This discussion focuses on calculating points in an elliptical orbit where the speed matches the local circular orbital speed and determining the corresponding flight path angle. The velocities are equal at the periapsis and apoapsis, with the flight path angle defined by the equation tan(γ) = (e sin(θ)) / (1 + e cos(θ)). At periapsis, the angle is 0, and at apoapsis, it is π. The conclusion confirms that these angles are indeed the solutions for the flight path angle at these specific points in the orbit.
Aerospace engineers, astrophysicists, and students studying orbital mechanics will benefit from this discussion, particularly those interested in the dynamics of elliptical orbits and flight path calculations.
dwsmith said:Determine the location of the point(s) on an elliptical orbit at which the speed is equal to the (local) circular orbital speed. Determine the flight path angle at this location.
What equation(s) should I be using or thinking about for this?
Sudharaka said:
dwsmith said:So the velocities are the same on the semi-major axis. That is, on the periapsis and apoapsis.
The flight path angle is giving by
$$
\tan\gamma = \frac{e\sin\theta}{1 + e\cos\theta}
$$
At periapsis, the angle is 0, and at apoapsis, the angle is pi.
So $\gamma = 0,\pi$? Is this really the solution?