Exercises — Orbit types — LEO, MEO, GEO, HEO, SSO, Molniya

WHAT: Match the three numbers to a family. Altitude $\approx 35786$ km with $e=0,i=0$ is the fingerprint of GEO (geostationary). WHY it matters: at this radius the period equals one sidereal day, and because it is circular ($e=0$) and equatorial ($i=0$), the satellite hangs motionless over one point on the equator. A ground dish never has to move. Answer: Geostationary orbit; it appears fixed in the sky.

WHY this reasoning: speed is $v=\sqrt{\mu /r}$ — a decreasing function of $r$. Larger $r$ ⇒ smaller $v$. A is at smaller $r$, so A is faster. Period grows with $a$ ($T\propto a^{3/2}$), so the higher one, B, has the longer period. Answer: A faster, B longer period.

WHAT: convert altitude to radius, plug into $v=\sqrt{\mu /r}$. $r=(6371+550)\text{km}=6921\text{km}=6.921\times 10^6$ m. WHY this tool: it's a circular orbit, so gravity = centripetal gives exactly $v=\sqrt{\mu /r}$; no ellipse machinery needed. $v=\sqrt{\frac{3.986\times 10^{14}}{6.921\times 10^6}}=\sqrt{5.759\times 10^7}\approx 7589\text{m/s}\approx 7.59\text{km/s}.$

WHAT: use $T=2\pi \sqrt{r^3/\mu }$ (circle ⇒ $a=r$). $r^3=(6.921\times 10^6{)}^3=3.316\times 10^{20}{\text{m}}^3$. $T=2\pi \sqrt{\frac{3.316\times 10^{20}}{3.986\times 10^{14}}}=2\pi \sqrt{8.319\times 10^5}=2\pi (912.1)\approx 5731\text{s}\approx 95.5\text{min}.$ WHY: period is one circumference $2\pi r$ divided by the speed — algebra collapses that into Kepler's 3rd law.

WHAT: invert Kepler's 3rd law to go from period → radius. $r={\left(\frac{\mu T^2}{4{\pi }^2}\right)}^{1/3}.$ WHY invert: we know the time we want (one Earth-spin) and are solving for distance. $\mu T^2=3.986\times 10^{14}\times (86164{)}^2=3.986\times 10^{14}\times 7.4242\times 10^9=2.9593\times 10^{24}$. Divide by $4{\pi }^2=39.478$: $7.4959\times 10^{22}$. Cube root: $r\approx 4.216\times 10^7$ m $=42164$ km. $h=42164-6371\approx 35793\text{km}$ (≈35 786 km at book precision).

WHAT: same inversion as L2.3, new $T$. $\mu T^2=3.986\times 10^{14}\times (43082{)}^2=3.986\times 10^{14}\times 1.8561\times 10^9=7.398\times 10^{23}$. $/4{\pi }^2=1.874\times 10^{22}$; cube root $r\approx 2.656\times 10^7$ m $=26560$ km. $h=26560-6371\approx 20190\text{km}$. WHY the check: $20190$ km is between 2000 and 35 786 km → MEO ✓. Notice halving the period doesn't halve the radius: $r\propto T^{2/3}$, so radius shrinks much more gently than period.

The figure shows the geometry we are using: perigee (near Earth) and apogee (far) sit at opposite ends of the long axis, Earth is at a focus, and the two radii $r_p$ and $r_a$ we relate below are drawn from that focus. Read $r_a=2a-r_p$ straight off the picture — the whole long axis is $2a$, and knocking off the short piece $r_p$ leaves the long piece $r_a$.

WHAT: use $a=\frac{1}{2}(r_p+r_a)$, i.e. $r_a=2a-r_p$. $r_p=6371+600=6971$ km. $2a=2\times 26560=53120$ km. $r_a=53120-6971=46149$ km $\Rightarrow h_a=46149-6371\approx 39780\text{km}$. WHY this works: the semi-major axis is by definition the average of the two extreme radii — that's the geometry of an ellipse, no forces needed. As a bonus, the eccentricity here is $e=\frac{r_a-r_p}{r_a+r_p}=\frac{46149-6971}{53120}\approx 0.74$ — a strongly stretched ellipse, exactly the Molniya signature.

WHAT: set the bracket to zero. ${\cos }^{2}i=1/5\Rightarrow \cos i=\pm 1/\sqrt{5}=\pm 0.4472$. Root 1 (positive cosine, prograde): $i=\arccos (+0.4472)\approx 63.43^{\circ }$. Root 2 (negative cosine, retrograde): $i=\arccos (-0.4472)\approx 116.57^{\circ }$. WHY this tool ($\arccos$): we know the cosine and want the angle back — $\arccos$ is exactly the "which angle has this cosine?" undo-button. Since the equation only involves ${\cos }^{2}i$, both a prograde tilt and its retrograde mirror ($180^{\circ }-63.43^{\circ }=116.57^{\circ }$) satisfy it. WHAT IT MEANS: both inclinations freeze the perigee — the physics genuinely permits either. Real Molniya orbits pick the prograde $63.43^{\circ }$ because it's cheaper to launch (you gain from Earth's eastward spin) and it parks apogee over the northern hemisphere; the retrograde $116.57^{\circ }$ twin would cost extra launch energy for no benefit here. This is the $J_2$ critical inclination that freezes the ellipse's orientation.

WHAT: track the sign of each factor. $J_2>0$, $(R_{\oplus }/p{)}^2>0$ (a square of a real length), and $n=\sqrt{\mu /a^3}>0$. The overall prefactor $-\frac{3}{2}{J}_2(\cdots )n$ is therefore negative. We need $\dot{\Omega }>0$ (positive/eastward). A negative prefactor times $\cos i$ must come out positive, so $\cos i$ must be negative. WHY it matters: $\cos i<0$ means $i>90^{\circ }$ — a retrograde orbit (typically ~98°). That "slightly backwards, near-polar" tilt is SSO's signature. Notice the point is Sun angle, not flying over the pole. Answer: retrograde, $i>90^{\circ }$.

WHAT: plug each $a$ (in metres) into $\epsilon =-\mu /(2a)$. LEO: $a=6.771\times 10^6$ m ⇒ $\epsilon =-\frac{3.986\times 10^{14}}{2(6.771\times 10^6)}=-2.943\times 10^7\text{J/kg}$. GEO: $a=4.2164\times 10^7$ m ⇒ $\epsilon =-\frac{3.986\times 10^{14}}{2(4.2164\times 10^7)}=-4.727\times 10^6\text{J/kg}$. WHY GEO costs more: GEO's $\epsilon$ is less negative (closer to zero), i.e. higher energy. Raising a satellite from LEO to GEO means adding energy. See Hohmann Transfer and Delta-v Budgets for the actual burns. WHY negative at all: a bound orbit has total energy below the escape threshold (which is $\epsilon =0$). Negative just means "gravitationally trapped." Answer: GEO needs more energy; energy is negative because the orbit is bound.

WHAT (step 1 — period): 15 orbits fit into one sidereal day, so $T=\frac{86164}{15}=5744.3\text{s}.$ WHY 15 into a sidereal day: the ground track repeats when the satellite's orbits and Earth's spin re-align in the star frame — that alignment period is the sidereal day. WHAT (step 2 — radius): invert Kepler. $\mu T^2=3.986\times 10^{14}\times (5744.3{)}^2=3.986\times 10^{14}\times 3.2997\times 10^7=1.3153\times 10^{22}$. $/4{\pi }^2=3.3318\times 10^{20}$; cube root $a\approx 6.936\times 10^6$ m $=6936$ km. WHAT (step 3 — altitude): $h=6936-6371\approx 565\text{km}$. WHY it lands in LEO: 565 km is inside 160–2000 km, exactly where imaging satellites live — deep enough for resolution, high enough to dodge the worst drag. See also Satellite Ground Tracks.

WHY vis-viva: the orbit is elliptical, so speed changes with position — the plain $\sqrt{\mu /r}$ (circles only) no longer applies. Vis-viva gives speed at any radius. Work in metres. Perigee: $\frac{2}{r_p}-\frac{1}{a}=\frac{2}{6.971\times 10^6}-\frac{1}{2.656\times 10^7}=2.869\times 10^{-7}-3.765\times 10^{-8}=2.493\times 10^{-7}$. $v_p=\sqrt{3.986\times 10^{14}\times 2.493\times 10^{-7}}=\sqrt{9.936\times 10^7}\approx 9968\text{m/s}\approx 9.97\text{km/s}.$ Apogee: $\frac{2}{r_a}-\frac{1}{a}=\frac{2}{4.6149\times 10^7}-3.765\times 10^{-8}=4.334\times 10^{-8}-3.765\times 10^{-8}=5.69\times 10^{-9}$. $v_a=\sqrt{3.986\times 10^{14}\times 5.69\times 10^{-9}}=\sqrt{2.268\times 10^6}\approx 1506\text{m/s}\approx 1.51\text{km/s}.$ WHAT IT MEANS: $v_p/v_a\approx 6.6$. The satellite screams through perigee and dawdles at apogee — that dawdle over the northern hemisphere is the whole point of Molniya, and it matches Kepler's 2nd law (equal areas swept in equal times ⇒ slow when far). Sanity: $r_p{v}_p=6.971\times 10^6\times 9968=6.95\times 10^{10}$ and $r_a{v}_a=4.6149\times 10^7\times 1506=6.95\times 10^{10}$ ✓ angular momentum conserved.