Exercises — Orbital perturbations — J2 effect (oblateness), derivation of nodal precession

The only inclination-dependent piece in $\dot{\Omega }$ is $-\cos i$, and $n,J_2,(R_E/p{)}^2$ are all positive. So the sign of $\dot{\Omega }$ is the opposite sign of $\cos i$ — follow the solid drift curve in the figure (labelled $\dot{\Omega }$).

• At $i=30°$: $\cos 30°=+0.866>0$, so $-\cos i<0$ → $\dot{\Omega }<0$ → westward (node regresses).
• At $i=120°$: $\cos 120°=-0.5<0$, so $-\cos i>0$ → $\dot{\Omega }>0$ → eastward (node advances).

The switch happens at $i=90°$ where $\cos i=0$ and there is zero nodal precession (the curve crosses the horizontal axis, marked by the dotted vertical line).

• Sun-synchronous exploits the node $\Omega$ (the whole plane swivels): Sun-synchronous orbits.
• Molniya exploits the argument of perigee $\omega$ (the ellipse rotates in its own plane), freezing it at critical inclination $i=63.4°$.

Why $n$ first? Every precession rate scales linearly with $n$, so it is the base timescale. $n=\sqrt{\frac{\mu }{a^3}}=\sqrt{\frac{3.986\times 10^{14}}{(7.0\times 10^6{)}^3}}.$ Denominator: $(7.0\times 10^6{)}^3=3.43\times 10^{20}$. Ratio $=1.1621\times 10^{-6}$. Square root: $n=1.0780\times 10^{-3}\text{rad/s}.$ Period $T=2\pi /n=6.2832/1.0780\times 10^{-3}=5828\text{s}=97.1\text{min}$. This links straight back to the Keplerian $T=2\pi \sqrt{a^3/\mu }$.

Use $n=1.0780\times 10^{-3}$ rad/s from L2.1. Since $e=0$, $p=a$, so ${\left(\frac{R_E}{p}\right)}^2={\left(\frac{6.378\times 10^6}{7.0\times 10^6}\right)}^2=(0.9111{)}^2=\mathrm{0.8302.}$ $\cos 55°=0.5736$. Then $\dot{\Omega }=-\frac{3}{2}(1.0780\times 10^{-3})(1.0826\times 10^{-3})(0.8302)(0.5736).$ Multiply step by step: $\frac{3}{2}\times 1.0780\times 10^{-3}=1.617\times 10^{-3}$; $\times 1.0826\times 10^{-3}=1.7506\times 10^{-6}$; $\times 0.8302=1.4535\times 10^{-6}$; $\times 0.5736=8.337\times 10^{-7}$. $\dot{\Omega }=-8.337\times 10^{-7}\text{rad/s}.$ Convert: $\times 86400\times (180/\pi )=\times 4.9536\times 10^6$: $\dot{\Omega }\approx -4.13°/\text{day}(\text{westward}).$

Both share $a$, hence the same $n$, and the same $i$, so the only difference is the $(R_E/p{)}^2$ factor, because $p=a(1-e^2)$. $\frac{{\dot{\Omega }}_B}{{\dot{\Omega }}_A}=\frac{(R_E/p_B{)}^2}{(R_E/p_A{)}^2}=\frac{p_A^2}{p_B^2}=\frac{a^2}{[a(1-e^2){]}^2}=\frac{1}{(1-e^2{)}^2}.$ With $e=0.3$: $1-e^2=0.91$, $(0.91{)}^2=0.8281$. $\frac{{\dot{\Omega }}_B}{{\dot{\Omega }}_A}=\frac{1}{0.8281}=\mathrm{1.2076.}$ The eccentric orbit precesses ~21% faster. This is exactly the "use $p$, not $a$" warning: eccentricity is not a small correction.

Why set $\dot{\omega }=0$? A vanishing apsidal rate means perigee is frozen — the whole point of a Molniya design. $\dot{\omega }\propto (5{\cos }^{2}i-1)$, so set the bracket to zero: $5{\cos }^{2}i-1=0\Rightarrow {\cos }^{2}i=\frac{1}{5}\Rightarrow \cos i=\pm \sqrt{0.2}=\pm \mathrm{0.4472.}$ $i=\arccos (0.4472)=63.43°,i=\arccos (-0.4472)=116.57°.$ Both are the critical inclination. Molniya uses the prograde one, $63.4°$.

Step 1 — mean motion. $n=\sqrt{\mu /a^3}=\sqrt{3.986\times 10^{14}/(7.2\times 10^6{)}^3}$. Cube: $(7.2\times 10^6{)}^3=3.7325\times 10^{20}$; ratio $=1.0679\times 10^{-6}$; root $n=1.0334\times 10^{-3}$ rad/s.

Step 2 — bulge factor. $e=0\Rightarrow p=a$, so $(R_E/p{)}^2=(6.378/7.2{)}^2=(0.8858{)}^2=0.7847$.

Step 3 — isolate $\cos i$. We want $i$, but $i$ sits inside $\cos i$, which is multiplied by the known constants $-\frac{3}{2}n{J}_2(R_E/p{)}^2$. So we divide both sides by that whole constant block to leave $\cos i$ alone on the right — a legal move because the block is a fixed nonzero number. Note the block is negative (the leading $-\frac{3}{2}$), and this sign will matter: $\cos i=\frac{\dot{\Omega }}{-\frac{3}{2}n{J}_2(R_E/p{)}^2}.$ Denominator: $-\frac{3}{2}(1.0334\times 10^{-3})(1.0826\times 10^{-3})(0.7847)=-1.3167\times 10^{-6}$. $\cos i=\frac{+1.991\times 10^{-7}}{-1.3167\times 10^{-6}}=-\mathrm{0.1512.}$ A positive target $\dot{\Omega }$ divided by a negative constant block gives a negative $\cos i$ — the sign flip is forced, and it tells you the orbit must be retrograde.

Step 4 — invert. $\cos i$ is negative, so $i$ must lie past $90°$: $i=\arccos (-0.1512)=98.70°$. The retrograde ($i>90°$) inclination is exactly what makes $\cos i$ negative, flipping $-\cos i$ positive so the node drifts eastward to chase the Sun. See the marked crossing point in the chart.

Rate is $0.9856°/\text{day}$ by construction. Over $2\times 365.25=730.5$ days: $\Delta \Omega =0.9856\times 730.5=719.98°\approx 720°.$ That is exactly two full turns ($2\times 360°$) — one per year — which is the defining property of a Sun-synchronous orbit: the plane keeps pace with the Sun, so the local solar time at each equator crossing stays fixed.

Mean motion. $a^3=(2.656\times 10^7{)}^3=1.8735\times 10^{22}$; $\mu /a^3=2.1275\times 10^{-8}$; $n=1.4586\times 10^{-4}$ rad/s. (This gives $T\approx 11.97$ h — the half-sidereal-day Molniya period.)

Semi-latus rectum. $p=a(1-e^2)=2.656\times 10^7(1-0.5476)=2.656\times 10^7\times 0.4524=1.2016\times 10^7$ m. $(R_E/p{)}^2=(6.378\times 10^6/1.2016\times 10^7{)}^2=(0.5308{)}^2=0.2818$.

(a) Node. $\cos 63.43°=0.4472$. $\dot{\Omega }=-\frac{3}{2}(1.4586\times 10^{-4})(1.0826\times 10^{-3})(0.2818)(0.4472).$ Chain: $\frac{3}{2}\times 1.4586\times 10^{-4}=2.1879\times 10^{-4}$; $\times 1.0826\times 10^{-3}=2.369\times 10^{-7}$; $\times 0.2818=6.676\times 10^{-8}$; $\times 0.4472=2.985\times 10^{-8}$ rad/s. $\dot{\Omega }=-2.985\times 10^{-8}\text{rad/s}\times 4.9536\times 10^6=-0.148°/\text{day}.$

(b) Perigee. $5{\cos }^{2}i-1=5(0.4472{)}^2-1=5(0.2)-1=0$. So $\dot{\omega }=0$ — perigee is frozen, exactly the Molniya design goal. Apogee stays parked over the northern hemisphere.

Convert the observation to rad/s. $-3.0°/\text{day}÷4.9536\times 10^6=-6.056\times 10^{-7}$ rad/s.

Mean motion. $a^3=(7.5\times 10^6{)}^3=4.2188\times 10^{20}$; $\mu /a^3=9.4479\times 10^{-7}$; $n=9.7200\times 10^{-4}$ rad/s.

Bulge factor ($e=0$, $p=a$): $(R_E/p{)}^2=(6.378/7.5{)}^2=(0.8504{)}^2=0.7232$.

Isolate $\cos i$. Same isolation move as L4.1: divide the observed $\dot{\Omega }$ by the constant block $-\frac{3}{2}n{J}_2(R_E/p{)}^2$ to strip everything off $\cos i$. Build the block: $-\frac{3}{2}n{J}_2(R_E/p{)}^2:\frac{3}{2}\times 9.7200\times 10^{-4}=1.458\times 10^{-3};\times 1.0826\times 10^{-3}=1.5784\times 10^{-6};\times 0.7232=1.1415\times 10^{-6}.$ With the leading minus the block is $-1.1415\times 10^{-6}$. Now divide: $\cos i=\frac{\dot{\Omega }}{-\frac{3}{2}n{J}_2(R_E/p{)}^2}=\frac{-6.056\times 10^{-7}}{-1.1415\times 10^{-6}}=\mathrm{0.5305.}$ Here both numerator and block are negative, so the ratio comes out positive — we are in the prograde regime ($i<90°$).

Invert. $\cos i=0.5305\Rightarrow i=\arccos (0.5305)=57.96°\approx 58.0°$. Since the observed $\dot{\Omega }<0$ (westward) forces $\cos i>0$, we take the prograde root $i\approx 58.0°$. (The retrograde alias $i=122.0°$ would give $\cos i<0$ hence $\dot{\Omega }>0$, contradicting the westward observation.)