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

What we're testing: reading the meaning of $e$ off the definition. $e$ is the shape number. When $e=0$ the conic is a circle. The orbit equation $r=\frac{a(1-e^2)}{1+e\cos \nu }$ with $e=0$ becomes $r=a$ for every $\nu$ — the radius never changes. Therefore $r_p=r_a=a$. A circle: apogee equals perigee equals $a$.

Changes with time: the ==true anomaly $\nu$== — it is the "where is the satellite right now?" angle. Constant: $a,e,i,\Omega ,\omega$. They fix the unchanging geometry (size, shape, tilt, swing, in-plane pointing) of the ellipse. Only the moving point needs a moving number.

$i$ is the tilt of the orbital plane away from the equator.

• $i=90^{\circ }$ → the plane stands straight up through the poles → a polar orbit.
• $i=0^{\circ }$ → the plane lies flat in the equatorial plane → an equatorial orbit. ($i>90^{\circ }$ would be retrograde — going against Earth's spin.)

Step 1 — altitude → radius. Why: elements are measured from Earth's centre (the focus), so add $R_E$. $r_p=6378+500=6878\text{km}$, $r_a=6378+2500=8878\text{km}$. Step 2 — semi-major axis. The full major axis is $r_p+r_a=2a$, so $a=\frac{r_p+r_a}{2}=\frac{6878+8878}{2}=7878\text{km}.$ Step 3 — eccentricity. From $r_a-r_p=2ae$ and $r_a+r_p=2a$, dividing kills $2a$: $e=\frac{r_a-r_p}{r_a+r_p}=\frac{2000}{15756}=\mathrm{0.1269.}$

Why vis-viva: it links speed to position using only $a$, because energy $\epsilon =-\mu /2a$ is constant. $v^2=\mu \left(\frac{2}{r}-\frac{1}{a}\right)\text{at}r=r_p=6878.$ $v_p^2=3.986\times 10^5\left(\frac{2}{6878}-\frac{1}{7878}\right)=3.986\times 10^5(2.9078\times 10^{-4}-1.2694\times 10^{-4}).$ $v_p^2=3.986\times 10^5(1.6384\times 10^{-4})=65.31\Rightarrow v_p=8.08\text{km/s}.$ Sensible: perigee is the fastest point, a bit above circular LEO speed.

Step 1 — semi-latus rectum. $p=a(1-e^2)=7878(1-0.1269^2)=7878(1-0.016103)=7751.1\text{km}$. Step 2 — orbit equation. Why this formula: $r=\frac{p}{1+e\cos \nu }$ is the ellipse in focus-centred polar form. $r=\frac{7751.1}{1+0.1269\cos 60^{\circ }}=\frac{7751.1}{1+0.1269(0.5)}=\frac{7751.1}{1.06345}=7288.9\text{km}.$

$p=a(1-e^2)=8000(1-0.04)=7680\text{km}$. Use $r=\frac{p}{1+e\cos \nu }$:

• $\nu =0^{\circ }$: $\cos =1$, $r=\frac{7680}{1.20}=6400\text{km}$ (perigee $=a(1-e)=6400$ ✓).
• $\nu =90^{\circ }$: $\cos =0$, $r=7680\text{km}$ ($=p$).
• $\nu =180^{\circ }$: $\cos =-1$, $r=\frac{7680}{0.80}=9600\text{km}$ (apogee $=a(1+e)=9600$ ✓).
• $\nu =270^{\circ }$: $\cos =0$, $r=7680\text{km}$ ($=p$). Symmetry: $\cos \nu =\cos (360^{\circ }-\nu )$, so $r(90^{\circ })=r(270^{\circ })$. The ellipse is mirror-symmetric about the major axis; the two points a quarter-turn either side of perigee are the same distance out.

RAAN. The south→north equator crossing is the ascending node. Its right ascension (angle from $X$/♈ in the equatorial plane) is $\Omega =70^{\circ }$. Argument of perigee. $\omega$ is the angle from the ascending node to perigee, in the orbital plane, along the motion — given as $\omega =50^{\circ }$. True anomaly. $\nu$ is measured from perigee to the satellite. The satellite is $50^{\circ }$ before perigee, i.e. it still has $50^{\circ }$ to go, so it sits at $\nu =360^{\circ }-50^{\circ }=310^{\circ }$ (equivalently $-50^{\circ }$). Argument of latitude. $u=\omega +\nu =50^{\circ }+310^{\circ }=360^{\circ }\equiv 0^{\circ }$. That means node→satellite is $0^{\circ }$: the satellite is at the ascending node right now — consistent with "it just crossed the equator." ✓

Energy first. $\epsilon =-\mu /2a$ depends only on $a$ — same for both. Equal energy, different shapes. Perigee speed. $r_p=a(1-e)$, so orbit 2's perigee is deeper (smaller $r_p$), and vis-viva says smaller $r$ → larger $v$. $r_{p1}=7500(0.95)=7125\text{km}$, $r_{p2}=7500(0.65)=4875\text{km}$. $v_p^2=\mu \left(\frac{2}{r_p}-\frac{1}{a}\right).$ $v_{p1}^2=3.986\times 10^5(2/7125-1/7500)=3.986\times 10^5(2.8070\times 10^{-4}-1.3333\times 10^{-4})=58.74$, $v_{p1}=7.664\text{km/s}$. $v_{p2}^2=3.986\times 10^5(2/4875-1/7500)=3.986\times 10^5(4.1026\times 10^{-4}-1.3333\times 10^{-4})=110.4$, $v_{p2}=10.51\text{km/s}$. Ratio $v_{p2}/v_{p1}=1.371$. Orbit 2 is ~37% faster at perigee — yet both have equal total energy. The eccentric orbit just trades speed between a fast perigee and a slow apogee.

Step 1 — radii. $r_p=6378+600=6978\text{km}$, $r_a=6378+35786=42164\text{km}$. Step 2 — $a$ and $e$. $a=\frac{r_p+r_a}{2}=\frac{6978+42164}{2}=24571\text{km},e=\frac{r_a-r_p}{r_a+r_p}=\frac{35186}{49142}=\mathrm{0.7160.}$ Step 3 — period. Why this formula: Kepler's third law, $T=2\pi \sqrt{a^3/\mu }$, comes from equating gravity to the centripetal need over one loop; it depends only on $a$. $T=2\pi \sqrt{\frac{24571^3}{3.986\times 10^5}}=2\pi \sqrt{\frac{1.4835\times 10^{13}}{3.986\times 10^5}}=2\pi \sqrt{3.7218\times 10^7}=2\pi (6100.7)=38331\text{s}\approx 10.65\text{h}.$ Step 4 — apogee speed. vis-viva at $r=r_a$: $v_a^2=\mu \left(\frac{2}{r_a}-\frac{1}{a}\right)=3.986\times 10^5\left(\frac{2}{42164}-\frac{1}{24571}\right)=3.986\times 10^5(4.7433\times 10^{-5}-4.0698\times 10^{-5}).$ $v_a^2=3.986\times 10^5(6.735\times 10^{-6})=2.6847\Rightarrow v_a=1.639\text{km/s}.$ Slow at apogee — exactly as Kepler's 2nd law demands.

Step 1 — radius at $\nu =120^{\circ }$. $p=a(1-e^2)=24571(1-0.7160^2)=24571(1-0.51266)=24571(0.48734)=11973\text{km}$. $r=\frac{p}{1+e\cos 120^{\circ }}=\frac{11973}{1+0.7160(-0.5)}=\frac{11973}{0.6420}=18650\text{km}.$ Step 2 — perifocal components. In the perifocal (P–Q–W) frame P points to perigee, Q is $90^{\circ }$ ahead: ${\vec{r}}_{PQW}=r\left(\begin{array}{c}\cos \nu \\ \sin \nu \\ 0\end{array}\right)=18650\left(\begin{array}{c}\cos 120^{\circ }\\ \sin 120^{\circ }\\ 0\end{array}\right)=18650\left(\begin{array}{c}-0.5\\ 0.8660\\ 0\end{array}\right)=\left(\begin{array}{c}-9325\\ 16151\\ 0\end{array}\right)\text{km}.$ Step 3 — the three rotations (words). To reach ECI, apply ${\vec{r}}_{ECI}=R_z(\Omega )R_x(i)R_z(\omega ){\vec{r}}_{PQW}$:

1. $R_z(\omega )$ — spin about $W/Z$ so perigee sits at the right in-plane angle from the node.
2. $R_x(i)$ — tilt the plane about the line of nodes by inclination $i$.
3. $R_z(\Omega )$ — swing the tilted plane about $Z$ so the node lands at right ascension $\Omega$.

Invert Kepler's third law. From $T=2\pi \sqrt{a^3/\mu }$, $a={\left(\frac{\mu T^2}{4{\pi }^2}\right)}^{1/3}={\left(\frac{3.986\times 10^5\times 43082^2}{4{\pi }^2}\right)}^{1/3}.$ Numerator: $3.986\times 10^5\times 1.85606\times 10^9=7.3982\times 10^{14}$. Divide by $4{\pi }^2=39.478$: $1.87401\times 10^{13}$. Cube root: $a=2.6607\times 10^4=26607\text{km}$. Circular altitude ($e=0$ → $r=a$): altitude $=a-R_E=26607-6378=20229\text{km}$. (This is essentially the GPS orbit.)

Energy and $a$. Specific energy is $\epsilon =-\mu /2a$. Setting $\epsilon =0$ forces $-\mu /2a=0$, which can only hold as $a\to \infty$ — the ellipse stretches into a parabola. That is precisely the $e=1$ boundary. ✓ Escape speed. At $\epsilon =0$: $\frac{v^2}{2}-\frac{\mu }{r}=0\Rightarrow v^2=\frac{2\mu }{r}\Rightarrow v_{esc}=\sqrt{\frac{2\mu }{r}}$. (Same as vis-viva with $1/a\to 0$.) Numbers at $r=6978$: $v_{esc}=\sqrt{\frac{2\times 3.986\times 10^5}{6978}}=\sqrt{114.24}=10.69\text{km/s}.$ GTO perigee speed (L4·Q1) was $v_p=\sqrt{\mu (2/r_p-1/a)}$ with $a=24571$: $v_p^2=3.986\times 10^5(2/6978-1/24571)=3.986\times 10^5(2.8661\times 10^{-4}-4.0698\times 10^{-5})=89.63$, $v_p=9.467\text{km/s}$. Comparison: $v_p=9.467<10.69=v_{esc}$. The GTO is bound (elliptic, $e<1$) — it does not escape, consistent with $a$ finite.

(a) Perigee vanishes. $\nu$ is measured from perigee. But when $e=0$ every point of the orbit is at the same radius $r=a$ — there is no unique closest point, so "perigee" is undefined, and therefore $\nu$ has no anchor. It is a degenerate coordinate. (b) Radius survives. $r=\frac{a(1-e^2)}{1+e\cos \nu }\to a$ for all $\nu$ as $e\to 0$. The geometry (a circle of radius $a$) is completely well-defined; only the labelling angle broke. (c) The fix — argument of latitude. The ascending node still exists (the plane still crosses the equator), so we can measure the satellite's angle from the node instead: $u=\omega +\nu$. Since $\omega$ (node→perigee) and $\nu$ (perigee→satellite) both individually lose meaning but their sum (node→satellite) does not, mission software uses $u$ directly for near-circular orbits. This is why $u$ is called a "non-singular" element for $e\to 0$.

When does $r$ blow up? $r\to \infty$ requires the denominator $1+e\cos \nu \to 0$, i.e. $\cos {\nu }_{\infty }=-\frac{1}{e}=-\frac{1}{1.2}=-\mathrm{0.8333.}$ Solve, both signs. ${\nu }_{\infty }=\pm \arccos (-0.8333)=\pm 146.44^{\circ }$.

• $+146.44^{\circ }$ is the outbound asymptote (leaving to infinity).
• $-146.44^{\circ }$ (equivalently $213.56^{\circ }$) is the inbound asymptote (arriving from infinity). Sign check: for a bound ellipse $e<1$, $-1/e<-1$, so $\cos \nu =-1/e$ has no solution — $r$ never reaches infinity, the orbit closes. Only for $e\ge 1$ does an asymptote exist. ✓ Physical meaning: $r\to \infty$ marks the direction along which the spacecraft approaches from / departs to infinite distance — the hyperbolic asymptotes. The angle between the two, $2(180^{\circ }-146.44^{\circ })=67.1^{\circ }$... more precisely the turn angle involves $\delta =2\arcsin (1/e)$; the asymptotes themselves lie at $\nu =\pm 146.44^{\circ }$ from perigee.

• Got L1–L2 clean ::: you can read and plug. Solid foundation.
• Got L3 ::: you handle signs and combine energy + geometry.
• Got L4 ::: you can run the full elements↔state pipeline.
• Got L5 ::: you reason about limits and edge cases — mastery.