Question

An earth satellite has the instantaneous position vector \(\vec{r}\) and velocity vector \(\vec{v}\) as given below. Here \(\hat{p}\) and \(\hat{q}\) denote the unit vectors along the \(x\) and \(y\) axes of the perifocal frame, respectively. Assume that the value of the gravitational parameter is 398600 \(\text{km}^3/\text{s}^2\). Which one of the following trajectories does the satellite follow?
\[ \vec{r} = (8000\hat{p} + 9000\hat{q}) \text{ km and } \vec{v} = (-6\hat{p} + 6\hat{q}) \text{ km/s} \]

• A. Circle
• B. Hyperbola
• C. Parabola
• D. Straight line

Hint:

Work out the specific mechanical energy \(\varepsilon = v^2/2 - \mu/r\); its sign alone tells you the conic section (positive means hyperbola).

Answer 1

The Correct Option is B

Step 1: Recall the eccentricity vector formula. For an orbit around a body with gravitational parameter $\mu$, the eccentricity vector is \[ \vec{e} = \frac{1}{\mu}\left[\left(v^2 - \frac{\mu}{r}\right)\vec{r} - (\vec{r}\cdot\vec{v})\vec{v}\right] \] and the magnitude of this vector is the orbit eccentricity. An eccentricity below 1 means an ellipse, exactly 1 means a parabola, and above 1 means a hyperbola.

Step 2: Work out the scalar quantities needed. $r^2 = 8000^2+9000^2 = 145000000$, so $r \approx 12041.6$ km. $v^2 = (-6)^2+6^2 = 72$ km$^2$/s$^2$. $\vec{r}\cdot\vec{v} = (8000)(-6)+(9000)(6) = -48000+54000 = 6000$ km$^2$/s. $v^2 - \mu/r = 72 - 398600/12041.6 = 72 - 33.10 = 38.90$

Step 3: Build the eccentricity vector. \[ \vec{e} = \frac{1}{398600}\Big[38.90(8000,9000) - 6000(-6,6)\Big] \] \[ = \frac{1}{398600}\big[(311200,350100) - (-36000,36000)\big] = \frac{1}{398600}(347200,314100) \] \[ \vec{e} = (0.871,\ 0.788) \]

Step 4: Find the eccentricity and read off the orbit type. \[ e = \sqrt{0.871^2+0.788^2} = \sqrt{0.759+0.621} = \sqrt{1.380} \approx 1.175 \] Since $e \approx 1.175 > 1$, the orbit is open and unbound. This matches what the energy check gives, and confirms the trajectory is hyperbolic, not an ellipse, circle, parabola or straight line. \[ \boxed{e \approx 1.175 > 1 \Rightarrow \text{Hyperbola}} \]