Dynamics in Differential Form (Variable Mass, Drag Forces)

The high-school statement of Newton's second law, $F = ma$, is a special case. The general statement is a differential equation for the momentum,

and two large classes of problem force you to take it seriously in that form. The first is motion with a velocity-dependent force — air resistance, viscous drag, the magnetic force — where $\vec{F}$ is a function of $\vec{v}$ and the equation of motion is a genuine first-order ODE that must be integrated, not merely evaluated. The second is variable-mass dynamics — rockets, falling raindrops that accrete water, chains piling on the floor — where $m$ itself changes and applying "$F = ma$" naively gives the wrong answer.

This chapter treats both with the standard tools: separation of variables for first-order ODEs, and careful momentum bookkeeping over an infinitesimal time interval for the variable-mass case. The reading is Morin's Introduction to Classical Mechanics, chapters on drag and on the rocket equation.

The discipline this chapter demands. Write the equation of motion first, as a differential equation, before substituting numbers. Identify its type (separable? linear? constant terminal state?), integrate it symbolically, and only then put in the data. Guessing the form of $v(t)$ is how sign errors and missing time constants creep in.

1. Newton's Second Law as a Differential Equation

For a particle of constant mass, $\vec{F} = \dfrac{d(m\vec{v})}{dt} = m\dfrac{d\vec{v}}{dt}$, so with a force that depends on position, velocity and time,

is a second-order ODE for $\vec{r}(t)$ (first-order for $\vec{v}(t)$). When the force depends only on velocity, the $\vec{v}$-equation decouples from position entirely and can be solved on its own — this is exactly the drag problem. Three integration strategies recur:

• $F$ depends on $t$ only: integrate twice directly.
• $F$ depends on $v$ only: separate variables as $\dfrac{m\,dv}{F(v)} = dt$, or use $\dfrac{m v\,dv}{F(v)} = dx$ if you want $v(x)$ instead of $v(t)$.
• $F$ depends on $x$ only: use $a = v\dfrac{dv}{dx}$ to get $m v\,dv = F(x)\,dx$, the work–energy route.

2. Separable First-Order ODEs in Mechanics

A one-dimensional equation of motion of the form $m\dot{v} = F(v)$ is separable: gather all $v$-dependence on one side,

Carrying out the left integral gives $t$ as a function of $v$, which is then inverted to get $v(t)$. To get the distance travelled, either integrate $v(t)$ once more, or restart from $m v\dfrac{dv}{dx} = F(v)$ and separate as $\dfrac{m v\,dv}{F(v)} = dx$.

3. Linear (Stokes) Drag

At low speeds — small objects, viscous fluids, low Reynolds number — the drag force is proportional to velocity:

For a body falling vertically under gravity with this drag, taking down as positive,

The terminal velocity is the steady state $\dot{v} = 0$, i.e. $v_t = \dfrac{mg}{b}$. Rewriting the equation as $\dot{v} = -\dfrac{b}{m}(v - v_t)$ and separating,

The body approaches terminal velocity exponentially, with time constant $\tau = m/b$. Integrating once more (with $v_0 = 0$) gives the distance

Worked example. A droplet of mass $m = 2.0\times10^{-8}~\text{kg}$ falls through air with $b = 4.0\times10^{-8}~\text{kg/s}$. Then $v_t = mg/b = (2.0\times10^{-8})(9.81)/(4.0\times10^{-8}) \approx 4.9~\text{m/s}$ and $\tau = m/b = 0.50~\text{s}$. After $t = 1.5~\text{s} = 3\tau$, $v = v_t(1 - e^{-3}) \approx 4.9(0.950) \approx 4.7~\text{m/s}$ — already within $5\%$ of terminal.

4. Quadratic (Newton) Drag

At everyday speeds and sizes — a skydiver, a baseball, a car — the drag is quadratic:

directed opposite to the velocity. For vertical fall (down positive),

Terminal velocity: $v_t = \sqrt{\dfrac{mg}{c}}$. Separating variables,

The integral is a standard $\text{arctanh}$ form, giving

for release from rest. The approach to $v_t$ is again asymptotic, but through $\tanh$ rather than a pure exponential; for $t \gg v_t/g$ the two forms both give $v \to v_t$. If instead you want $v(x)$ — useful when time is not of interest — separating $m v\,dv/(mg - cv^2) = dx$ yields $v^2 = v_t^2\left(1 - e^{-2cx/m}\right)$.

Worked example. A skydiver of mass $80~\text{kg}$ with $c = 0.25~\text{kg/m}$ (belly-to-earth) has $v_t = \sqrt{(80)(9.81)/0.25} \approx 56~\text{m/s} \approx 200~\text{km/h}$. The characteristic time is $v_t/g \approx 5.7~\text{s}$; after about $11~\text{s}$ ($\approx 2 v_t/g$) the diver is at $\tanh(2) \approx 0.96$ of terminal speed.

5. The Variable-Mass Equation

When a body's mass changes because it ejects or accretes matter, you cannot write $F = m\dot{v}$ with the instantaneous $m$. Instead, apply momentum conservation to the whole system (body plus the matter it is about to exchange) over an interval $dt$.

Let a body have mass $m$ and velocity $\vec{v}$. In time $dt$ it accretes mass $dm$ (so $dm > 0$; for a rocket $dm < 0$) which had velocity $\vec{u}$ before joining. The total momentum before is $m\vec{v} + dm\,\vec{u}$; after, it is $(m + dm)(\vec{v} + d\vec{v})$. The change equals the external impulse $\vec{F}_{\text{ext}}\,dt$:

Expanding, dropping the second-order term $dm\,d\vec{v}$, and writing $\vec{v}_{\text{rel}} = \vec{u} - \vec{v}$ for the velocity of the accreted matter relative to the body:

The extra term $\vec{v}_{\text{rel}}\,\dfrac{dm}{dt}$ is the thrust. For a rocket, $dm/dt < 0$ and $\vec{v}_{\text{rel}}$ points backward, so the thrust points forward. For a body sweeping up stationary matter, $\vec{u} = 0$ so $\vec{v}_{\text{rel}} = -\vec{v}$ and the term is a retarding force $-\vec{v}\,\dot{m}$.

6. The Tsiolkovsky Rocket Equation

Take a rocket in free space ($\vec{F}_{\text{ext}} = 0$) ejecting exhaust at constant speed $v_e = |\vec{v}_{\text{rel}}|$ relative to itself. The 1D equation is

Integrating from initial mass $m_0$ (velocity $v_0$) to final mass $m_f$:

The velocity gain depends only on the exhaust speed and the mass ratio, logarithmically — which is why staging and high exhaust velocity matter so much. If gravity acts (vertical launch, uniform $g$), the same integration with $F_{\text{ext}} = -mg$ over burn time $t_b$ gives the "gravity loss": $\Delta v = v_e\ln(m_0/m_f) - g t_b$.

Worked example. A rocket with $v_e = 3.0~\text{km/s}$ needs $\Delta v = 8.0~\text{km/s}$ to reach orbit (ignoring losses). The required mass ratio is $m_0/m_f = e^{8.0/3.0} = e^{2.67} \approx 14.4$: the vehicle must be more than $90\%$ propellant by mass. Raising $v_e$ to $4.5~\text{km/s}$ drops the ratio to $e^{1.78} \approx 5.9$.

7. Accreting Mass: The Falling Raindrop

A spherical raindrop of radius $r$ falls through a stationary cloud, sweeping up all the water in its path, so $\dot m = k\,r^2 v$ for some constant $k$ (the swept volume per unit time is $\pi r^2 v$); since $m \propto r^3$ this is $\dot m \propto m^{2/3} v$. With $\vec{u} = 0$, the equation of motion is

— here it is legitimate to use $d(mv)/dt$, because the accreted matter starts at rest and contributes no momentum. Substitute a trial power-law solution $v = At$, $m = Bt^n$. The accretion law gives $nBt^{n-1} \propto t^{2n/3}\cdot t$, so $n - 1 = 2n/3 + 1$ and $n = 6$ (mass growing as $t^6$). Matching the $mg$ equation then gives $A(n+1) = g$, so the drop falls with constant acceleration $g/7$ in the regime where its initial mass is negligible. The factor $1/7$ (not $1/2$, not $1$) is the signature that this is a genuine variable-mass problem and not ordinary free fall — a favourite exam check that you have set the momentum balance up correctly. (A cruder law, $\dot m = kv$ with no dependence on the drop's size, instead forces $n = 2$ and gives $g/3$.)

8. Common Pitfalls

• Writing $F = m\dot{v}$ for a variable-mass body. The correct equation is $m\dot{v} = F_{\text{ext}} + v_{\text{rel}}\dot{m}$. Whether $d(mv)/dt = F_{\text{ext}}$ happens to work depends on the accreted matter being at rest.
• Getting the sign of the thrust term wrong. $v_{\text{rel}} = u - v$ is the exhaust velocity relative to the rocket; for rearward exhaust it is negative, and with $\dot{m} < 0$ the product is a positive (forward) thrust.
• Treating the exhaust speed as measured in the ground frame. $v_e$ in the rocket equation is relative to the rocket; it is essentially constant even as the rocket speeds up.
• Forgetting that quadratic drag is not linear in $v$. You cannot superpose solutions, and the falling and rising phases of a projectile with quadratic drag obey different equations (drag always opposes motion).
• Using terminal velocity as if it were reached instantly. It is an asymptote; quote the characteristic time ($\tau = m/b$ for linear drag, $v_t/g$ for quadratic) alongside it.
• Assuming an object dropped with drag ever exceeds $v_t$. From rest it approaches $v_t$ from below and never reaches it; only if launched downward faster than $v_t$ does it decelerate toward it.

Summary

• The general second law is $\vec{F}_{\text{ext}} = d\vec{p}/dt$; for constant mass this is $m\dot{\vec{v}} = \vec{F}$, a first-order ODE in $\vec{v}$ when $\vec{F} = \vec{F}(\vec{v})$.
• Separable ODEs: $\displaystyle\int \dfrac{m\,dv}{F(v)} = t$ for $v(t)$; use $m v\,dv/F(v) = dx$ for $v(x)$.
• Linear drag $-b v$: $v(t) = v_t + (v_0 - v_t)e^{-t/\tau}$ with $v_t = mg/b$, $\tau = m/b$ — exponential approach to terminal velocity.
• Quadratic drag $-c v^2$: $v(t) = v_t\tanh(gt/v_t)$ with $v_t = \sqrt{mg/c}$; approach governed by $\tanh$, timescale $v_t/g$.
• Variable mass: $m\dot{\vec{v}} = \vec{F}_{\text{ext}} + \vec{v}_{\text{rel}}\dot{m}$, derived from momentum balance over $dt$; the second term is the thrust.
• Rocket equation: $\Delta v = v_e\ln(m_0/m_f)$ in free space — logarithmic in the mass ratio, linear in exhaust speed; subtract $g t_b$ for a vertical launch.
• Accreting raindrop ($\dot{m} \propto r^2 v$, matter at rest): falls at constant acceleration $g/7$, a diagnostic that the momentum bookkeeping is right.