Orbital Elements and Types Questions and Answers

Answer: B

The satellite forms an elliptic orbit.

At injection the radius is \(r = 6378 + 350 = 6728\) km. Compare the injection speed with the circular and escape speeds at that radius:

\[ \begin{aligned} v_{c} &= \sqrt{\mu_{E}/r} = \sqrt{\frac{3.986 \times 10^{14}}{6.728 \times 10^{6}}} \\ &\approx 7.70\ \text{km/s} \\ v_{esc} &= \sqrt{2}\,v_{c} \approx 10.89\ \text{km/s} \end{aligned} \]

The injection speed of 8.0 km/s, horizontal, lies between the two, so the orbit is an ellipse with the injection point as perigee. Exactly 7.70 km/s would give a circle, 10.89 km/s or more would escape, and a speed below circular at this height would make injection the apogee of a lower orbit, which is how a satellite might fall back.

Answer: B

The velocity is \(V = \sqrt{\mu/R}\).

In a circular orbit, gravity provides exactly the centripetal force:

\[ \begin{aligned} \frac{mV^{2}}{R} &= \frac{GMm}{R^{2}} = \frac{\mu m}{R^{2}} \\ V &= \sqrt{\frac{\mu}{R}} \end{aligned} \]

\(\sqrt{\mu/2R}\) is smaller than the circular speed by a factor \(\sqrt{2}\); it equals half the escape speed \(\sqrt{2\mu/R}\). \(\dfrac{2\pi}{\sqrt{\mu}}R^{3/2}\) is the orbital period, a time, not a velocity, and the version with \(R^{2/3}\) is that period with the exponent inverted.

Answer: B

The satellite follows a hyperbola.

The type of orbit follows from the sign of the specific orbital energy \(\varepsilon = v^{2}/2 - \mu/r\). With the given vectors:

\[ \begin{aligned} r &= \sqrt{8000^{2} + 9000^{2}} \approx 12041.6\ \text{km} \\ v^{2} &= 6^{2} + 6^{2} = 72\ \text{km}^{2}/\text{s}^{2} \\ \varepsilon &= 36 - \frac{398600}{12041.6} \\ &\approx 36 - 33.10 = 2.90\ \text{km}^{2}/\text{s}^{2} \end{aligned} \]

The energy is positive, so the orbit is open and hyperbolic. Equivalently, the speed \(\sqrt{72} \approx 8.49\) km/s exceeds the local escape speed \(\sqrt{2\mu/r} \approx 8.14\) km/s.

A parabola needs exactly zero energy, and a circle needs negative energy with the speed equal to \(\sqrt{\mu/r} \approx 5.75\) km/s. A straight line would need zero angular momentum, but \(h = r_p v_q - r_q v_p = 8000(6) - 9000(-6) = 102000\ \text{km}^{2}/\text{s}\), which is not zero.