Question

An unpowered space vehicle at altitude \(600\ \text{km}\) (above mean sea level) is moving at speed \(9\ \text{km s}^{-1}\) relative to an Earth-centered frame. Take Earth's radius \(R_E=6400\ \text{km}\) and \(GM_E=3.98\times10^{14}\ \text{m}^3\text{s}^{-2}\). The trajectory is:

• A. Circular
• B. Elliptic
• C. Parabolic
• D. Hyperbolic

Hint:

Memorize the hierarchy at radius \(r\):

• \(V < V_c \;\Rightarrow\;\) elliptic (apogee side)
• \(V = V_c \;\Rightarrow\;\) circular
• \(V_c < V < V_{\mathrm{esc}} \;\Rightarrow\;\) elliptic (perigee side)
• \(V = V_{\mathrm{esc}} \;\Rightarrow\;\) parabolic
• \(V < V_{\mathrm{esc}} \;\Rightarrow\;\) hyperbolic

Answer 1

The Correct Option is B

Step 1: Orbit radius at the given point.
Altitude \(=600\ \text{km}\Rightarrow r = R_E+600 = 6400+600 = 7000\ \text{km} = 7.0\times10^{6}\ \text{m}.\) 

Step 2: Circular and escape speeds at \(r\).
Circular speed: \(V_c=\sqrt{\mu/r}\), \(\mu=GM_E=3.98\times10^{14}\). \[ V_c=\sqrt{\frac{3.98\times10^{14}}{7.0\times10^{6}}} =\sqrt{5.6857\times10^{7}}\approx 7.54\times10^{3}\ \text{m/s}=7.54\ \text{km/s}. \] Escape speed: \(V_{\mathrm{esc}}=\sqrt{2\mu/r}=\sqrt{2}\,V_c\approx 10.67\ \text{km/s}.\) 

Step 3: Classify using speed range (or energy sign).
Given \(V=9\ \text{km/s}\) satisfies \(V_c(7.54) < V(9) < V_{\mathrm{esc}}(10.67)\).
Equivalently, specific mechanical energy \(\varepsilon=\tfrac{V^2}{2}-\tfrac{\mu}{r}\) is negative (bound) but not the special circular case.
Hence the conic is a bounded ellipse (non-circular). \[ \boxed{\text{Elliptic trajectory}} \]