What information can I get if an elliptical orbit intersecting a circular orbit of radius 𝑟 and velocity of ellipse is equal to velocity of circular?

Question

How can I approach solving the question below?

Consider an elliptical orbit intersecting a circular orbit of radius $r$ as shown below. If $v_{ellipse} = v_{circular} = v $, find:

1. True anomaly ($\theta$) at $r$ ONLY in terms of the elliptical orbit eccentricity ($\epsilon$).
2. Flight path angle ($\gamma$) at $r$ ONLY in terms of the elliptical orbit eccentricity ($\epsilon$).
3. The $\Delta v$ required to transfer from the circular orbit to the elliptical orbit at $r$ ONLY in terms of the elliptical orbit eccentricity ($\epsilon$) and radius $r$.

Answer 1

All right, I'd approach this in the following manner.

Part 1: True Anomaly $\theta$ at $r$ in terms of Elliptical Orbit Eccentricity $e$

The first place I'd start is the Vis-Viva Equation, which, for all orbits around a particular body with a particular gravitational parameter $\mu$ links relative speed $v$ with radial distance $r$ and semimajor-axis $a$.

$$v^2 = \mu\left ( \frac{2}{r}-\frac{1}{a} \right )$$

From this we know that if $v_{ellipse}$ = $v_{circular}$ = $v$ at $r$, then $r$ = $a$ for both the ellipse and the circle.

Your provided equations for True Anomaly $\theta$ for elliptical orbits goes to the following once we substitute in $r$ for $a$. We'll just grab the one in the first or second quadrant.

$$\theta = \cos^{-1} \left ( \frac{r(1-e^2)}{re} - \frac{1}{e}\right )=\cos^{-1} \left ( \frac{(1-e^2)}{e} - \frac{1}{e}\right )$$

$$= \cos^{-1} \left ( -\frac{e^2}{e} \right ) = \cos^{-1} (-e)$$

Part 2: Flight Path Angle $\gamma$ at $r$ in terms of the Elliptical Orbit Eccentricity $e$

Flight path angle, $\gamma$ (at the link it's "Angle of velocity relative to the perpendicular to the radial direction):

$$\gamma = \tan^{-1}\frac{e \sin \theta}{1 + e \cos\theta}$$

And using trigonometric identities, I can get to (though I'm being a bit sloppy with my $\pm$):

$$\gamma = \tan^{-1}\frac{e \sqrt{1-\cos^2 \theta}}{1 + e \cos\theta} = \tan^{-1}\frac{e \sqrt{1-e^2}}{1 - e^2} = \tan^{-1}\frac{e }{\sqrt{1-e^2}}$$

And thanks to HopDavid for reminding me of inverse trigonometric identities, because that means for the flight path angle, $$\gamma = \sin^{-1}e$$ Part 3: $\Delta v$ required to go from circular to elliptical orbit at $r$

And, as mentioned by David Hammen in the comments, you can't express the delta-v solely in terms of $e$ and $r$, because the delta-V is absolutely going to be a function of $v$, and that's going to be dependent on what you're orbiting, and how far you are from it.

If you have the Flight path angle, the speed, you can then call on the Law of Cosines to acquire the delta-V, as we're just looking at vector addition with two legs of magnitude $v$ separated by an angle $\gamma$

$$\Delta v^2 = 2v^2 - 2v^2 \cos\gamma = 2v^2 - 2v^2 \cos\left({\sin^{-1}e}\right) $$ $$ = 2v^2 \left(1 - \cos\left(\sin^{-1}e\right)\right) $$

Inverse trigonometric identity to the rescue again, and: $$ \Delta v^2 = 2v^2(1 - \sqrt{1-e^2}) $$

Thanks again to HopDavid for pointing out my missing square on the Law of Cosines. That brings us to $$ \Delta v = \sqrt{2v^2(1 - \sqrt{1-e^2})} $$ At the very least, if we are allowed to bring in the gravitational parameter $\mu$, we can go to the Circular Orbit Velocity equation: $$v = \sqrt{\frac{\mu}{r}}$$

And wind up with the following.

$$\Delta v = \sqrt{2\frac{\mu}{r} \left(1 - \sqrt{1-e^2}\right)}$$

but I'm uncertain if that's allowed by the letter of the question. This agrees with HopDavid's answer and I thank him for his assistance.

Answer 2

In his excellent answer Notovy used the vis viva equation to demonstrate same speeds implies r = a.

A basic property points on an ellipse: sum of distance from one focus plus distance to the other focus is 2a.

So if the radius vector has the same length as semi major axis a, that implies the end of the radius vector lies on the end of the ellipse's semi-minor axis.

Without loss of generality we can choose our units so the radius of the circle as well as the semi major axis a is one unit.

Using these units the distance from ellipse center to the focus at the center of the system is e, the eccentricity of the ellipse.

Using basic definitions of trigonometry based on the unit circle we can see cos(θ) = -e. So:

θ = acos(-e)

Now let's put in the velocity vectors:

A tangent to an ellipse at the semi minor axis is parallel to the major axis. We construct a parallelogram using radius and velocity vector as sides. Opposite corners of the parallelogram are congruent angles.

A velocity vector in a circular orbit is at 90º to the radius vector. If a line cuts two parallel lines, opposite agles are congruent. So we know the velocity vecotr from the circular orbit also cross the parallelogram edge opposite the position vector at a right angle.

The three angles of a triangle sum to 180º. So we know that flight path angle λ is the same as angle cpf1.

Another basic way to define trig functions is using a right triangle with hypotenuse 1. We can see sin(λ) = e. So:

λ = asin(e)

Now for dv in terms of r and e. Like everyone says it's also necessary to use µ.

dV is the base of an isosceles triangle both sides having the velocity of the circular orbit, (µ/r)^(1/2). Like notovny I use the law of cosines to get dV:

Notovny refers to trigonometric identities to get cos(λ) = (1-e^2)^1/2. I use the Pythagorean Theorem to show that.

dV = (2(µ/r) (1 - (1-e^2 )^1/2 ))^1/2

My answer is pretty much the same as notovny's but I use a visual approach and high school geometry. So hopefully my explanation is more accessible to visual thinkers like myself.