Exercises — Coordinate systems — Earth-Centered Inertial (ECI), Earth-Centered Earth-Fixed (ECEF), North-East-Down (NED), launch, bo

Answer: ECEF (Earth-Centered Earth-Fixed). ECEF's axes are glued to the spinning Earth. Since the pad rides along with the ground, it never moves relative to those axes — so its coordinates stay fixed. In ECI (whose axes stay pointed at the distant stars) the same pad traces a big circle once per day, so its ECI coordinates change constantly.

Answer: ECI. $\vec{F}=m\vec{a}$ only holds in a frame that is neither rotating nor accelerating — an inertial frame. Of our five, only ECI qualifies (its axes point at fixed stars). ECEF spins; NED and body ride a curving, turning trajectory; the launch frame is approximately inertial but only for a short flight.

• (a) → body frame — sensors are bolted to the rocket, so they measure along $x_b,y_b,z_b$.
• (b) → NED — "how high" and "which way is the horizon" are literally the Down and North/East directions.
• (c) → ECI — the only inertial frame, where $\vec{F}=m\vec{a}$ needs no fictitious forces.

Why this formula: only the part of the pad's position perpendicular to the spin axis actually sweeps around. That perpendicular distance is $R_{\oplus }\cos \varphi$ (the radius of the circle of latitude). Speed = angular rate × radius = ${\omega }_e{R}_{\oplus }\cos \varphi$. Step 1 — cosine: $\cos 28.5°=0.8788$. Step 2 — multiply: $v=7.292\times 10^{-5}\times 6.378\times 10^6\times 0.8788\approx 408.7\text{m/s}.$ So launching eastward from the Cape hands the rocket a free $\approx 409\text{m/s}$ head start. (Compare $465\text{m/s}$ at the equator — the boost shrinks as you go north because $\cos \varphi$ shrinks.)

Step 1 — seconds: $t=4\times 3600=14400\text{s}$. Step 2 — angle: $\theta =7.292\times 10^{-5}\times 14400=1.0500\text{rad}$. Step 3 — degrees: $1.0500\times 57.30\approx 60.16°$. Sanity check: $15°/\text{hr}\times 4\text{hr}=60°$. ✔️

With $\theta =90°$: $\cos \theta =0$, $\sin \theta =1$.

Why "Down" is that row: "Down" points straight to Earth's center, i.e. opposite the outward radial direction $\hat{r}$. The outward radial at $(\varphi ,\lambda )$ has ECEF components $(\cos \varphi \cos \lambda ,\cos \varphi \sin \lambda ,\sin \varphi )$, so Down is the negative of that. Step 1 — plug in $\varphi =45°,\lambda =0$: $\cos 45°=\sin 45°=0.7071$, $\cos 0=1,\sin 0=0$. $\hat{D}=(-0.7071,0,-0.7071).$ Step 2 — check length: $\sqrt{0.7071^2+0+0.7071^2}=\sqrt{0.5+0.5}=1$. ✔️ What it looks like (figure below): the lavender quarter-circle is Earth's surface drawn in the $X_E$–$Z_E$ plane. The coral dot is the 45°N point; the mint arrow is the outward radial $\hat{r}$ (up); the coral arrow is $\hat{D}=-\hat{r}$, tilted halfway between $-X_E$ (toward the equatorial plane) and $-Z_E$ (toward the south pole). The butter arc marks $\varphi =45°$. Both components being equal in size is exactly what "45°" should produce.

Edge cases (cover every latitude):

• North pole $\varphi =90°$: $\cos 90°=0,\sin 90°=1$, so $\hat{D}=(0,0,-1)=-{\hat{Z}}_E$. Down points straight toward the south pole — correct, since standing on the north pole "down" is along the axis toward Earth's center.
• South pole $\varphi =-90°$: $\sin (-90°)=-1$, so $\hat{D}=(0,0,+1)=+{\hat{Z}}_E$. Down now points toward $+Z$ (north pole) — again correct: from the south pole, center-ward is the $+Z$ direction. Note the sign of the $z$-component flipped with the sign of $\varphi$; the formula handles it automatically through $-\sin \varphi$.
• Southern hemisphere $\varphi =-45°,\lambda =0$: $\cos (-45°)=0.7071,\sin (-45°)=-0.7071$, giving $\hat{D}=(-0.7071,0,+0.7071)$. Compared to the $45°$N case only the $z$-component changed sign — Down now tilts toward $+Z_E$ (northward-and-inward), the mirror image across the equator. Length is still $1$. So the single expression is valid for all $\varphi \in [-90°,90°]$ and all $\lambda$; nothing special happens at the poles.

If ${\vec{v}}_E=0$, the naive rule gives ${\vec{v}}_I=0$ — "the pad isn't moving." But the transport theorem keeps ${\vec{\omega }}_e\times {\vec{r}}_E$. Let us compute that cross product component by component, not just quote its size. Set up the vectors in ECEF: the spin is along $Z$, ${\vec{\omega }}_e=(0,0,{\omega }_e)$, and an equatorial pad on the $X_E$ axis is ${\vec{r}}_E=(R_{\oplus },0,0)$. Cross product (using $\hat{i}\times \hat{j}=\hat{k}$ etc.):

(a) $\cos 30°=0.8660$, $\sin 30°=0.5$.

Step 1 — Kourou: $\cos 5.2°=0.99589$, so $v_K=7.292\times 10^{-5}\times 6.378\times 10^6\times 0.99589=463.2\text{m/s}$. Step 2 — Cape: from Ex 2.1, $v_C=408.7\text{m/s}$. Step 3 — difference: $\Delta v=463.2-408.7=54.5\text{m/s}$. Interpretation: being $23°$ closer to the equator buys Kourou about $55\text{m/s}$ of free velocity — one concrete reason equatorial launch sites are prized.

With ${\vec{v}}_E=0$, ${\vec{v}}_I=R_z^T({\vec{\omega }}_e\times {\vec{r}}_E)$. Rotation ($R_z^T$) preserves magnitude, so $|{\vec{v}}_I|=|{\vec{\omega }}_e\times {\vec{r}}_E|={\omega }_e(R_{\oplus }\cos \varphi ),$ because (as shown in Ex 3.2) the cross product keeps only the perpendicular-to-axis part $R_{\oplus }\cos \varphi$. $|{\vec{v}}_I|=7.292\times 10^{-5}\times 6.378\times 10^6\times \cos 28.5°=408.7\text{m/s}.$ Direction: ${\vec{\omega }}_e\times {\vec{r}}_E$ with ${\vec{\omega }}_e$ up the spin axis points eastward (tangent to the circle of latitude). So the "motionless" pad is really sailing east at $\approx 409\text{m/s}$ in inertial space — identical to Ex 2.1, confirming the boost is this inertial velocity.

Step 1 — period: $T=2\pi /{\omega }_e=6.2832/7.292\times 10^{-5}=86166\text{s}$. Step 2 — hours: $86166/3600=23.935\text{h}\approx 23\text{h}56\text{min}$. Why under 24 h: a solar day (24 h) is the time for the Sun to return overhead. But Earth also orbits the Sun, so it must spin a bit extra each day to catch up to the moving Sun. The pure star-relative (sidereal) spin is therefore about 4 minutes shorter. ECI is tied to the stars, so ${\omega }_e$ is the sidereal rate.

3rd row $=(-\cos \varphi \cos \lambda ,-\cos \varphi \sin \lambda ,-\sin \varphi )$. Step 1 — evaluate the trig. With $\varphi =0$: $\cos \varphi =1,\sin \varphi =0$. With $\lambda =90°$: $\cos 90°=0,\sin 90°=1$. Step 2 — substitute: $\hat{D}=(-1\cdot 0,-1\cdot 1,-0)=(0,-1,0).$ Step 3 — check length: $\sqrt{0^2+(-1{)}^2+0^2}=1$. ✔️ Step 4 — interpret the direction. This is exactly $-{\hat{Y}}_E$. A point at $90°$E on the equator sticks out along $+Y_E$; "straight down" from there means heading toward Earth's center, i.e. in the $-Y_E$ direction. There is no $Z$ component because on the equator "down" lies entirely in the equatorial plane — consistent with the pole cases in Ex 3.1, where instead the $Z$ component carried everything. So the same 3rd-row formula smoothly interpolates from all-$Z$ at the poles to all-horizontal on the equator.

L1: name the frame. L2: plug the boost/rotation formula. L3: explain the signs & the extra transport term. L4: chain and invert transforms correctly. L5: use sidereal timing and the transport theorem to get true inertial velocity.