Terminal velocity

Examples

Based on air resistance, for example, the terminal speed of a skydiver in a belly-to-earth (i.e., face down) free fall position is about 55 m/s.{{cite web | editor-last=Elert | editor-first=Glenn

Higher speeds can be attained if the skydiver pulls in their limbs (see also freeflying). In this case, the terminal speed increases to about 90 m/s, which is almost the terminal speed of the peregrine falcon diving down on its prey. The same terminal speed is reached for a typical .30-06 bullet dropping downwards—when it is returning to the ground having been fired upwards or dropped from a tower—according to a 1920 U.S. Army Ordnance study.

Competition speed skydivers fly in a head-down position and can reach speeds of 150 m/s. The current record is held by Felix Baumgartner who jumped from an altitude of 127582 ft and reached 380 m/s, though he achieved this speed at high altitude where the density of the air is much lower than at the Earth's surface, producing a correspondingly lower drag force.

The biologist J. B. S. Haldane wrote,

Physics

For terminal velocity in falling through air, where viscosity is negligible compared to the drag force, and without considering buoyancy effects, terminal velocity is given by V_t= \sqrt\frac{2 m g}{\rho A C_d} where

• V_t represents terminal velocity,
• m is the mass of the falling object,
• g is the acceleration due to gravity,
• C_d is the drag coefficient,
• \rho is the density of the fluid through which the object is falling, and
• A is the projected area of the object.

In reality, an object approaches its terminal speed asymptotically.

Buoyancy effects, due to the upward force on the object by the surrounding fluid, can be taken into account using Archimedes' principle: the mass m has to be reduced by the displaced fluid mass \rho V, with V the volume of the object. So instead of m use the reduced mass m_r = m-\rho V in this and subsequent formulas.

The terminal speed of an object changes due to the properties of the fluid, the mass of the object and its projected cross-sectional surface area.

Air density increases with decreasing altitude, at about 1% per 80 m (see barometric formula). For objects falling through the atmosphere, for every 160 m of fall, the terminal speed decreases 1%. After reaching the local terminal velocity, while continuing the fall, speed decreases to change with the local terminal speed.

Using mathematical terms, defining down to be positive, the net force acting on an object falling near the surface of Earth is (according to the drag equation):

F_\text{net} = m a = m g - \frac{1}{2} \rho v^2 A C_d,

with v(t) the velocity of the object as a function of time t.

At equilibrium, the net force is zero (Fnet = 0) and the velocity becomes the terminal velocity :

m g - {1 \over 2} \rho V_t^2 A C_d = 0.

Solving for V**t yields:

The drag equation is—assuming ρ, g and C**d to be constants: m a = m \frac{\mathrm{d}v}{\mathrm{d}t} = m g - \frac{1}{2} \rho v^2 A C_d.

Although this is a Riccati equation that can be solved by reduction to a second-order linear differential equation, it is easier to separate variables.

A more practical form of this equation can be obtained by making the substitution .

Dividing both sides by m gives \frac{\mathrm{d}v}{\mathrm{d}t} = g \left( 1 - \alpha^2 v^2 \right).

The equation can be re-arranged into \mathrm{d}t = \frac{\mathrm{d}v}{g ( 1 - \alpha^2 v^2)}.

Taking the integral of both sides yields \int_0^t {\mathrm{d}t'} = {1 \over g}\int_0^v \frac{\mathrm{d}v'}{1-\alpha^2 v^{\prime 2}}.

After integration, this becomes t - 0 = {1 \over g}\left[{\ln(1 + \alpha v') \over 2\alpha} - \frac{\ln(1 - \alpha v')}{2\alpha} + C \right]{v'=0}^{v'=v} ={1 \over g} \left[{\ln \frac{1 + \alpha v'}{1 - \alpha v'} \over 2\alpha} + C \right]{v'=0}^{v'=v}

or in a simpler form t = {1 \over 2\alpha g} \ln \frac{1 + \alpha v}{1 - \alpha v} = \frac{\mathrm{artanh}(\alpha v)}{\alpha g}, with artanh the inverse hyperbolic tangent function.

Alternatively, \frac{1}{\alpha}\tanh(\alpha g t) = v, with tanh the hyperbolic tangent function. Assuming that g is positive (which it was defined to be), and substituting α back in, the speed v becomes v = \sqrt\frac{2mg}{\rho A C_d} \tanh \left(t \sqrt{\frac{g \rho A C_d }{2m}}\right).

Using the formula for terminal velocity V_t = \sqrt\frac{2mg}{\rho A C_d} the equation can be rewritten as v = V_t \tanh \left(t \frac{g}{V_t}\right).

As time tends to infinity (t → ∞), the hyperbolic tangent tends to 1, resulting in the terminal speed V_t = \lim_{t \to \infty} v(t) = \sqrt\frac{2mg}{\rho A C_d}.

For very slow motion of the fluid, the inertia forces of the fluid are negligible (assumption of massless fluid) in comparison to other forces. Such flows are called creeping or Stokes flows and the condition to be satisfied for the flows to be creeping flows is the Reynolds number, Re \ll 1. The equation of motion for creeping flow (simplified Navier–Stokes equation) is given by:

{\mathbf \nabla} p = \mu \nabla^2 {\mathbf v}

where:

• \mathbf v is the fluid velocity vector field,
• p is the fluid pressure field,
• \mu is the liquid/fluid viscosity.

The analytical solution for the creeping flow around a sphere was first given by Stokes in 1851. From Stokes' solution, the drag force acting on the sphere of diameter d can be obtained as

where the Reynolds number, Re = \frac{\rho d}{\mu} V. The expression for the drag force given by equation () is called Stokes' law.

When the value of C_d is substituted in the equation (), we obtain the expression for terminal speed of a spherical object moving under creeping flow conditions:

V_t = \frac{g d^2}{18 \mu} \left(\rho_s - \rho \right), where \rho_s is the density of the object.

Applications

The creeping flow results can be applied in order to study the settling of sediments near the ocean bottom and the fall of moisture drops in the atmosphere. The principle is also applied in the falling sphere viscometer, an experimental device used to measure the viscosity of highly viscous fluids, for example oil, paraffin, tar etc.

Terminal velocity in the presence of buoyancy force

When the buoyancy effects are taken into account, an object falling through a fluid under its own weight can reach a terminal velocity (settling velocity) if the net force acting on the object becomes zero. When the terminal velocity is reached the weight of the object is exactly balanced by the upward buoyancy force and drag force. That is where

• W is the weight of the object,
• F_b is the buoyancy force acting on the object, and
• D is the drag force acting on the object.

If the falling object is spherical in shape, the expression for the three forces are given below: where

• d is the diameter of the spherical object,
• g is the gravitational acceleration,
• \rho is the density of the fluid,
• \rho_s is the density of the object,
• A = \frac{1}{4} \pi d^2 is the projected area of the sphere,
• C_d is the drag coefficient, and
• V is the characteristic velocity (taken as terminal velocity, V_t ).

Substitution of equations (–) in equation () and solving for terminal velocity, V_t to yield the following expression

In equation (), it is assumed that the object is denser than the fluid. If not, the sign of the drag force should be made negative since the object will be moving upwards, against gravity. Examples are bubbles formed at the bottom of a champagne glass and helium balloons. The terminal velocity in such cases will have a negative value, corresponding to the rate of rising up.

References

References

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