Question bank — Kepler's first law — orbits are conic sections

False. An ellipse has two foci and a separate center; the Sun occupies one focus. The other focus and the center are both empty points in space.

True. The focus–center distance is $ae$, so the focus–focus separation is $2ae\to 0$ as $e\to 0$; both foci and the center coincide — that limiting shape is the circle, which is why the "Sun-at-center" illusion holds for small $e$.

True. Setting $e=0$ in $r=p/(1+e\cos \theta )$ gives $r=p$, a constant — a circle. The circle is a genuine special case of the conic family, not a separate law.

True. The single derivation ($u^{\prime \prime }+u=\mu /h^2$) produces all of them; which one you get is fixed by the initial energy, giving $e=\sqrt{1+2\mathcal{E}{h}^2/{\mu }^2}$.

False. Hyperbolic ($e>1$) trajectories are real unbound flybys; the interstellar object ʻOumuamua had $e\approx 1.2$. "Bound vs unbound" is set by energy sign, not by any cap on $e$.

False. Only the circle ($e=0$) has constant speed. For $e>0$ the body is fastest at perihelion and slowest at aphelion — this is Kepler's second law in action.

True. At $\theta =\pm 90^{\circ }$, $\cos \theta =0$, so $r=p$ exactly. That is the geometric meaning of $p$: the focus-to-curve distance measured perpendicular to the major axis.

True. From $e=\sqrt{1+2\mathcal{E}{h}^2/{\mu }^2}$ with $\mathcal{E}<0$, the term $2\mathcal{E}{h}^2/{\mu }^2$ is negative and grows more negative as $h$ increases, so $e$ decreases toward $0$ — the orbit becomes rounder (a circle is the maximum-$h$ orbit for given energy).

Error in the middle: $\cos \pi =-1$ makes the denominator smallest ($1-e$), which makes $r$ largest. The conclusion (aphelion at $\theta =\pi$) is right, but the reason is inverted — small denominator, big $r$.

Error: $r$ is measured from the focus (the origin of the polar frame), because that origin is where the attracting mass sits. Measuring from the center would not give this clean single-focus equation.

Error: $e=1$ is not bound. It is the exact threshold $\mathcal{E}=0$ — the object reaches infinity with zero leftover speed and never returns. Bound orbits require $e<1$ strictly.

Error, backwards: the planet moves fastest near perihelion (equal areas in equal times), so it spends less time there. It lingers near aphelion where it crawls.

Error: conservation has nothing to do with strength. $h$ is conserved because gravity is central (points along $\hat{r}$), giving zero torque about the focus regardless of how strong it is.

Error: $u=1/r$ is a general polar-coordinate change of variables (with $u^{\prime \prime }$ meaning $d^{2}u/d{\theta }^2$); it converts any central-force radial equation into a form in $u$. It's especially clean for $1/r^2$ (giving linear driven SHM), but it's a tool, not a coincidence of gravity.

Error: $\mathcal{E}>0$ gives $e>1$, a hyperbola — an open, unbound curve. No positive-energy orbit closes into any ellipse, no matter how stretched.

Because the derivation forces it: solving $u^{\prime \prime }+u=\mu /h^2$ gives $u=\mu /h^2+A\cos \theta$, i.e. $r=p/(1+e\cos \theta )$ — a form whose polar origin (where $r$ is measured from) is the attracting mass. That origin is a focus, not the center, and it sits off-center by $ae$ on the perihelion ($\theta =0$) side because the $\cos \theta$ term makes $r$ smallest there. The asymmetry is built into the equation, not just a picture.

Its general solution is $u=\mu /h^2+A\cos \theta$ (with $u^{\prime \prime }=d^{2}u/d{\theta }^2$), and inverting $r=1/u$ gives exactly $r=p/(1+e\cos \theta )$ — the polar form of a conic. No other function type can come out of driven simple harmonic motion.

Only $1/r^2$ (and the linear spring $\propto r$) yields orbits that close after one revolution without precessing — this is Bertrand's theorem; other powers give rosette paths that never quite repeat. (Inverse-square law and Bertrand's theorem)

Because $e$ and $\mathcal{E}$ carry the same information via $e=\sqrt{1+2\mathcal{E}{h}^2/{\mu }^2}$: $e<1\iff \mathcal{E}<0$ (bound), $e=1\iff \mathcal{E}=0$, $e>1\iff \mathcal{E}>0$. Knowing $e$ already fixes the sign of energy.

We want the shape $r(\theta )$, not the timetable $r(t)$. Angular momentum $h=r^2\dot{\theta }$ lets us eliminate time cleanly, delivering the geometry of the orbit directly. (Conservation of angular momentum in central forces)

Negative energy means the body is trapped in the gravitational potential well — it lacks the energy to reach infinity, so its distance $r$ stays finite and the orbit closes. (Escape velocity and orbital energy)

A circle of radius $r=p$; perihelion and aphelion coincide and the speed is constant everywhere. It's the degenerate ellipse with both foci at the center.

$r\to \infty$: the body runs off to infinity along an asymptote. For $e>1$ there are two such angles (the hyperbola's asymptotic directions); for $e=1$ there is one, at $\theta =\pi$.

No. Since $e<1$, the smallest the denominator gets is $1-e>0$ (at $\theta =\pi$), so $r$ stays finite and bounded — the orbit never escapes.

There is no aphelion. As $\theta \to \pi$ the denominator $1+\cos \theta \to 0$ and $r\to \infty$; the object recedes forever, reaching infinity with exactly zero speed.

The formula degenerates: $p=h^2/\mu =0$, so the "orbit" collapses to a radial plunge straight into the center. With no sideways motion there is no conic — just free-fall along a line.

Marginally unbound. It's the exact dividing line ($\mathcal{E}=0$): the body escapes but arrives at infinity with zero remaining speed — the slowest possible escape.

No. They share the same latus-rectum width ($r=p$ at $\theta =\pm 90^{\circ }$), but different $e$ gives very different perihelion $p/(1+e)$ and aphelion $p/(1-e)$ — hence different overall extent and shape.