Exercises — Atmospheric drag — exponential atmosphere model, orbit decay

WHAT: $B$ is drag-area per unit mass, $B=\frac{C_{D}A}{m}$. WHY: It bundles the three "how draggy" numbers into one, so decay formulas need only $B$. $B=\frac{0.30}{15}=0.020{\text{m}}^2/\text{kg}$

Tool choice: We want a magnitude, so use $a_{drag}=\frac{1}{2}B\rho v^2$ (the $v^2$ form, no vector needed). $a_{drag}=\frac{1}{2}(0.020)(5\times 10^{-12})(7670{)}^2$ $=\frac{1}{2}(0.020)(5\times 10^{-12})(5.883\times 10^7)\approx 2.94\times 10^{-6}{\text{m/s}}^2$ That is about $3\times 10^{-7}g$ — invisible per second, ruinous over months.

Tool: $\rho (h)={\rho }_0{e}^{-(h-h_0)/H}$. Dropping means $h<h_0$, so $(h-h_0)$ is negative and density rises.

• Drop $50$ km $=1H$: factor $e^{+1}\approx 2.718$ — density multiplies by $\approx 2.72$.
• Drop $150$ km $=3H$: factor $e^{+3}\approx 20.09$ — density multiplies by $\approx 20.1$. Reading: every scale height down multiplies $\rho$ by $e$. Three of them = a 20× thicker soup.

Tool: $\frac{da}{dt}=-B\rho \sqrt{\mu a}$ (rate of shrinkage of the orbit). First the root: $\mu a=(3.986\times 10^{14})(6.778\times 10^6)=2.702\times 10^{21}$ $\sqrt{\mu a}=5.198\times 10^{10}{\text{m}}^2/\text{s}$ Then: $\frac{da}{dt}=-(0.020)(5\times 10^{-12})(5.198\times 10^{10})=-5.198\times 10^{-3}\text{m/s}$ Convert to per day ($86400$ s): $-5.198\times 10^{-3}\times 86400\approx -449\text{m/day}$ So the orbit drops roughly $0.45$ km each day — at this altitude.

Tool: $H=\frac{k_{B}T}{m_{mol}g}$ — derived from hydrostatic equilibrium plus ideal gas (see parent). $H=\frac{(1.380\times 10^{-23})(1000)}{(2.66\times 10^{-26})(8.7)}=\frac{1.380\times 10^{-20}}{2.314\times 10^{-25}}\approx 5.96\times 10^4\text{m}\approx 60\text{km}$ Reading: hot ($T$ big) and light ($m_{mol}$ small) air puffs up — a fat scale height, exactly why the thermosphere is so extended.

Tools: $\epsilon =-\frac{\mu }{2a}$ and $v=\sqrt{\frac{\mu }{a}}$ (from Two-Body Problem / Vis-Viva Equation). $\epsilon =-\frac{3.986\times 10^{14}}{2(6.778\times 10^6)}=-2.941\times 10^7\text{J/kg}$ $v=\sqrt{\frac{3.986\times 10^{14}}{6.778\times 10^6}}=\sqrt{5.882\times 10^7}\approx 7670\text{m/s}$ Consistent with the speed we assumed back in Problem 1.2 — good.

Step 1 — power = force · velocity. Drag acceleration is ${\vec{a}}_{drag}=-\frac{1}{2}B\rho v\vec{v}$ (magnitude $\frac{1}{2}B\rho v^2$, direction $-\hat{v}$). Power per unit mass: $\frac{d\epsilon }{dt}={\vec{a}}_{drag}\cdot \vec{v}=-\frac{1}{2}B\rho v(\vec{v}\cdot \vec{v})=-\frac{1}{2}B\rho v\cdot v^2=-\frac{1}{2}B\rho v^3.$ WHY negative: ${\vec{a}}_{drag}$ is anti-parallel to $\vec{v}$, so the dot product is $-|\vec{a}||\vec{v}|$ — energy always leaves. Step 2 — cross-check via $a$. Differentiate $\epsilon =-\mu /2a$: $\frac{d\epsilon }{dt}=\frac{\mu }{2a^2}\frac{da}{dt}.$ Substitute $\frac{da}{dt}=-B\rho \sqrt{\mu a}$: $\frac{d\epsilon }{dt}=\frac{\mu }{2a^2}(-B\rho \sqrt{\mu a})=-\frac{B\rho }{2}\frac{\mu \sqrt{\mu a}}{a^2}=-\frac{1}{2}B\rho \frac{{\mu }^{3/2}}{a^{3/2}}.$ Since $v=\sqrt{\mu /a}$, we have $v^3=(\mu /a{)}^{3/2}={\mu }^{3/2}/a^{3/2}$, so this is exactly $-\frac{1}{2}B\rho v^3$. ✓ The two formulas agree.

WHAT dominates: in $\left|\frac{da}{dt}\right|=B\rho \sqrt{\mu a}$, the density $\rho$ changes exponentially while $\sqrt{\mu a}$ changes only as a gentle square root. So the ratio at two altitudes is essentially the density ratio. WHY: dropping altitude $\Delta h$ multiplies density by $e^{+\Delta h/H}$ (from Problem 1.3's sign logic).

• One full scale height ($\Delta h=H$): factor $e^1\approx 2.72$.
• $\Delta h=30$ km, $H=60$ km: factor $e^{30/60}=e^{0.5}\approx 1.65$. Reading: every $60$ km of descent nearly triples the shrink rate, and thirty km already boosts it by $65\%$. That compounding is the "runaway" — the lower you fall, the faster you fall.

WHERE the density is largest: perigee is the lowest point, so $\rho (r_p)\gg \rho (r_a)$ — exponentially larger. Drag magnitude $\propto \rho v^2$, and $v$ is also largest at perigee, so the drag "kick" is overwhelmingly concentrated in the brief perigee pass. WHAT it does to the shape: a drag impulse at perigee removes energy where the satellite is deepest, which lowers the apogee (the far side) most strongly, while perigee barely moves. Over many orbits the apogee walks down toward the perigee — the ellipse circularizes. See Perturbations in Orbital Mechanics. Then: once nearly circular and low, every point sits in thick air, $\rho$ is uniformly high, and the final plunge follows.

Start from $\epsilon =-\mu /2a$. Take the natural log of the magnitude: $\ln |\epsilon |=\ln (\mu /2)-\ln a$. Differentiate in time: $\frac{1}{\epsilon }\frac{d\epsilon }{dt}=-\frac{1}{a}\frac{da}{dt}.$ Interpretation: $\frac{da}{dt}<0$ (orbit shrinks), so the right side is positive; but $\epsilon <0$, so a positive $\frac{1}{\epsilon }\frac{d\epsilon }{dt}$ means $\frac{d\epsilon }{dt}<0$ — energy decreases. Consistent: as $a$ falls, $\epsilon$ becomes more negative (deeper well), and speed rises. This is the drag paradox stated as a clean identity. See Orbital Energy & Specific Mechanical Energy.

Local estimate: at constant $449$ m/day, $t=\frac{\Delta a}{|da/dt|}=\frac{50000\text{m}}{449\text{m/day}}\approx 111\text{days}$ to fall the first $50$ km. WHY the true lifetime is shorter: the rate is not constant. As $a$ drops, altitude drops, $\rho$ climbs exponentially (Problem 3.2), so $|da/dt|$ grows and each successive $50$ km takes less time. Summing a shrinking sequence of times gives a total far below $111\text{days}\times (\text{number of 50-km steps})$. The constant-rate number is only an upper-bound feel for the first step.

Tool: $\frac{da}{dt}=-B\rho \sqrt{\mu a}$. Same orbit ⇒ same $\rho$, $\mu$, $a$. So the ratio of decay rates is just the ratio of $B$: $\frac{(da/dt{)}_Y}{(da/dt{)}_X}=\frac{B_Y}{B_X}=\frac{0.5}{0.005}=100.$ Y decays $100\times$ faster instantaneously, so (to first order) Y's lifetime is roughly $1/100$ of X's. Reading: big-area/light-mass bodies (sails, balloons) are natural deorbit devices — this is exactly the drag-sail deorbit principle. See Reentry Aerodynamics & Ballistic Coefficient.

From $\frac{da}{dt}=-B\rho (a)\sqrt{\mu a}$, separate variables: $dt=\frac{da}{-B\rho (a)\sqrt{\mu a}}\Rightarrow t={\int }_{a_1}^{a_2}\frac{-da}{B{\rho }_1{e}^{-(a-a_1)/H}\sqrt{\mu a}}.$ Since $a_2<a_1$ (decaying), flip limits to make it positive: $t={\int }_{a_2}^{a_1}\frac{e^{+(a-a_1)/H}}{B{\rho }_1\sqrt{\mu a}}da.$ Which end dominates: the integrand carries $e^{+(a-a_1)/H}$, which is largest near $a=a_1$ (the high, thin-air start) where $\rho$ is smallest, so $dt$ per unit $da$ is largest there. Almost the entire lifetime is spent slowly bleeding down through the top few scale heights. By the time the satellite is deep, $\rho$ is huge, $dt$ is tiny, and the final kilometres flash by. This matches the "long quiet decay, sudden fiery end" picture.

• (a) $\rho \to 0$: $\frac{da}{dt}\to 0$ — no air, no decay. The orbit is eternal, exactly the frictionless Two-Body Problem limit. ✓
• (b) $B\to 0$: $\frac{da}{dt}\to 0$ — a body with zero drag-area-per-mass (a cannonball of infinite density) ignores the atmosphere. ✓ This is why heavy compact objects survive reentry: tiny $B$.
• (c) $a\to R_{\oplus }$: $\sqrt{\mu a}$ stays finite, but the model's $\rho (a)={\rho }_1{e}^{+(a_1-a)/H}$ blows up as $a$ falls — decay rate diverges. Physically: the satellite is no longer orbiting, it's plowing through dense air and reenters. The circular-orbit assumption ($v=\sqrt{\mu /a}$) has broken; you must switch to a ballistic-reentry model. The divergence is the model politely telling you it has expired. See Reentry Aerodynamics & Ballistic Coefficient.

(i) Speed: $v=\sqrt{\mu /a_1}=\sqrt{3.986\times 10^{14}/6.778\times 10^6}\approx 7670\text{m/s}$ (as in 2.3). (ii) Initial rate: $|da/dt|=B{\rho }_1\sqrt{\mu a_1}=(0.02)(5\times 10^{-12})(5.198\times 10^{10})\approx 5.20\times 10^{-3}\text{m/s}\approx 449\text{m/day}$ (as in 2.1). (iii) After $\Delta h=120$ km $=2H$: density rises by $e^2\approx 7.389$: $\rho =5\times 10^{-12}\times e^2\approx 3.69\times 10^{-11}{\text{kg/m}}^3.$ New $a\approx 6.658\times 10^6$ m, so $\sqrt{\mu a}\approx \sqrt{(3.986\times 10^{14})(6.658\times 10^6)}=\sqrt{2.654\times 10^{21}}\approx 5.152\times 10^{10}$. Then: $|da/dt|=(0.02)(3.69\times 10^{-11})(5.152\times 10^{10})\approx 3.80\times 10^{-2}\text{m/s}\approx 3287\text{m/day}.$ The rate jumped from $\sim 0.45$ km/day to $\sim 3.3$ km/day — roughly $7\times$ faster, essentially the $e^2$ density factor (the $\sqrt{\mu a}$ barely moved). (iv) One sentence: because the decay rate itself accelerates exponentially as the orbit sinks, a single boost only resets the clock — without repeated reboosts the satellite always slides back into the thickening air and reenters. See Perturbations in Orbital Mechanics.