Question bank — Atmospheric drag — exponential atmosphere model, orbit decay

False — locally it removes energy, but a smaller orbit ($v=\sqrt{\mu /a}$) means a higher circular speed, so over an orbit the satellite ends up faster, not slower.

True — $a$ shrinks, so $-\mu /2a$ becomes more negative (smaller); drag is a sink of mechanical energy.

False — ${\vec{v}}^2$ is not a legal vector; the force is $-\frac{1}{2}B\rho v\vec{v}$, so its magnitude scales as $v^2$ while its direction stays $-\hat{v}$.

True — $B=C_{D}A/m$ shrinks as $m$ grows, and $da/dt=-B\rho \sqrt{\mu a}$ is proportional to $B$, so more mass per unit area resists drag better.

False — $H$ is the vertical distance over which density falls by a factor $e\approx 2.72$; the atmosphere never sharply "ends," it just keeps thinning exponentially.

True — $H=k_{B}T/(m_{mol}g)$, so $H\propto 1/m_{mol}$; heavier molecules settle lower and the density profile drops off faster.

False — as $a$ drops, altitude drops, and $\rho$ climbs exponentially, so $|da/dt|\propto \rho$ blows up; decay is strongly nonlinear and ends in a rapid plunge.

False — drag is not central and always points along $-\vec{v}$, so its work ${\vec{a}}_{drag}\cdot \vec{v}=-\frac{1}{2}B\rho v^3$ is negative every instant; it never restores energy.

True — it starts from $dP=-\rho gdh$ (pressure balances weight); combined with the ideal gas law at constant $T$ it gives the exponential, exactly as in Hydrostatic Equilibrium (Atmospheres & Stars).

False — $v=\sqrt{\mu /a}$, so halving $a$ multiplies $v$ by $\sqrt{2}\approx 1.41$, not by 2; speed scales as $a^{-1/2}$.

The magnitude is tiny per second, but it acts continuously for months; the cumulative energy loss is enormous, so drag ultimately dominates the fate of any LEO satellite.

Wrong — $\rho (h)={\rho }_0{e}^{-(h-h_0)/H}$ changes by orders of magnitude as altitude drops; a fixed $\rho$ badly underestimates the runaway final phase. You must use $\rho (a)$.

Backwards — lower energy means a smaller orbit, and a smaller orbit is faster. Energy and speed move in opposite directions here; only the altitude and energy fall.

$H\propto 1/g$, so larger $g$ gives a smaller $H$; but near-surface warming/cooling changes $T$ far more, so real profiles are dominated by the temperature structure, not $g$.

No violation — kinetic energy rises but potential energy $-\mu /a$ falls twice as fast, so total $\epsilon =-\mu /2a$ decreases. The lost energy goes into heating the atmosphere and satellite.

For a near-circular orbit drag mainly circularizes it — it bites hardest at perigee (densest air), lowering apogee and pulling the orbit rounder before it decays; it does not pump up eccentricity.

$B$ depends entirely on the satellite — its drag coefficient, cross-sectional area, and mass. The atmosphere enters separately through $\rho$. (See Reentry Aerodynamics & Ballistic Coefficient.)

Because hydrostatic balance plus the ideal gas law gives $d\rho /\rho =-dh/H$, whose only solution is exponential; density falls by a constant fraction per fixed height, not a constant amount.

Density is exponential, so even a modest altitude difference between perigee and apogee makes $\rho$ enormously larger at perigee; drag $\propto \rho v^2$ concentrates there, like a brief "brake tap" each pass.

They're identical ($\vec{v}=v\hat{v}$), but $v\vec{v}$ is cleaner to differentiate/integrate in equations of motion and makes the $v^2$ magnitude and $-\hat{v}$ direction both explicit.

As $a$ shrinks the satellite drops into far denser air; $\rho$ grows exponentially while $\sqrt{\mu a}$ barely changes, so $|da/dt|=B\rho \sqrt{\mu a}$ runs away — the last few kilometers happen in minutes. See Perturbations in Orbital Mechanics.

Drag only tells us the rate of energy loss; to convert that into altitude change we need $\epsilon =-\mu /2a$ and $v=\sqrt{\mu /a}$ from the Two-Body Problem and Vis-Viva Equation to link energy, speed, and orbit size.

$H=k_{B}T/(m_{mol}g)$; the thermosphere is very hot ($T\sim 1000$ K) and made of light atomic species, so it "puffs up" — large $T$ and small $m_{mol}$ both raise $H$.

Because a bound orbit's energy is negative; making it more negative shrinks the orbit and (via $v=\sqrt{\mu /a}$) speeds the body up — it can only "stop" by hitting the atmosphere densely enough to reenter, not by coasting to rest.

Zero — with no atmosphere there's no drag and the orbit is stable forever; the whole decay story is driven entirely by $\rho >0$.

$\rho$ has grown by many $e$-foldings and is huge, so $|da/dt|$ is enormous; physically the satellite has already reached the dense lower atmosphere and burns up — the near-circular model breaks down here.

Infinite — $B=C_{D}A/m=0$ makes $a_{drag}=0$, so no energy is lost and the orbit never decays, regardless of density.

Decay is astronomically slow but not exactly zero; also at extreme altitudes the mean free path is huge (free-molecular flow), so the continuum drag formula $\frac{1}{2}B\rho v^2$ becomes only an approximation.

Then $B=C_{D}A/m$ is not constant; the effective drag averages over the tumble, and an unpredictable spinning body has a much harder-to-forecast lifetime than a stable, fixed-attitude one.

Kinetic energy rises ($v$ increases as $a$ falls), even though total energy falls; the extra KE comes from the far larger drop in potential energy $-\mu /a$.