Exercises — Kepler's first law — orbits are conic sections

WHAT we check: where $e=0.6$ falls on the eccentricity ruler.

• $e=0$ → circle
• $0<e<1$ → ellipse (bound)
• $e=1$ → parabola (marginally escapes)
• $e>1$ → hyperbola (unbound)

Since $0<0.6<1$, this is an ellipse, and it is bound — the body returns forever.

$r$ is measured from the focus, which sits at the origin of our polar coordinates. That is exactly why the Sun (at the focus) — not the empty geometric center — is the reference point. Setting $r$ from the center would give a different, more complicated equation with no $1/(1+e\cos \theta )$ form.

Fastest at perihelion ($\theta =0$, smallest $r$); slowest at aphelion ($\theta =\pi$, largest $r$). By Kepler's second law, the body sweeps equal areas in equal times, so when it is close (small $r$) it must move fast to sweep the same area.

Step 1 — pick the cleanest point. Perihelion has $\theta =0$, so $\cos \theta =1$ and the equation collapses to $r_{\min }=\frac{p}{1+e}.$ Why here? We want to solve for $e$, but the equation carries two unknowns, $e$ and the angle $\theta$. Perihelion is the one place where $\cos \theta$ takes a known value ($+1$), erasing the angle and leaving $e$ as the only unknown — a single equation in a single variable, which we can always invert.

Step 2 — solve for $e$. $1+e=\frac{p}{r_{\min }}=\frac{8000}{6000}=1.3333,e=\mathrm{0.3333.}$ The orbit is an ellipse ($0<e<1$). ✓

For an ellipse the semi-major axis relates to the extremes by $r_{\min }=a(1-e)$ and $r_{\max }=a(1+e)$.

Why these formulas? The perihelion and aphelion lie on opposite ends of the major axis, at distances $a(1-e)$ and $a(1+e)$ from the focus; their average is $a$ (the center-to-vertex distance) and their difference is $2ae$ (twice the focus offset).

Perihelion: $r_{\min }=1.496\times 10^{11}(1-0.0167)=1.4710\times 10^{11}$ m. Aphelion: $r_{\max }=1.496\times 10^{11}(1+0.0167)=1.5210\times 10^{11}$ m.

The difference is only about $1.7\%$ of $a$ — Earth's orbit is nearly circular.

At $\theta =90^{\circ }$: $\cos 90^{\circ }=0$, so $r=\frac{p}{1+e\cdot 0}=p=8000\text{ km}.$ Why this angle is special: at $90^{\circ }$ the $\cos$ term switches off entirely, so $r$ equals $p$ regardless of eccentricity. This is why $p$ is called the semi-latus rectum — it is literally the orbital radius when you stand $90^{\circ }$ from perihelion, straight "up" from the focus.

At $\theta =180^{\circ }$ (aphelion): $\cos 180^{\circ }=-1$, so $r_{\max }=\frac{p}{1-e}=\frac{8000}{1-0.333}=\frac{8000}{0.667}=11994\text{ km}\approx 1.199\times 10^4\text{ km}.$ Why $1-e$ here: $\cos 180^{\circ }=-1$ makes the denominator $1+e(-1)=1-e$, its smallest possible value, so $r$ is at its largest — exactly the aphelion, as expected.

(a) By definition $p=h^2/\mu$. Why this relation? It comes straight out of the orbit derivation (parent note, Step 6): the constant driving term in the linearized equation $u^{\prime \prime }+u=\mu /h^2$ sets the size scale, and inverting $u=1/r$ makes $p=h^2/\mu$ the natural length. So knowing $h$ and $\mu$ gives $p$ with no extra data. $p=\frac{(4.0\times 10^{15}{)}^2}{1.327\times 10^{20}}=\frac{1.6\times 10^{31}}{1.327\times 10^{20}}=1.2057\times 10^{11}\text{m}.$

(b) At perihelion, $r_{\min }=p/(1+e)$. Why perihelion again? Same reason as Exercise 2.1: it is the only point where $\cos \theta$ is known ($+1$), so it turns the two-unknown orbit equation into one solvable equation for $e$. $1+e=\frac{p}{r_{\min }}=\frac{1.2057\times 10^{11}}{8.0\times 10^{10}}=1.5071,e=\mathrm{0.5071.}$ An ellipse — this comet is bound and periodic (a "returning" comet).

Step 1 — compute the correction term. $\frac{2\mathcal{E}{h}^2}{{\mu }^2}=\frac{2(5.0\times 10^6)(3.0\times 10^{15}{)}^2}{(1.327\times 10^{20}{)}^2}.$ Numerator: $2\times 5.0\times 10^6\times 9.0\times 10^{30}=9.0\times 10^{37}$. Denominator: $(1.327\times 10^{20}{)}^2=1.7609\times 10^{40}$. Ratio: $9.0\times 10^{37}/1.7609\times 10^{40}=5.111\times 10^{-3}$.

Step 2 — take the root. $e=\sqrt{1+5.111\times 10^{-3}}=\sqrt{1.005111}=\mathrm{1.00255.}$

Why the sign of $\mathcal{E}$ matters: from the derivation above, $e^2-1=2\mathcal{E}{h}^2/{\mu }^2$, so positive energy forces $e^2>1$, i.e. $e>1$. This is a hyperbola — an unbound flyby that escapes with leftover speed. Barely, but unmistakably.

Step 1 — relate $p$ and $a$. For an ellipse, $p=a(1-e^2)$. Why this relation, seen on the picture? Look at Solution figure s01: the major axis stretches from perihelion to aphelion, a total length $r_{\min }+r_{\max }=2a$. Add the two extreme distances: $r_{\min }+r_{\max }=\frac{p}{1+e}+\frac{p}{1-e}=p\frac{(1-e)+(1+e)}{(1+e)(1-e)}=\frac{2p}{1-e^2}=2a.$ So $a=\frac{p}{1-e^2}$, equivalently $p=a(1-e^2)$. It is just "the two vertices average to $a$," turned into algebra — nothing mysterious. $a=\frac{p}{1-e^2}=\frac{1.2057\times 10^{11}}{1-0.5071^2}=\frac{1.2057\times 10^{11}}{1-0.2571}=\frac{1.2057\times 10^{11}}{0.7429}=1.6229\times 10^{11}\text{m}.$

Step 2 — aphelion. $r_{\max }=a(1+e)=1.6229\times 10^{11}(1.5071)=2.4459\times 10^{11}\text{m}.$ Check: $r_{\max }$ should also equal $p/(1-e)=1.2057\times 10^{11}/0.4929=2.446\times 10^{11}$ m. ✓

(a) At perihelion the velocity is purely tangential (the radial rate $\dot{r}=0$ there, since $r$ is at a minimum). So $h=r_{\min }{v}_{\min }$: $h=(7.0\times 10^6)(9.0\times 10^3)=6.30\times 10^{10}{\text{m}}^2/\text{s}.$ Why $\dot{r}=0$? At the closest point, $r$ momentarily stops shrinking and starts growing — a minimum — so its time-derivative is zero, leaving only the sideways (tangential) motion.

(b) $p=h^2/\mu =(6.30\times 10^{10}{)}^2/3.986\times 10^{14}=3.969\times 10^{21}/3.986\times 10^{14}=9.9573\times 10^6\text{m}.$

(c) At perihelion $r_{\min }=p/(1+e)$: $1+e=\frac{p}{r_{\min }}=\frac{9.9573\times 10^6}{7.0\times 10^6}=1.4225,e=\mathrm{0.4225.}$

(d) $0<e<1$ → ellipse, a bound satellite orbit. ✓

Step 1 — energy at perihelion. $\frac{1}{2}{v}_{\min }^2=\frac{1}{2}(9.0\times 10^3{)}^2=4.05\times 10^7\text{J/kg}.$ $\frac{\mu }{r_{\min }}=\frac{3.986\times 10^{14}}{7.0\times 10^6}=5.694\times 10^7\text{J/kg}.$ $\mathcal{E}=4.05\times 10^7-5.694\times 10^7=-1.644\times 10^7\text{J/kg}.$ Negative → bound → ellipse. Consistent. ✓

Step 2 — recover $e$. $\frac{2\mathcal{E}{h}^2}{{\mu }^2}=\frac{2(-1.644\times 10^7)(6.30\times 10^{10}{)}^2}{(3.986\times 10^{14}{)}^2}.$ Numerator: $2\times (-1.644\times 10^7)\times 3.969\times 10^{21}=-1.3050\times 10^{29}$. Denominator: $(3.986\times 10^{14}{)}^2=1.5888\times 10^{29}$. Ratio: $-0.8214$. $e=\sqrt{1-0.8214}=\sqrt{0.1786}=\mathrm{0.4226.}$ Matches part (c)'s $e=0.4225$ (rounding). Two independent routes, same answer — a strong check. See Vis-viva equation for the general speed–radius relation behind this.

Derivation. $r_{\min }=\frac{p}{1+e},r_{\max }=\frac{p}{1-e}.$ Multiply: $r_{\min }{r}_{\max }=\frac{p}{1+e}\cdot \frac{p}{1-e}=\frac{p^2}{(1+e)(1-e)}=\frac{p^2}{1-e^2}.$ Now substitute $p=a(1-e^2)$ once: $\frac{p^2}{1-e^2}=\frac{p\cdot a(1-e^2)}{1-e^2}=ap$. And since $p=a(1-e^2)$, that also equals $a\cdot a(1-e^2)=a^2(1-e^2)$. All three forms agree. ∎

Numeric check. Using $r_{\min }=p/(1+e)=1.2057\times 10^{11}/1.5071=8.000\times 10^{10}$ m and $r_{\max }=2.4459\times 10^{11}$ m: $r_{\min }{r}_{\max }=(8.000\times 10^{10})(2.4459\times 10^{11})=1.9567\times 10^{22}{\text{m}}^2.$ Compare $ap=(1.6229\times 10^{11})(1.2057\times 10^{11})=1.9567\times 10^{22}{\text{m}}^2$. ✓

(a) $p=r_{\min }(1+e)=3.0\times 10^{10}(1+1.20)=3.0\times 10^{10}\times 2.20=6.60\times 10^{10}$ m.

(b) $r\to \infty$ means the denominator $1+e\cos \theta \to 0$, i.e. $\cos {\theta }_{\infty }=-\frac{1}{e}=-\frac{1}{1.20}=-\mathrm{0.8333.}$ The principal value is ${\theta }_{\infty }=\arccos (-0.8333)=146.44^{\circ }=2.5559\text{rad}.$ Why $\arccos$? We know the cosine of the escape angle and want the angle itself — $\arccos$ is exactly the "which angle has this cosine?" undo-operation, and it returns the single value in $[0^{\circ },180^{\circ }]$. But cosine is even: $\cos (-\theta )=\cos \theta$, so $\theta =-146.44^{\circ }$ satisfies the same equation. A hyperbola therefore has two asymptote angles, ${\theta }_{\infty }=\pm 146.44^{\circ }$ — the incoming and outgoing directions the object travels along. Physically it sweeps only the wedge $-146.44^{\circ }<\theta <+146.44^{\circ }$; beyond it the denominator $1+e\cos \theta$ would be negative (forbidden, as flagged in the edge-case callout at the top).

(c) Geometric reason (see figure s02). For an ellipse $e<1$, so $-1/e<-1$. But $\cos \theta$ can never dip below $-1$ — it lives in $[-1,+1]$. So the equation $\cos \theta =-1/e$ has no solution: the denominator $1+e\cos \theta$ stays $\ge 1-e>0$ for every angle, $r$ is always finite, and the curve loops back on itself — a closed orbit with no escape direction. For a hyperbola $e>1$, so $-1/e$ lies inside $[-1,0)$, an achievable cosine. At that angle the denominator collapses to zero and $r$ blows up; the orbit runs off to infinity along two straight-line asymptotes at $\pm {\theta }_{\infty }$. In short: whether infinity is reachable at a real angle is decided entirely by whether $1/e\le 1$, i.e. by whether $e\ge 1$.

Cleanest inversion point for finding $e$ ::: Perihelion, where $\cos \theta =1$ gives $r_{\min }=p/(1+e)$. $h$ in terms of perihelion data ::: $h=r_{\min }{v}_{\min }$ (velocity is purely tangential there). Escape (asymptote) angles of a hyperbola ::: $\cos {\theta }_{\infty }=-1/e$, giving $\theta =\pm {\theta }_{\infty }$; exists only when $e\ge 1$. Sign of energy for a bound ellipse ::: $\mathcal{E}<0$. Product of perihelion and aphelion distances ::: $r_{\min }{r}_{\max }=ap=a^2(1-e^2)$. Where does $e=\sqrt{1+2\mathcal{E}{h}^2/{\mu }^2}$ come from ::: Evaluate $\mathcal{E}=\frac{1}{2}{v}^2-\mu /r$ at perihelion; it rearranges to $e^2-1=2\mathcal{E}{h}^2/{\mu }^2$.