^{2}$$ |
(1.25) |
To obtain the angle $\nu $ and radius r after a time t, we integrate equation (1.25) with respect to time. This (x,y) location of the smaller mass, with respect to the center of the larger mass, is then
$$\begin{array}{l}x=r\mathrm{cos}\nu \\ y=r\mathrm{sin}\nu \end{array}$$ |
(1.26) |