Exercises — Converting between orbital elements and state vectors (r, v)

WHAT: We are matching an element to the quantity that pins it. The ==semi-major axis $a$ sets the size. It is fixed by the specific energy $\epsilon =v^2/2-\mu /r$== via $a=-\mu /(2\epsilon )$. WHY: Energy depends only on how big the ellipse is (vis-viva), not on where the body sits — so one energy number is enough to recover $a$.

$\vec{n}=\hat{z}\times \vec{h}$. Since $\vec{h}$ is perpendicular to the orbit plane and $\hat{z}$ is the reference-frame vertical, their cross product lies along the line where the orbit plane cuts the reference plane — the line of nodes. By convention $\vec{n}$ points to the ascending node (where the body crosses upward). If $\vec{n}=\vec{0}$ the orbit is equatorial and $\Omega$ is undefined.

False. $\vec{e}=\frac{\vec{v}\times \vec{h}}{\mu }-\frac{\vec{r}}{r}$ points toward periapsis (closest approach). Its magnitude is the scalar $e$.

WHAT: energy → size, one plug-in. $\epsilon =\frac{v^2}{2}-\frac{\mu }{r}=\frac{7.8^2}{2}-\frac{398600}{6800}=30.42-58.62=-28.20{\text{km}}^2/{\text{s}}^2$ $a=-\frac{\mu }{2\epsilon }=-\frac{398600}{2(-28.20)}=7067\text{km}$ WHY $a>0$: $\epsilon <0$ means a bound (elliptical) orbit, and the minus signs cancel to give a positive size.

WHAT: cross product then a cosine. $\vec{h}=\vec{r}\times \vec{v}=(0\cdot 3-0\cdot 6.5,0\cdot 0-8000\cdot 3,8000\cdot 6.5-0)=(0,-24000,52000)$ $h=\sqrt{0+24000^2+52000^2}=57271{\text{km}}^2/\text{s}$ $\cos i=\frac{h_z}{h}=\frac{52000}{57271}=0.9080\Rightarrow i=24.83^{\circ }$ WHY: $\vec{h}$ is the axle of the orbit; its tilt away from $\hat{z}$ is the inclination.

$r_p=a(1-e)=7500(0.95)=7125\text{km}$; $r_a=a(1+e)=7500(1.05)=7875\text{km}$. WHY: periapsis and apoapsis are the two endpoints of the major axis; $a$ is the midpoint, $ae$ is the offset of the focus.

WHAT: build $\vec{e}$ from the recipe, read its geometry. $r=7000$, $v^2=7.5^2+1.0^2=57.25$, $\vec{r}\cdot \vec{v}=0$. $\vec{e}=\frac{1}{\mu }[(v^2-\frac{\mu }{r})\vec{r}-(\vec{r}\cdot \vec{v})\vec{v}]=\frac{1}{398600}(57.25-56.943)(7000,0,0)$ $=\frac{0.30714\cdot 7000}{398600}(1,0,0)=(0.005393,0,0)$ So $e=0.00539$ — nearly circular. Sign check: $\vec{r}\cdot \vec{v}=0$ means the body is at an apse (radial velocity zero). Since $\vec{e}$ points along $+\hat{x}$ and $\vec{r}$ is also along $+\hat{x}$, $\vec{r}$ and $\vec{e}$ are parallel → $\cos \nu =+1$ → the body sits at periapsis ($\nu =0$). WHY the parallel test: $\nu$ is the angle between periapsis ($\vec{e}$) and current position ($\vec{r}$); parallel ⇒ zero angle ⇒ periapsis.

WHAT: cosine gives a raw angle; the companion sign disambiguates. $n=\sqrt{3000^2+4000^2}=5000$. $\cos \Omega =n_x/n=-3000/5000=-0.6\Rightarrow {\Omega }_{\text{raw}}=126.87^{\circ }$. Quadrant fix: $n_y<0$, so $\Omega =360^{\circ }-126.87^{\circ }=233.13^{\circ }$. WHY: ${\cos }^{-1}$ only returns $[0^{\circ },180^{\circ }]$. The node here points into the third quadrant ($n_x<0,n_y<0$), which lives beyond $180^{\circ }$; the $n_y<0$ rule reflects it below the $x$-axis. The figure below shows the raw and corrected angles as mirror images across the $x$-axis.

Read it as: same $x$-projection, opposite $y$-sign — so the cosine can't tell them apart, but $n_y$ can.

Step 1 — magnitudes. WHY first: every downstream formula ($\epsilon$, $\cos i$, $\vec{e}$) needs the scalar lengths $r=|\vec{r}|$ and $v=|\vec{v}|$, so we compute them once up front. $r=\sqrt{6524^2+6862^2+6449^2}=\sqrt{42562576+47087044+41589601}=\sqrt{131239221}=11455.1\text{km}$ $v=\sqrt{4.902^2+5.533^2+1.976^2}=\sqrt{24.0296+30.6141+3.9046}=\sqrt{58.5483}=7.6517\text{km/s}$ Step 2 — energy → $a$. WHY energy: the specific energy $\epsilon$ depends only on orbit size, so it is the cleanest single quantity that hands us $a$ (via $a=-\mu /2\epsilon$) without needing any angles. $\epsilon =\frac{7.6517^2}{2}-\frac{398600}{11455.1}=29.274-34.797=-5.523{\text{km}}^2/{\text{s}}^2$ $a=-\frac{398600}{2(-5.523)}=36090\text{km}$ Step 3 — $\vec{h}$ and $i$. WHY $\vec{h}$: the angular-momentum vector is the axle perpendicular to the orbit plane, so its tilt from $\hat{z}$ is the inclination — one cosine reads it off. $\vec{h}=\vec{r}\times \vec{v}=(6862\cdot (-1.976)-6449\cdot 5.533,6449\cdot 4.902-6524\cdot (-1.976),6524\cdot 5.533-6862\cdot 4.902)$ $=(-49246.8,44510.8,2469.5){\text{km}}^2/\text{s},h=\sqrt{49246.8^2+44510.8^2+2469.5^2}=66436{\text{km}}^2/\text{s}$ $\cos i=\frac{h_z}{h}=\frac{2469.5}{66436}=0.03717\Rightarrow i=87.87^{\circ }(\text{near-polar})$ Step 4 — $\vec{e}$ then $e$. WHY $\vec{e}$: the eccentricity vector is the only quantity carrying shape ($e=|\vec{e}|$) and the periapsis direction at once, so it finishes the shape half of the conversion (and would seed $\omega ,\nu$ if asked). We first need $\vec{r}\cdot \vec{v}$, the radial-motion indicator: $\vec{r}\cdot \vec{v}=6524(4.902)+6862(5.533)+6449(-1.976)=31980.6+37967.5-12743.2=57204.9$. $\vec{e}=\frac{1}{\mu }[(v^2-\frac{\mu }{r})\vec{r}-(\vec{r}\cdot \vec{v})\vec{v}]$ $v^2-\mu /r=58.548-34.797=23.751$. Then $\vec{e}=\frac{1}{398600}[23.751(6524,6862,6449)-57204.9(4.902,5.533,-1.976)]$ $=\frac{1}{398600}[(154951,162978,153170)-(280418,316505,-113037)]=\frac{(-125467,-153527,266207)}{398600}$ $\vec{e}=(-0.3148,-0.3852,0.6679),e=|\vec{e}|=0.8329$ So this is a highly eccentric, near-polar orbit with $a\approx 36090$ km, $e\approx 0.833$, $i\approx 87.9^{\circ }$.

Step 1 — perifocal position. WHY the semi-latus rectum and orbit equation: in the perifocal frame periapsis lies on the $\hat{p}$ axis, so position is trivially $r(\cos \nu ,\sin \nu ,0)$ — but we first need the scalar distance $r$ at this true anomaly. The orbit equation $r=p/(1+e\cos \nu )$ is exactly the conic written from the focus, and $p$ is its natural size-scale (the width at $\nu =90^{\circ }$). Using $p=a(1-e^2)$ converts our given size element $a$ into that scale: $p=a(1-e^2)=8000(0.99)=7920\text{km}$ $r=\frac{p}{1+e\cos \nu }=\frac{7920}{1+0.1\cos 30^{\circ }}=\frac{7920}{1.08660}=7288.8\text{km}$ ${\vec{r}}_{pqw}=r(\cos \nu ,\sin \nu ,0)=7288.8(0.86603,0.5,0)=(6312.3,3644.4,0)\text{km}$ Step 2 — rotate PQW → ECI. Apply the composite matrix $\tilde{R}$ displayed at the top of this page. Since $z_{pqw}=0$, only the first two columns act, giving (with $c=\cos ,s=\sin$): $x=(c\Omega c\omega -s\Omega s\omega ci)x_{pqw}+(-c\Omega s\omega -s\Omega c\omega ci)y_{pqw}$ WHY these exact cosine–sine combinations? $\tilde{R}$ is three stacked rotations, read innermost→outermost:

• $R_3(-\omega )$ first swings periapsis (the $\hat{p}$ axis) forward by $\omega$ inside the orbit plane — this is why $\cos \omega ,\sin \omega$ appear paired with the in-plane coordinates.
• $R_1(-i)$ then tips the whole plane up by the inclination about the node line — this is why every term that "leaves the plane" is multiplied by $\cos i$ or $\sin i$ (the $ci$, $si$ factors).
• $R_3(-\Omega )$ finally swivels the node line to its sky longitude about $\hat{z}$ — the source of the $\cos \Omega ,\sin \Omega$ factors. Each entry is a product of one factor from each rotation, so a term like $s\Omega c\omega ci$ literally reads "swivelled by $\Omega$, spun by $\omega$, tilted by $i$." The negatives live inside because we transform components from PQW into ECI, the inverse of turning the frame.

Numbers: $c\Omega =0.76604,s\Omega =0.64279,c\omega =0.5,s\omega =0.86603,ci=0.86603,si=0.5$. Row 1 coefficients: $0.76604(0.5)-0.64279(0.86603)(0.86603)=0.38302-0.48209=-0.09907$; $-0.76604(0.86603)-0.64279(0.5)(0.86603)=-0.66342-0.27835=-0.94177$. $x=-0.09907(6312.3)+(-0.94177)(3644.4)=-625.4-3432.3=-4057.7\text{km}$ Row 2 coefficients: $s\Omega c\omega +c\Omega s\omega ci=0.64279(0.5)+0.76604(0.86603)(0.86603)=0.32140+0.57451=0.89590$; $-s\Omega s\omega +c\Omega c\omega ci=-0.64279(0.86603)+0.76604(0.5)(0.86603)=-0.55668+0.33172=-0.22496$. $y=0.89590(6312.3)+(-0.22496)(3644.4)=5655.2-819.9=4835.3\text{km}$ Row 3 coefficients: $s\omega si=0.86603(0.5)=0.43301$; $c\omega si=0.5(0.5)=0.25$. $z=0.43301(6312.3)+0.25(3644.4)=2733.3+911.1=3644.4\text{km}$ Answer: ${\vec{r}}_{\text{ECI}}=(-4058,4835,3644)\text{km}$. Check: $|\vec{r}|=\sqrt{4058^2+4835^2+3644^2}=7289\text{km}$ = $r$ ✓ (rotations preserve length).

WHAT breaks: With $i=0$ the orbit plane is the reference plane, so it never "cuts up through" it — the line of nodes vanishes: $\vec{n}=\hat{z}\times \vec{h}=\vec{0}$. Then $\cos \Omega =n_x/n$ is $0/0$ → $\Omega$ is undefined, and since $\omega$ is measured from the (missing) node, $\omega$ is undefined too. The fix: eccentricity direction is still perfectly good ($e\ne 0$), so use the true longitude of periapsis $\varpi =\Omega +\omega$ measured directly from the $\hat{x}$ axis, computed from $\vec{e}$: $\cos \varpi =e_x/e$ (quadrant by $e_y$). More generally, equinoctial elements remove all such $0/0$ cases. WHY this works: the combination $\Omega +\omega$ stays observable even when each piece individually is not — the ambiguity is only in where you split the angle, not in the total.

WHAT breaks: With $e=0$ the eccentricity vector collapses, $\vec{e}=\vec{0}$. There is now no periapsis direction, so:

• $\omega$ (measured to periapsis from the node) is undefined — $\cos \omega =\frac{\vec{n}\cdot \vec{e}}{ne}$ is $0/0$.
• $\nu$ (measured from periapsis) is undefined too — $\cos \nu =\frac{\vec{e}\cdot \vec{r}}{er}$ is $0/0$. The node still exists ($i\ne 0$ ⇒ $\vec{n}\ne \vec{0}$), so $\Omega$ is fine. The fix: measure position from the node instead of from the missing periapsis. Use the argument of latitude $u=\omega +\nu$, computed directly from the node and position vectors: $\cos u=\frac{\vec{n}\cdot \vec{r}}{nr}(\text{quadrant: if }{r}_z<0,u=2\pi -u)$ WHY this works: the sum $\omega +\nu$ is the angle from the node to the body, which is perfectly well-defined regardless of where periapsis "would" be. Only the split into $\omega$ and $\nu$ is ambiguous. (If the orbit is both circular and equatorial, even $\Omega$ dies and you fall back on the true longitude $\ell =\Omega +\omega +\nu$, or on equinoctial elements.)

State the companion sign for each quadrant-ambiguous element. $\Omega :::n_y$ $\omega :::e_z$ $\nu :::\vec{r}\cdot \vec{v}$