At time t0 the state vector of an earth satellite is r0 = 1600I + 5310 J + 3800K(km) (a) v0 = −7.350I + 0.4600J + 2.470K(km/s) (b) Determine the position and velocity 3200 seconds later and plot the orbit in three dimensions.

Question

At time [latex]t_{0}[/latex] the state vector of an earth satellite is

[latex]r _{0}=1600 \hat{ I }+5310 \hat{ J }+3800 \hat{ K }( km )[/latex] (a)

[latex]v _{0}=-7.350 \hat{ I }+0.4600 \hat{ J }+2.470 \hat{ K }( km / s )[/latex] (b)

Determine the position and velocity 3200 seconds later and plot the orbit in three dimensions.

Accepted Answer

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:

[latex]\alpha=1.4613 imes 10^{-4} km ^{-1}[/latex]. Since this is positive, the orbit is an ellipse.
Step 2:

[latex]\chi=294.42 km ^{\frac{1}{2}}[/latex].

Step 3:

f = −0.94843 and [latex]g=-354.89 s ^{-1}[/latex].

Step 4:

[latex]\underline{r =1090.9 \hat{ I }-5199.4 \hat{ J }-4480.6 \hat{ K }( km )} [/latex] r = 6949.8km.
Step 5:

[latex]\dot{f}=0.00045324 s ^{-1}, \quad \dot{g}=-0.88479[/latex].

Step 6:

[latex]\underline{ v=7.2284 \hat{ I }+1.9997 \hat{ J }-0.46311 \hat{ K }( km / s ) }[/latex]

To plot the orbit, we observe that one complete revolution means a change in the eccentric anomaly E of 2π radians. According to Equation [latex]3.54_{2}[/latex], the corresponding change in the universal anomaly is

[latex]\chi=\sqrt{a} E=\sqrt{\frac{1}{\alpha}} E=\sqrt{\frac{1}{0.00014613}} \cdot 2 \pi=519.77 km ^{\frac{1}{2}}[/latex]

Letting χ vary from 0 to 519.77 in small increments, we employ the Lagrange coefficient formulation (Equation 3.64 plus 3.66a and 3.66b) to compute

[latex]r =f r _{0}+g v _{0}[/latex] (3.64)

[latex]f=1-\frac{\chi^{2}}{r_{0}} C\left(\alpha \chi^{2} ight)[/latex] (3.66a)

[latex]g=\Delta t-\frac{1}{\sqrt{\mu}} \chi^{3} S\left(\alpha \chi^{2} ight)[/latex] (3.66b)

[latex]r =\left[1-\frac{\chi^{2}}{r_{0}} C\left(\alpha \chi^{2} ight) ight] r _{0}+\left[\Delta t-\frac{1}{\sqrt{\mu}} \chi^{3} S\left(\alpha \chi^{2} ight) ight] v _{0}[/latex]

where Δt for a given value of χ is given by Equation 3.45. Using a computer to plot the points obtained in this fashion yields Figure 4.6, which also shows the state vectors at [latex]t_{0} ext { and } t_{0}+3200 s[/latex].

[latex]a=\frac{1}{\frac{2}{r}-\frac{v^{2}}{\mu}}[/latex] (3.45)

Question 4.2: At time t0 the state vector of an earth satellite is r0 = 16...

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