Question 4.2: At time t0, the state vector of an earth satellite is r0 = 1......
At time t_{0} , the state vector of an earth satellite is
\mathrm{r}_{0}=1600\hat{i}+5310\hat{j}+3800\hat{\mathrm{k}}\left(\mathrm{km}\right) (a)
\mathrm{v}_{0}=-7.350\mathrm{\hat{i}}+0.4600\mathrm{\hat{j}}+2.470\mathrm{\hat{k}}\left(\mathrm{km/s}\right) (b)
Determine the position and velocity 3200 s later and plot the orbit in three dimensions .
We will use the universal variable formulation and Algorithm 3.4, which was illustrated in detail in Example 3.7. Therefore, only the results of each step are presented here.
Step 1:
(α here is not to be confused with the right ascension.)
α = 1.4613 × 10^{-4} km^{-1} . Since this is positive, the orbit is an ellipse.
Step 2:
\chi=294.42{\mathrm{~km}}^\frac{1}{2}
Step 3:
f=-0.94843~{\mathrm{and}}~g=-354.89~s^{-1}
Step 4:
r = 1090.9 \hat{i} – 5199.4 \hat{j} – 4480.6 \hat{k}\ (km) ⇒ r = 6949.8 km
Step 5:
\dot{f}=0.00045324\,\mathrm{s}^{-1},\,\,\,\dot{\mathrm{g}}=-0.88479
Step 6:
v = 7.2284 \hat{i} + 1.9997 \hat{j} – 0.46311\ \hat{k} (km/s)
To plot the elliptical orbit, we observe that one complete revolution means a change in the eccentric anomaly E of 2π radians. According to Eqn (3.57), the corresponding change in the universal anomaly is
\chi = \begin{cases} \frac{h}{\sqrt{μ}} \tan \frac{θ}{2} parabola \\ {\sqrt{a}}E\qquad{\mathrm{elilipse}} (t_{0}=0,\mathrm{~at~periapse}) (3.57)\\ \sqrt{-a}{ F}~~~~\mathrm{hyperbola} \end{cases}
\chi={\sqrt{a}}E={\sqrt{\frac{1}{\alpha}}}E={\sqrt{\frac{1}{0.00014613}}}.2\pi=519.77{\mathrm{~km}}^{\frac{1}{2}}
Letting \chi vary from 0 to 519.77 in small increments, we employ the Lagrange coefficient formulation (Eqn (3.67) plus (3.69a) and (3.69b)) to compute
\mathbf{r}=f\mathbf{r}_{0}+g\mathbf{v}_{0} (3.67)
f=1-{\frac{\chi^{2}}{r_{0}}}C(\alpha\chi^{2}) (3.69a)
g=\Delta t-{\frac{1}{\sqrt{\mu}}}\chi^{3}S(\alpha\chi^{2}) (3.69b)
\mathbf{r}=\left[1-{\frac{\chi^{2}}{r_{0}}}C{\Big(}\alpha\,\chi^{2}{\Big)}\right]\mathbf{r}_{0}+\left[\Delta t-{\frac{1}{\sqrt{\mu}}}\chi^{3}S{\Big(}\alpha\,\chi^{2}{\Big)}\right]\mathbf{v}_{0}
where Δt for a given value of \chi is given by Eqn (3.49). Using a computer to plot the points obtained in this fashion yields Figure 4.6, which also shows the state vectors at t_{0} and t_{0} + 3200 s .
\sqrt{\mu}\Delta t={\frac{r_{0}v_{ r_{0}}}{\sqrt{\mu}}}\chi^{2}C{\bigl(}\alpha\chi^{2}{\bigr)}+(1-\alpha r_{0})\chi^{3}S{\bigl(}\alpha\chi^{2}{\bigr)}+r_{0}\chi (3.49)
