Exercises — Physical meaning of each orbital element

Passing over both poles means the orbit plane is tilted a full quarter-turn from the equator. That tilt is the inclination $i$. $i=90^{\circ }$ Why $i$ and not $\Omega$? $\Omega$ only says which compass direction the tilt faces; $i$ says how steep it is. Polar = maximum steepness = $90^{\circ }$.

Only ==$\nu$== (true anomaly) changes. The other five are constant because energy, the orbit plane, and the ellipse's shape/orientation are all conserved in the pure two-body problem. (Real perturbations — see Orbital perturbations — slowly drift the others, but that is beyond the ideal case.)

$e=\frac{r_a-r_p}{r_a+r_p}=\frac{0}{14400}=0$ A circle. And since $a=(r_p+r_a)/2=7200$ km equals the constant radius, here (and only here) $a$ literally is the distance to the satellite.

Convert altitudes to focal distances (add Earth's radius, because $r$ is measured from Earth's centre = the focus): $r_p=500+6378=6878\text{km},r_a=5500+6378=11878\text{km}$ $a=\frac{r_p+r_a}{2}=\frac{18756}{2}=9378\text{km}$ $e=\frac{r_a-r_p}{r_a+r_p}=\frac{5000}{18756}=0.2666$ Why add $R_{\oplus }$? Altitude is measured from the surface, but every orbital formula uses distance from the focus (Earth's centre).

Kepler's third law (Kepler's Laws): $T=2\pi \sqrt{\frac{a^3}{\mu }}=2\pi \sqrt{\frac{9378^3}{398600}}$ $a^3=8.248\times 10^{11}{\text{km}}^3,\frac{a^3}{\mu }=2.0693\times 10^6{\text{s}}^2$ $T=2\pi \sqrt{2.0693\times 10^6}=2\pi (1438.5)=9038\text{s}\approx 150.6\text{min}$ Why only $a$ appears? Period depends on energy, and energy depends only on $a$ — the eccentricity $e$ never enters $T$.

The Vis-viva equation gives speed at any radius: $v=\sqrt{\mu \left(\frac{2}{r}-\frac{1}{a}\right)}=\sqrt{398600\left(\frac{2}{6878}-\frac{1}{9378}\right)}$ $\frac{2}{6878}=2.9078\times 10^{-4},\frac{1}{9378}=1.0663\times 10^{-4}$ $v=\sqrt{398600\times 1.8415\times 10^{-4}}=\sqrt{73.40}=8.567\text{km/s}$ Why fastest here? Periapsis is the closest point, so kinetic energy is highest (Kepler's Laws, equal-areas → fast when near).

Compute the specific angular momentum $\vec{h}=\vec{r}\times \vec{v}$ (Angular momentum in orbits): $\vec{h}=\left|\begin{array}{ccc}\hat{I}& \hat{J}& \hat{K}\\ 7000& 0& 0\\ 0& 6& 4\end{array}\right|=(0\cdot 4-0\cdot 6,0\cdot 0-7000\cdot 4,7000\cdot 6-0)$ $\vec{h}=(0,-28000,42000){\text{km}}^2/\text{s}$ Magnitude: $h=\sqrt{0^2+28000^2+42000^2}=\sqrt{2.548\times 10^9}=50478{\text{km}}^2/\text{s}$. Inclination uses the $z$-component of $\vec{h}$ against its length: $\cos i=\frac{h_z}{h}=\frac{42000}{50478}=0.8320\Rightarrow i=33.69^{\circ }$ Why $h_z/h$? $\vec{h}$ is perpendicular to the orbit plane; $\hat{K}$ is perpendicular to the equator. The angle between the two normals is the angle between the two planes.

$\vec{n}=\hat{K}\times \vec{h}=(h\text{-based})=(-h_y,h_x,0)=(28000,0,0)$ $|\vec{n}|=28000$. $\cos \Omega =\frac{n_x}{|\vec{n}|}=\frac{28000}{28000}=1\Rightarrow \Omega =0^{\circ }\text{ or }360^{\circ }$ Quadrant check: $n_y=0$ (not $<0$), so $\Omega =0^{\circ }$. Why the sign of $n_y$? $\cos$ gives the same value for $\Omega$ and $360^{\circ }-\Omega$; only the $y$-sign tells us which side of the vernal-equinox axis the node lies on.

$\epsilon =\frac{v^2}{2}-\frac{\mu }{r}=\frac{100}{2}-\frac{398600}{8000}=50-49.825=+0.175{\text{km}}^2/{\text{s}}^2$ $\epsilon >0$, so the orbit is hyperbolic (unbound — it escapes). Check the escape speed at this radius: $v_{esc}=\sqrt{2\mu /r}=\sqrt{2\cdot 398600/8000}=\sqrt{99.65}=9.982$ km/s. Since $10.0>9.982$, the object is (just barely) escaping. ✓

Rearrange the orbit equation $r=\frac{a(1-e^2)}{1+e\cos \nu }$ for $\cos \nu$: $1+e\cos \nu =\frac{a(1-e^2)}{r}\Rightarrow \cos \nu =\frac{1}{e}\left(\frac{a(1-e^2)}{r}-1\right)$ Compute the semi-latus rectum $p=a(1-e^2)=9378(1-0.07108)=8711.4$ km. $\cos \nu =\frac{1}{0.2666}\left(\frac{8711.4}{8000}-1\right)=\frac{1}{0.2666}(0.08893)=0.3336$ $\nu =\arccos (0.3336)=70.51^{\circ }\text{or}289.49^{\circ }$ Since the satellite moves outward ($\vec{r}\cdot \vec{v}>0$), it is between periapsis and apoapsis on the ascending half, so $0<\nu <180^{\circ }$: $\boxed{\nu =70.51^{\circ }}$ Why the direction matters: the same $r$ occurs twice per orbit (once climbing out, once falling in). Only the radial-velocity sign distinguishes them.

$\cos \omega =\frac{\vec{n}\cdot \vec{e}}{|\vec{n}||\vec{e}|}$ $\vec{n}\cdot \vec{e}=1(0.2)+0(0.2)+0(0.1)=0.2$. $|\vec{e}|=\sqrt{0.2^2+0.2^2+0.1^2}=\sqrt{0.09}=0.3$. $\cos \omega =\frac{0.2}{1\times 0.3}=0.6667\Rightarrow \omega =48.19^{\circ }$ Quadrant check: $e_z=0.1>0$, so periapsis is above the reference plane and $\omega <180^{\circ }$. Keep $\omega =48.19^{\circ }$. Why $\vec{e}$? Its direction is the line toward periapsis and its magnitude is $e$ — so it carries both shape and in-plane pointing in one vector (Eccentricity vector).

Step 1 — energy → $a$. $r=8000$, $v=7.5$. $\epsilon =\frac{v^2}{2}-\frac{\mu }{r}=\frac{56.25}{2}-\frac{398600}{8000}=28.125-49.825=-21.70{\text{km}}^2/{\text{s}}^2$ Negative ⇒ bound. Then $a=-\frac{\mu }{2\epsilon }=-\frac{398600}{2(-21.70)}=9184\text{km}$.

Step 2 — eccentricity vector. $\vec{e}=\frac{1}{\mu }\left[\left(v^2-\frac{\mu }{r}\right)\vec{r}-(\vec{r}\cdot \vec{v})\vec{v}\right]$ Here $\vec{r}\cdot \vec{v}=8000(0)+0(7.5)+0=0$ (velocity purely tangential ⇒ we are at an apse!). $v^2-\mu /r=56.25-49.825=6.425$. $\vec{e}=\frac{1}{398600}\left[6.425(8000,0,0)-0\right]=\frac{(51400,0,0)}{398600}=(0.1290,0,0)$ So $e=0.1290$.

Step 3 — true anomaly. $\vec{e}$ points along $+x$, and $\vec{r}$ also points along $+x$, so $\vec{r}$ is at periapsis: $\cos \nu =\frac{\vec{e}\cdot \vec{r}}{|\vec{e}|r}=\frac{0.1290\cdot 8000}{0.1290\cdot 8000}=1\Rightarrow \nu =0^{\circ }$ Confirmed by $\vec{r}\cdot \vec{v}=0$ with speed high ⇒ periapsis. Answers: $a=9184$ km, $e=0.129$, $\nu =0^{\circ }$, bound. ✓

Invert Kepler III for $a$: $T=2\pi \sqrt{\frac{a^3}{\mu }}\Rightarrow a^3=\mu {\left(\frac{T}{2\pi }\right)}^2$ $\frac{T}{2\pi }=\frac{7200}{6.28319}=1145.92\text{s},{\left(\frac{T}{2\pi }\right)}^2=1.31313\times 10^6$ $a^3=398600\times 1.31313\times 10^6=5.2341\times 10^{11}{\text{km}}^3$ $a=(5.2341\times 10^{11}{)}^{1/3}=8059\text{km}$ Altitude $=a-R_{\oplus }=8059-6378=1681\text{km}$. Why invert Kepler III? Period is the design goal; $a$ is the knob that sets it, and for a circle altitude follows directly since $r=a$.

$a_A=\frac{7000+11000}{2}=9000\text{km},a_B=\frac{8000+10000}{2}=9000\text{km}$ Equal $a$ ⇒ equal energy ⇒ equal period: $T=2\pi \sqrt{\frac{9000^3}{398600}}=2\pi \sqrt{1.8290\times 10^6}=8497\text{s}\approx 141.6\text{min (both)}$ Eccentricities differ: $e_A=\frac{11000-7000}{18000}=0.2222,e_B=\frac{10000-8000}{18000}=0.1111$ Takeaway: identical $a$ fixes the clock; $e$ only reshapes the path the satellite takes during that fixed time. This is the concrete pay-off of "energy depends only on $a$."

Which element answers "how long is the lap"? ::: $a$ (via $T=2\pi \sqrt{a^3/\mu }$) Which sign test disambiguates $\nu$? ::: sign of $\vec{r}\cdot \vec{v}$ (outward $>0$ ⇒ $\nu <180^{\circ }$) What is the numerator of the orbit equation? ::: the semi-latus rectum $p=a(1-e^2)$, not $a$ Positive specific energy $\epsilon$ means what orbit? ::: hyperbolic (unbound / escaping) Two orbits, same $a$, different $e$ — same or different period? ::: same period (energy depends only on $a$)