Question

F and G denote two points on a spacecraft’s orbit around a planet, as indicated in the figure. O is the center of the planet, P is the periapsis, and the angles are as indicated in the figure. If \( OF = 8000 \, {km} \), \( OG = 10000 \, {km} \), \( \theta_F = 0^\circ \), and \( \theta_G = 60^\circ \), the eccentricity of the spacecraft's orbit is \_\_\_\_\_\_\_\_ (rounded off to two decimal places).
\begin{figure} \centering \includegraphics[width=0.5\linewidth]{53.png} \end{figure}

Hint:

The eccentricity of an orbit is the ratio of the difference between the aphelion and perihelion distances to the sum of these distances. This gives an idea of how elongated the orbit is.

Answer 1

Polar Equation of an Elliptical Orbit \[ r = \frac{a(1 - e^2)}{1 + e \cos \theta} \] Step 1: At Periapsis (\( F \)) \[ 8000 = a(1 - e) \quad {(1)} \] Step 2: At Point \( G \) (\( \theta = 60^\circ \)) \[ 10000 = \frac{a(1 - e^2)}{1 + 0.5e} \quad {(2)} \] Step 3: Solve for \( e \)
From (1): \( a = \frac{8000}{1 - e} \). Substitute into (2): \[ 10000 = \frac{8000 (1 + e)}{1 + 0.5e} \] Simplify: \[ 5 = \frac{4(1 + e)}{1 + 0.5e} \implies e = \frac{2}{3} \approx 0.6667 \] Final Answer The eccentricity is \(\boxed{0.67}\).