Calculating Elliptical Orbit Points & Flight Path Angle

[Dustinsfl]

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]

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?

Hi dwsmith, :)

The velocity of an object in elliptical orbit is given >>here<< and that of a circular orbit is given >>here<<. So by equating two speeds you will be able to find values for \(r\). The equation for the flight path angle is given >>here<<.

Kind Regards,
Sudharaka.

 

[Dustinsfl]

Hi dwsmith, :)

The velocity of an object in elliptical orbit is given >>here<< and that of a circular orbit is given >>here<<. So by equating two speeds you will be able to find values for \(r\). The equation for the flight path angle is given >>here<<.

Kind Regards,
Sudharaka.

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?

 

[Sudharaka]

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?

Assuming you have done the algebra correctly, the answer is yes. I am not too confident about the method used since my knowledge about these kind of problems related to physics is quite limited. Hope some other member will be able to provide more insight on this problem. :)

 

[CellCoree]

To calculate the location of the point(s) on an elliptical orbit where the speed is equal to the circular orbital speed, you can use the vis-viva equation:

v = √(GM(2/r - 1/a))

where v is the speed, G is the gravitational constant, M is the mass of the central body, r is the distance from the central body to the orbiting object, and a is the semi-major axis of the elliptical orbit.

To determine the flight path angle at this location, you can use the following equation:

tan(γ) = (e sin(θ)) / (1 + e cos(θ))

where γ is the flight path angle, e is the eccentricity of the orbit, and θ is the true anomaly (the angle between the periapsis and the current position of the orbiting object).

It is important to note that these equations assume a two-body system and do not take into account any external forces or perturbations. For more accurate calculations, additional equations and factors may need to be considered.