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: Find the magnitude of the position and velocity vectors.
The position vector is \(\vec{r} = 8000\hat{p} + 9000\hat{q}\) km, so its magnitude is
\[ r = \sqrt{8000^2 + 9000^2} = \sqrt{145000000} \approx 12041.6 \text{ km} \]
The velocity vector is \(\vec{v} = -6\hat{p} + 6\hat{q}\) km/s, so its magnitude is
\[ v = \sqrt{(-6)^2 + 6^2} = \sqrt{72} \approx 8.485 \text{ km/s} \]

Step 2: Use the vis-viva energy equation to find the specific mechanical energy.
For any two-body orbit the specific mechanical energy is
\[ \varepsilon = \frac{v^2}{2} - \frac{\mu}{r} \]
This single number tells us the shape of the orbit: negative energy gives an ellipse (or circle), zero energy gives a parabola, and positive energy gives a hyperbola.

Step 3: Substitute the numbers.
\[ \varepsilon = \frac{72}{2} - \frac{398600}{12041.6} = 36 - 33.10 = 2.90 \text{ km}^2/\text{s}^2 \]
The energy comes out positive.

Step 4: Check that the orbit is not degenerate (rule out the straight line option).
The specific angular momentum is
\[ h = r_p v_q - r_q v_p = (8000)(6) - (9000)(-6) = 48000 + 54000 = 102000 \text{ km}^2/\text{s} \]
Since \(h \neq 0\), the position and velocity vectors are not parallel, so the satellite is not falling on a straight radial line. This rules out option (D).

Step 5: Interpret the sign of the energy.
Positive specific energy means the satellite has more kinetic energy than it needs to stay bound to the earth. Its semi-major axis works out negative, \(a = -\mu/(2\varepsilon) \approx -68765\) km, which is the signature of a hyperbolic orbit. A circle needs \(\varepsilon < 0\) with zero eccentricity, and a parabola needs \(\varepsilon = 0\) exactly, so neither fits here.

Final Answer:
The satellite follows a hyperbolic trajectory. \[ \boxed{\text{Hyperbola}} \]