Discussion
At a point in the trajectory of an unpowered space vehicle moving about the Earth, the altitude above the mean sea level is 600 km, and the speed with reference to a coordinate system fixed to the center of mass of the Earth is 9 km/s. Assume that the Earth is a sphere with a radius 6400 km and \(G M_{Earth} = 3.98 \times 10^{14}\) m3/s2, where, G is the universal gravitational constant and \(M_{Earth}\) is mass of the Earth. The trajectory is:
Answer: B
The trajectory is elliptic.
The orbit type depends on how the speed compares with the circular and escape speeds at that radius. Here \(r = 6400 + 600 = 7000\) km.
\[ \begin{aligned} v_{c} &= \sqrt{\frac{GM}{r}} = \sqrt{\frac{3.98 \times 10^{14}}{7 \times 10^{6}}} \approx 7.54\ \text{km/s} \\ v_{esc} &= \sqrt{2}\,v_{c} \approx 10.66\ \text{km/s} \end{aligned} \]
The speed of 9 km/s is above the circular speed and below the escape speed, so the specific orbital energy is negative and the vehicle is on a closed, elliptic path.
A circular orbit would need exactly 7.54 km/s, a parabolic one exactly 10.66 km/s, and a hyperbolic one more than 10.66 km/s.
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