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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 $r$.

enter image description here enter image description here

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    $\begingroup$ Well, by the Vis-Viva equation, if two orbits around the same body have the same velocity at the same radial distance, they must have the same semi-major axis. $\endgroup$ – notovny Nov 9 '19 at 22:48
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    $\begingroup$ Note that part (c) is unanswerable as asked. The $\Delta v$ required to make this transfer cannot be expressed in terms of eccentricity $e$ and radius $r$ only. It can however be expressed in terms of eccentricity $e$ and circular orbit velocity $v$ only. $\endgroup$ – David Hammen Nov 10 '19 at 17:11
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    $\begingroup$ Hello, I just added a formula sheet that includes most of what I learned. My idea for the (a) is to use mean anomaly, eccentric anomaly and Position as a func. of time equation to solve this problem. But after that, I got an equation which looks very complex and seems difficult to simplify to something like theta = ... Then, I thought it might be a wrong assumption. $\endgroup$ – Yibowen Zhao Nov 11 '19 at 9:07
  • $\begingroup$ Hi @YibowenZhao your edit looks great, thanks! $\endgroup$ – uhoh Nov 11 '19 at 9:56
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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.

enter image description here

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.

enter image description here

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:

enter image description here

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.

enter image description here

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:

enter image description here

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.

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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.

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    $\begingroup$ We agree that θ = acos(-e). And I agree that λ = atan(e/sqrt(1-e^2). However λ = asin(e) is much preferable in my opinion. The diagrams in my answer are based on high school geometry. Basic properties of an ellipse and very basic trig. $\endgroup$ – HopDavid Nov 13 '19 at 3:42
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    $\begingroup$ @HopDavid - Ah, the inverse trigonometric identities were ones that didn't get drilled into my head very solidly in high school and as a result,didn't fall out when I was writing my answer. Will update. $\endgroup$ – notovny Nov 13 '19 at 3:56
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    $\begingroup$ @notovy If you look at my triangle with hypotenuse 1 you will notice the unlabeled leg is sqrt(1-e^2). So by definition sin(λ) = e and cos(λ) = sqrt(1-e^2). So of course tan(λ) = e/sqrt(1-e^2) $\endgroup$ – HopDavid Nov 13 '19 at 4:13
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    $\begingroup$ When you start the law of cosines to get dV the term on the left side should be squared. I believe if not for that our answers would be the same. But I used a more visual approach. Check my answer, I'm not confident it's correct as there are several steps and opportunities for error. $\endgroup$ – HopDavid Nov 14 '19 at 19:03
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    $\begingroup$ @HopDavid Oops, you're right. Thanks again. $\endgroup$ – notovny Nov 14 '19 at 19:13

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