Question bank — Orbital elements (Keplerian) — semi-major axis a, eccentricity e, inclination i, RAAN Ω, argument of perigee ω, true ano

False. A circle has $e=0$, so there is no perigee — no "closest point" to measure from — and $\omega$ is undefined; engineers use the argument of latitude $u=\omega +\nu$ instead.

True. Period depends only on $a$ via $T=2\pi \sqrt{a^3/\mu }$; eccentricity, tilt, and orientation do not enter, so a nearly circular and a very elongated orbit with equal $a$ share one period.

False. Energy is $\epsilon =-\mu /2a$, which depends only on $a$. Changing $e$ reshapes the ellipse but leaves the energy — and hence $a$ — untouched (see Vis-viva equation).

True. Inclination is the tilt of the plane from the equator; at $90^{\circ }$ the plane contains the $Z$ (spin) axis, so the ground track sweeps over the North and South poles.

False. With $i=0$ the plane never leaves the equator, so there is no ascending node to point to; $\Omega$ becomes undefined (a degenerate/singular case).

False. By Kepler's 2nd law the satellite sweeps equal areas in equal times, so $\nu$ races through perigee and crawls at apogee; the uniform angle is the mean anomaly $M$ (see Kepler's equation and mean anomaly).

True. Since $r_a=a(1+e)$ and $r_p=a(1-e)$, adding gives $r_a+r_p=2a$, so $a=(r_a+r_p)/2$ — of radii from the focus, not altitudes.

False. Inclination is defined on $[0^{\circ },180^{\circ }]$; retrograde motion is encoded by $i>90^{\circ }$ (the $\vec{h}$ vector leans below the equator), never by a negative angle.

True. Six state numbers carry exactly the same information as six elements; the conversion computes $\vec{h}=\vec{r}\times \vec{v}$ for the plane, the eccentricity vector for $e$ and perigee direction, and energy for $a$ — done by State vectors to orbital elements, which is the standard state→elements recipe (see also Angular momentum vector h, Eccentricity vector).

The error is forgetting the focus is Earth's center: $r_p=R_E+\text{altitude}=6378+400=6778$ km. Altitude is measured from the surface, radius from the center.

Eccentricity sets shape, not size; a tiny orbit and a giant orbit can both have $e=0.9$. Only $a$ (and hence $r_a,r_p$) sets the size scale.

Only $\omega$ starts at the ascending node (to perigee). $\nu$ starts at perigee (to the satellite), so their sum $u=\omega +\nu$ is the node-to-satellite angle.

Backwards: larger $r$ makes $2/r$ smaller, so $v$ (the scalar speed) is smallest at apogee. Speed peaks at perigee where $r$ is smallest.

It is the angle between $\vec{h}$ (perpendicular to the plane) and $Z$, which equals the angle between the orbit plane and the equatorial plane — not between the plane and the axis.

The correct order is $R_z(\Omega )R_x(i)R_z(\omega )$ acting on ${\vec{r}}_{PQW}$. These are active rotations (they move the vector, holding axes fixed), and in matrix form the rightmost matrix touches the vector first — so reading right-to-left we swing perigee within the plane ($\omega$), then tilt the plane about the node line ($i$), then swing the node to its longitude ($\Omega$). Doing $i$ first would tilt before perigee is placed, giving a different, wrong orientation (see Perifocal coordinate frame, which defines the P–Q–W axes this starts from).

For a parabola $\epsilon =0$ (so $a\to \infty$), and for a hyperbola $\epsilon >0$ forcing $a<0$. The formula holds but $a$ is not a positive "size" anymore.

Because the full state is 3 position + 3 velocity = six independent numbers; the elements are a repackaging of the same six degrees of freedom (see Two-body problem, which shows the motion is fixed by these six).

The ellipse's shape and orientation are fixed by conserved quantities (energy fixes $a$; $\vec{h}$ fixes the plane; the Eccentricity vector — a fixed arrow pointing to perigee — fixes $e$ and $\omega$); only the satellite's position along the fixed curve moves, and that is $\nu$.

Elements separate the unchanging geometry (5 constants) from the one moving number, making the orbit's shape, tilt, and orientation instantly readable rather than buried in six wiggling coordinates.

$J_2$ is the largest zonal harmonic — a dimensionless number ($J_2\approx 1.083\times 10^{-3}$ for Earth) measuring how much Earth's equatorial bulge makes gravity depart from a perfect $1/r^2$. That extra pull slowly drifts $\Omega$ and $\omega$, so the five "constants" become slowly varying (see Orbital perturbations (J2)).

The orbit equation $r=p/(1+e\cos \nu )$ gives $\cos 90^{\circ }=0$, so $r=p$ exactly a quarter-orbit past perigee — $p$ is that "sideways" radius.

Otherwise the sign is ambiguous and $r=p/(1+e\cos \nu )$ would not track the satellite correctly through apogee; motion direction fixes whether $\nu$ increases toward apogee or back toward perigee.

Perigee vanishes, so $\omega$ (and with it the reference for $\nu$) becomes undefined; the fix is the argument of latitude $u=\omega +\nu$, which stays well-defined.

The node line disappears (plane never leaves the equator), so both lose their reference; the fix is the true longitude $\ell =\Omega +\omega +\nu$, measured straight from the $X$-axis.

It is a parabolic escape trajectory with exactly zero total energy $\epsilon =0$; since $\epsilon =-\mu /2a$, this forces $a\to \infty$ — the "ellipse" opens up and never closes.

$i=180^{\circ }$: the plane is equatorial but the motion is fully retrograde, the extreme opposite of a prograde equatorial orbit at $i=0^{\circ }$.

The orbit is a circle, so $e=0$ exactly, and $\omega$ is undefined because there is no unique closest point.

No — the trajectory is open, so there is no farthest point; only perigee (closest approach, $\nu =0$) exists, and $\nu$ is bounded by the asymptote angles where $r\to \infty$.