MHB 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. :)