Question bank — Physical meaning of each orbital element

False — with $e=0$ there is no periapsis (every point is equidistant), so the "direction the ellipse points" is undefined; $\omega$ becomes degenerate and is replaced by the argument of latitude $u$, the in-plane angle from the ascending node straight to the satellite itself.

False — if the orbit never leaves the reference plane there is no crossing point, so the node line vanishes and $\Omega$ is undefined; the longitude of periapsis $\varpi =\Omega +\omega$ (measured from the vernal equinox to periapsis) is used instead.

True — period $T=2\pi \sqrt{a^3/\mu }$ depends only on $a$, so shape is irrelevant to how long one lap takes even though the speeds differ wildly (see Kepler's Laws).

True — specific energy $\epsilon =-\mu /(2a)$ depends only on $a$; eccentricity redistributes energy between kinetic and potential but never changes the sum.

False — $90^{\circ }$ is the steepest (polar); past $90^{\circ }$ the orbit tilts back down but now runs retrograde, so $100^{\circ }$ is only $80^{\circ }$ from the plane again, just going the other way (see figure s03).

False — vis-viva gives $v^2=\mu (2/r-1/a)$, so at a fixed $r$ and fixed $a$ the speed is identical regardless of $e$; eccentricity changes which radii the satellite actually visits.

False — five are constant only in the ideal two-body Kepler orbit; real perturbations (oblateness, drag, third bodies) slowly change $\Omega$, $\omega$, and the others, per Orbital perturbations.

False — [[Eccentricity vector|$\vec{e}$]] points from the focus toward periapsis (closest approach); its magnitude equals $e$ and its direction defines the major axis (red arrow in figure s01).

Wrong — $a$ is half the major axis, a constant; the true distance $r$ swings between $r_p=a(1-e)$ and $r_a=a(1+e)$. Only when $e=0$ does $r=a$ everywhere.

Wrong — recall $\vec{n}$ is the node vector along the plane-intersection line (figure s02). Since $\cos (\Omega )=\cos (-\Omega )$, two possible angles share the same $n_x$; the sign of $n_y$ breaks the tie: $n_y\ge 0\Rightarrow \Omega =\arccos (n_x/|\vec{n}|)$, but $n_y<0\Rightarrow \Omega =360^{\circ }-\arccos (n_x/|\vec{n}|)$ (i.e. $\Omega >180^{\circ }$). Figure s04 walks this quadrant test.

Wrong — $\arccos$ collapses $\nu$ and $360^{\circ }-\nu$ together; you must check $\vec{r}\cdot \vec{v}$: outward ($>0$) means $0<\nu <180^{\circ }$, inward ($<0$) means $\nu >180^{\circ }$ (see the moving dot in figure s01).

Wrong — that is $\Omega$ (RAAN). Inclination $i$ tells you how steep the tilt is; a polar orbit is always $i=90^{\circ }$ but can face any $\Omega$.

Wrong — instantaneous speed does vary with $e$, but total specific energy $\epsilon =-\mu /(2a)$ is set purely by $a$; the $e$-driven speed changes are exactly cancelled by potential-energy changes.

Wrong — $\vec{h}=\vec{r}\times \vec{v}$ is perpendicular to the plane (both $\vec{r}$ and $\vec{v}$ lie in it), which is exactly why its direction encodes the plane's tilt via $\cos i=h_z/h$ (figure s02). See Angular momentum in orbits.

Wrong — $\epsilon =0$ means $a\to \infty$, which is a parabola (escape trajectory, $e=1$), not a circle; the satellite just barely reaches infinity with zero leftover speed.

Because the node line is the intersection of the orbital plane and the reference plane, and a vector perpendicular to both plane-normals ($\hat{K}$, the reference-plane normal, and $\vec{h}$, the orbit-plane normal) must lie along that intersection — the cross product delivers exactly that (figure s02).

The five ($a,e,i,\Omega ,\omega$) describe the fixed geometry and orientation of the ellipse, which the two-body law preserves; $\nu$ is the moving satellite's position on that frozen ellipse, so only it advances — figure s01 shows the ellipse standing still while just the red dot sweeps around, its angle $\nu$ ticking up.

The equinox lives in the reference plane, but periapsis lives in the tilted orbital plane; the ascending node is the shared reference point in the orbital plane from which in-plane angles like $\omega$ can be measured consistently.

More energy (less negative $\epsilon$) puts the satellite on a larger, more distant orbit where gravity is weaker, so its mean speed and mean motion drop — Kepler III makes $T$ grow as $a^{3/2}$.

Energy fixes only $a$; orbits with equal $a$ but any $e$, $i$, $\Omega$, or $\omega$ all share $\epsilon =-\mu /(2a)$, so shape and orientation are free to differ entirely.

Its magnitude is $e$ (the shape) and its direction is the major axis toward periapsis (the in-plane orientation), so a single vector delivers both element $e$ and the reference direction for $\omega$ and $\nu$.

Both become undefined — no periapsis and no node line exist; the orbit is described by the true longitude $\ell =\Omega +\omega +\nu$ measured straight from the vernal equinox.

$e=1$ exactly, and $a\to \infty$ (the "ellipse" no longer closes); the energy $\epsilon =-\mu /(2a)$ correspondingly goes to zero, the escape threshold.

$a$ becomes negative for a hyperbola, so $\epsilon =-\mu /(2a)$ turns positive — correctly signalling an unbound orbit with leftover speed at infinity; there is no apoapsis ($r_a=a(1+e)$ would be negative and meaningless), and the true anomaly $\nu$ is restricted to $|\nu |<\arccos (-1/e)$, the asymptote angle, beyond which the object has flown off to infinity.

Yes — the plane still tilts, still crosses the reference plane at a node, and $\vec{e}$ still points to periapsis, so $i$, $\Omega$, $\omega$ are all well-defined exactly as before; only the shape elements $a<0$, $e>1$ and the range of $\nu$ change to describe an open, one-pass trajectory.

$i=90^{\circ }$ exactly, and $\Omega$ remains fully meaningful — the plane still crosses the equator at a definite ascending node, so its compass orientation is well-defined even though the tilt is maximal (figure s03).

Outward — at $\nu =90^{\circ }$ it is past periapsis and climbing toward apoapsis, so $\vec{r}\cdot \vec{v}>0$; the sign flips to inward only after it passes apoapsis at $\nu =180^{\circ }$.

It means retrograde motion — the satellite circles opposite to Earth's spin; the plane isn't upside down, it just crosses the equator descending-to-ascending in the reverse sense (figure s03).

It is still a bound ellipse (closed, periodic), but $r_a=a(1+e)$ grows enormous — a long thin cigar that skims periapsis then flings far out before returning.