Exercises — Orbit shape from eccentricity — circle (e=0), ellipse (0 - e - 1), parabola (e=1), hyperbola (e - 1)

Read straight off Fact 3.

• (a) $e=0$: circle, bound.
• (b) $0<e<1$: ellipse, bound.
• (c) $e=1$: parabola, marginal escape (bound⇄unbound boundary).
• (d) $e>1$: hyperbola, unbound. What this looks like: see the family of curves in the figure below — as $e$ climbs the curve "opens up" from a closed loop to an open arm.

WHY a ratio: $e=c/a$ (Fact 4) is dimensionless — units cancel, so absolute size is irrelevant. Note "foci apart" is the full separation $2c$, so $c$ is half of it.

• Tiny: $c=1\text{ cm}$, $a=5\text{ cm}$, $e=1/5=0.2$.
• Giant: $c=100,000\text{ km}$, $a=500,000\text{ km}$, $e=100000/500000=0.2$. They are equally eccentric. Same shape, different scale.

WHAT / WHY: Fact 4 gives the two turning points directly. $r_{\min }=a(1-e)=4(1-0.75)=4(0.25)=1\text{ AU}$ $r_{\max }=a(1+e)=4(1+0.75)=4(1.75)=7\text{ AU}$ Check: their average $\frac{1+7}{2}=4=a$. ✓ (the semi-major axis is always the mean of the two extremes — that is what "semi-major axis" is.)

WHY $p=h^2/\mu$: $p$ is the value of $r$ at $\theta =90^{\circ }$, where $\cos \theta =0$ so $r=p$. It is fixed purely by angular momentum (Fact 1), independent of energy. $p=\frac{h^2}{\mu }=\frac{(7.0\times 10^{10}{)}^2}{3.986\times 10^{14}}=\frac{4.9\times 10^{21}}{3.986\times 10^{14}}\approx 1.229\times 10^7\text{m}$ About $12,290\text{ km}$ — comfortably above Earth's surface, so this is a real orbit.

Use Fact 1 with $e=1$: $r=\frac{p}{1+\cos \theta }$ (recall $\theta =0$ is perihelion).

• At $\theta =90^{\circ }$: $\cos 90^{\circ }=0$, so $r=\frac{p}{1+0}=p$.
• At $\theta =120^{\circ }$: $\cos 120^{\circ }=-\frac{1}{2}$, so $r=\frac{p}{1-\frac{1}{2}}=\frac{p}{1/2}=2p$. What it looks like: the comet is already twice as far at $120^{\circ }$ as at $90^{\circ }$, and heading toward infinity as $\theta \to 180^{\circ }$ (where $1+\cos \theta \to 0$).

WHY Fact 2: eccentricity is fixed by the combination of energy and angular momentum — this formula packages both. $\frac{2\epsilon h^2}{{\mu }^2}=\frac{2(-2.0\times 10^6)(6.0\times 10^{10}{)}^2}{(4.0\times 10^{14}{)}^2}$ Numerator: $2\times (-2.0\times 10^6)\times 3.6\times 10^{21}=-1.44\times 10^{28}$. Denominator: $1.6\times 10^{29}$. Ratio $=-0.09$. Then $e=\sqrt{1+(-0.09)}=\sqrt{0.91}\approx \mathrm{0.954.}$ Since $0<e<1$, it is an ellipse (bound) — consistent with $\epsilon <0$. Highly eccentric but still closed.

Algebra: put $\epsilon =0$ into Fact 2: $e=\sqrt{1+\frac{2(0)h^2}{{\mu }^2}}=\sqrt{1+0}=1.$ The $h$ term is multiplied by $\epsilon =0$, so it vanishes for every value of $h$. Physics — WHY $h$ is irrelevant here: $\epsilon =0$ means the body reaches infinity with exactly zero speed (all kinetic energy spent climbing out of the well). That "barely escapes" condition is a statement about energy alone; how much sideways swing ($h$) it has only changes the width of the parabola (its $p=h^2/\mu$), not the fact that it's a parabola. Every zero-energy orbit is a parabola. See Escape Velocity.

WHY Fact 5: the object is on the orbit only where $r$ is positive and finite; $r$ blows up where the denominator hits zero. (a) $\cos \theta =-1/e=-1/2.5=-0.4$. So $\theta =\arccos (-0.4)\approx 113.6^{\circ }$ (and by symmetry $-113.6^{\circ }$). (b) The physical arc is $-113.6^{\circ }<\theta <113.6^{\circ }$. Outside this, $1+e\cos \theta$ turns negative → $r$ negative → unphysical. Those two limiting angles are the asymptotes: the straight-line directions the object comes in on and leaves on. What it looks like: the deflected flyby path in the figure — a single open swing, never a loop.

WHY start with $a$: it's the average of the extremes, so it inverts Fact 4 cleanly. (a) Add the two relations of Fact 4: $r_{\min }+r_{\max }=a(1-e)+a(1+e)=a[(1-e)+(1+e)]=2a$. The $\pm e$ cancels, so $a=\frac{r_{\min }+r_{\max }}{2}=\frac{7000+21000}{2}=14000\text{ km}.$ (b) Subtract them: $r_{\max }-r_{\min }=a(1+e)-a(1-e)=a[(1+e)-(1-e)]=2ae$. The constant parts cancel, leaving only the $e$ term: $e=\frac{r_{\max }-r_{\min }}{2a}=\frac{r_{\max }-r_{\min }}{r_{\max }+r_{\min }}=\frac{21000-7000}{28000}=\frac{14000}{28000}=\mathrm{0.5.}$ (c) $p=a(1-e^2)=14000(1-0.25)=14000(0.75)=10500\text{ km}$. (d) $0<e<1$ → ellipse, bound. ✓ Cross-check: $r_{\min }=a(1-e)=14000(0.5)=7000$ ✓ and $r_{\max }=a(1+e)=14000(1.5)=21000$ ✓.

WHY link $r_p$ to $p$: for a parabola ($e=1$), Fact 1 at $\theta =0$ (perihelion) gives $r_p=\frac{p}{1+1}=\frac{p}{2}$, so $p=2r_p$. Then $p=h^2/\mu \Rightarrow h=\sqrt{\mu p}=\sqrt{\mu \cdot 2r_p}$. $h=\sqrt{(3.0\times 10^{13})(2)(5.0\times 10^6)}=\sqrt{3.0\times 10^{20}}\approx 1.732\times 10^{10}{\text{m}}^2/\text{s}.$ Sanity: larger required perihelion ⇒ larger $h$ ⇒ more sideways swing, which matches the geometry.

Part 1 — sign of $\epsilon$: The quantity under the root is $1+\frac{2\epsilon h^2}{{\mu }^2}$. Since $h^2/{\mu }^2>0$:

• $\epsilon <0$ ⇒ added term negative ⇒ root $<1$ ⇒ $e<1$ (bound: circle/ellipse).
• $\epsilon =0$ ⇒ root $=\sqrt{1}=1$ ⇒ $e=1$ (parabola, the boundary).
• $\epsilon >0$ ⇒ root $>1$ ⇒ $e>1$ (hyperbola, unbound). The three cases exhaust all real $\epsilon$, so the classification is complete and gap-free.

Part 2 — the circle condition: $e=0$ requires the whole square root to be $0$: $1+\frac{2\epsilon h^2}{{\mu }^2}=0\Rightarrow \epsilon =-\frac{{\mu }^2}{2h^2}.$ This is the most negative $\epsilon$ compatible with a given $h$: for fixed angular momentum, the deepest (most bound) energy gives the roundest orbit. Any less-negative $\epsilon$ stretches it into an ellipse. What it looks like: the innermost, perfectly round loop in the L1 figure — the ground state of shape.

(a) $e=0.5<1$ → A is an ellipse (bound); $e=2>1$ → B is a hyperbola (unbound). (b) Fact 1 with $\cos 60^{\circ }=0.5$ (angles measured from perihelion, $\theta =0$):

• A: $r_A=\frac{p}{1+0.5(0.5)}=\frac{p}{1.25}=0.8p$.
• B: $r_B=\frac{p}{1+2(0.5)}=\frac{p}{2}=0.5p$. (Interesting: at $60^{\circ }$ the hyperbola is actually closer — its perihelion is tighter because a bigger $e$ with the same $p$ means a smaller $r_p=p/(1+e)$.) (c) As $\theta$ grows past $90^{\circ }$, $\cos \theta$ turns negative. For B the term $e\cos \theta =2\cos \theta$ swings twice as hard, so the denominator races toward zero and hits it at $\cos \theta =-0.5$ ($\theta =120^{\circ }$) — $r$ diverges. For A, $0.5\cos \theta$ can never drag the denominator to zero (its most negative value is $1-0.5=0.5>0$), so $r$ stays finite forever. This is exactly why $e\ge 1$ escapes and $e<1$ does not — the whole bound/unbound distinction lives in that one denominator.