Radioactive Decay: A Journey to Stability

Two forces, one of them constant, one growing

A falling object feels gravity pulling down at a fixed mg, and air resistance pushing back up, growing as it speeds up. In the regime everyday objects fall through, drag is well modelled as proportional to the square of speed:

F_drag = ½ · ρ · v² · Cd · A

ρ  = density of the fluid (air ≈ 1.225 kg/m³ at sea level)
v  = speed relative to the fluid
Cd = drag coefficient (shape-dependent: ≈0.47 sphere, ≈1.0-1.3 skydiver belly-down)
A  = frontal (cross-sectional) area facing the flow

Where the two forces meet

Because drag scales with v² while gravity is fixed, there is exactly one speed at which they balance and net force drops to zero: the terminal velocity. Setting F_drag = mg and solving for v:

½ρv_t²CdA = mg v_t = √( 2mg / (ρ·Cd·A) ) Everything here is fixed by the object's properties — v_t doesn’t depend on how the object started or how far it has already fallen, only on its mass-to-area ratio and shape. Once it reaches v_t, net force is zero, so by Newton’s first law it simply continues at that constant speed until something else changes (like deploying a parachute, which suddenly increases A and demands a whole new, much lower v_t).

½ρv_t²CdA = mg

v_t = √( 2mg / (ρ·Cd·A) )

Рівняння диференціювання та його точне розв’язання

Записуємо k = ½ρCdA/m для зручності, другий закон Ньютона для падіння перетворюється на першорядну нелінійну ОДУ, яка, на відміну від маятника, має чисте замкнене розв’язання:

dv/dt = g - k·v² початкове значення з відпочинку, v(0) = 0 v(t) = v_t · tanh( g·t / v_t ) де v_t = √(g/k) tanh починається з 0 з нахилом g (чисте вільне падіння, як очікувалося, коли v ще мале і тертя незначне) та асимптотично вирівнюється до 1 зі зростанням t — тобто v плавно наближається до v_t і ніколи не перевищує її з відпочинку. Це саме форма, яку простежує графік часу-швидкості симуляції.

dv/dt = g - k·v²          starting from rest, v(0) = 0

v(t) = v_t · tanh( g·t / v_t )      where v_t = √(g/k)

Why a feather, a ball and a skydiver disagree so much

v_t depends on m/A — mass per unit frontal area — not on mass alone. A feather has almost no mass but a relatively huge area, so its m/A is tiny and it reaches a very low terminal velocity, drifting rather than falling. A dense compact ball has a much higher m/A and needs a much higher speed before drag can catch up. A skydiver belly-down maximises A relative to their mass (roughly 50 m/s, about 180 km/h) but reorients head-down to shrink A and can exceed 90 m/s — the same body, same mass, wildly different terminal velocities purely from changing shape.

Calculating Terminal Velocity

Terminal velocity is defined as the constant speed that an object approaches when falling through a fluid (like air). It’s not simply about gravity; it’s a balance between the force of gravity pulling the object down and the drag force exerted by the fluid resisting its motion. The higher the velocity, the greater the drag force.

To calculate terminal velocity, we need to consider two key factors: the object's mass (m) and the drag coefficient (Cd). The drag coefficient depends on the shape of the object and the properties of the fluid it’s moving through. A streamlined object will have a lower Cd than a blunt one. The drag force itself is calculated using the following formula: Fd = 0.5 * ρ * Cd * A * v^2, where ρ is the density of the fluid, A is the cross-sectional area of the object perpendicular to the flow, and v is the velocity.

The key equation for terminal velocity (Vt) arises when the drag force equals the gravitational force: Fg = Fd. Therefore, mg = 0.5 * ρ * Cd * A * v^2. Solving for v, we get Vt = sqrt((2mg)/(ρCdA)). This shows that terminal velocity is directly proportional to the square root of the object’s mass and inversely proportional to the drag coefficient and the fluid density.

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