Atmospheric drag — exponential atmosphere model, orbit decay

WHY does drag matter and WHAT is it?

WHAT: Drag is a force opposing motion through a fluid (here, the residual upper atmosphere).

WHY it's subtle: In orbit, kinetic and potential energy are linked. Removing energy shrinks the orbit, and a smaller orbit means a higher speed. So drag = decay = eventual reentry.

HOW: the drag force from first principles

Imagine the satellite (cross-section $A$) sweeping through air of density $\rho$ at speed $v$. In time $dt$ it sweeps a tube of length $v\,dt$, containing mass:

$dm_{air} = \rho \, A \, v \, dt$

If it imparts (roughly) speed $v$ to that gas, momentum given away per time is:

$F \sim \frac{dm_{air}}{dt}\,v = \rho A v^2$

Real bodies transfer only a fraction, captured by the drag coefficient $C_D$ (and the factor $\tfrac12$ from the standard aerodynamic definition):

Why this step? We wrote $v\,\vec v$ (not $\vec v^2$) so the direction is $-\hat v$ while the magnitude scales as $v^2$.

HOW: the exponential atmosphere model

WHY exponential? In hydrostatic equilibrium a slab of air is held up by pressure difference against gravity:

$dP = -\rho\, g\, dh$

Treating the upper air as an ideal gas at (roughly) constant temperature $T$:

$P = \frac{\rho k_B T}{m_{mol}} \quad\Rightarrow\quad dP = \frac{k_B T}{m_{mol}}\,d\rho$

Substitute:

$\frac{k_B T}{m_{mol}}\,d\rho = -\rho g\, dh \;\Rightarrow\; \frac{d\rho}{\rho} = -\frac{m_{mol}\,g}{k_B T}\,dh$

Integrate → exponential:

Why this step? The clean $d\rho/\rho = -dh/H$ integrates directly to an exponential — density falls off fast, which is why drag is essentially a surface phenomenon near perigee.

HOW: orbit decay — energy & altitude loss

For a near-circular orbit, total specific energy:

$\varepsilon = -\frac{\mu}{2a}, \qquad v = \sqrt{\frac{\mu}{a}} \;(\text{circular})$

where $\mu = GM_\oplus$ and $a$ = semi-major axis.

Drag does negative work per unit time (power):

$\frac{d\varepsilon}{dt} = \vec a_{drag}\cdot\vec v = -\tfrac12 B\rho v^2 \cdot v = -\tfrac12 B \rho v^3$

Why this step? Power = force · velocity; drag is anti-parallel to $\vec v$, so the work is negative.

Now link $\varepsilon$ to $a$: since $\varepsilon = -\mu/2a$,

$\frac{d\varepsilon}{dt} = \frac{\mu}{2a^2}\frac{da}{dt}$

Set equal and use $v^2 = \mu/a$, so $v^3 = (\mu/a)^{3/2}$:

$\frac{\mu}{2a^2}\frac{da}{dt} = -\tfrac12 B \rho \left(\frac{\mu}{a}\right)^{3/2}$

Worked examples

Common mistakes

Active recall

Connections

• Two-Body Problem — provides $v=\sqrt{\mu/a}$ and $\varepsilon=-\mu/2a$ used here.
• Vis-Viva Equation — general speed–radius relation behind the drag paradox.
• Orbital Energy & Specific Mechanical Energy — the quantity drag removes.
• Perturbations in Orbital Mechanics — drag as a non-conservative perturbing force.
• Reentry Aerodynamics & Ballistic Coefficient — where decay ends.
• Hydrostatic Equilibrium (Atmospheres & Stars) — origin of the exponential model.

Concept Map

Hinglish (regional understanding)