Terminal Velocity

In Book II of Newton’s Principia Newton argues that the velocity of an object acted upon by a drag force is proportional to the square of the object’s velocity. Empirically, physicists have shown that $$F_{drag} \propto v^n$$ where $$1 \le n \le 2$$ depending on the medium and the magnitude of the object’s velocity.

This problem will consider some object in free fall near the earth’s surface in which the air resistance is proportional to the velocity of the object. Moreover, the model will be built using the convention that “up” is positive and that “down” is negative.

There are two forces acting on the the object. The force of gravity acting downward and air resistance, the drag force, acting upward. By using Newton’s second law the interaction of these two forces can be formulated mathematically as

$$F_{drag} – F_{gravity} = F_{net}$$

As stated previously, the magnitude of the air resistance is assumed to be proportional to the velocity. Mathematically,

$$F_{drag} \propto v$$

The proportionality statement can be converted into a statement of equality by by multiplying the velocity by a proportionality constant $$\lambda$$. The value of the proportionality constant will be dependent on the shape of the object and the medium through which it is falling.

$$F_{drag} = \lambda v$$

The force of gravity is simply the weight of the object which is the mass of the object times the acceleration due to gravity near the surface of the earth.

$$F_{gravity} = mg$$

The net force can be expressed as

$$F_{net} = ma_{net} = m \frac{dv}{dt}$$

Substituting into $$F_{drag} – F_{gravity} = F_{net}$$ gives

$$\lambda v – mg = m \frac{dv}{dt}$$

The above differential equation is a first order linear and a solution for the velocity can be found using the appropriate integrating factor. Rearranging the equation into the standard form yields

$$\frac{dv}{dt} – \frac{ \lambda}{m} v = -g$$

From here the integrating factor $$\mu$$ can be determined.

$$\mu = e^{ \int \! \frac{ \lambda}{m} \ dt } = e^{ \frac{ \lambda}{m}t }$$

Substituting into the general solution for a first order linear differential equation gives

$$v \left( t \right) = e^{ -\frac{ \lambda}{m}t } \int \! -g e^{ \frac{ \lambda}{m}t } \ dt = e^{ -\frac{ \lambda}{m}t } \left( \frac{-mg}{ \lambda } e^{ \frac{ \lambda}{m}t } + C \right)$$

Simplifying

$$v \left( t \right) = \frac{-mg}{ \lambda } + C e^{ -\frac{ \lambda}{m}t }$$

Assuming that the object begins its free fall at $$t =0$$ with initial velocity $$v \left( 0 \right) = v_0$$ it is possible to determine the value of the constant $$C$$

$$v \left( 0 \right) = v_0 = \frac{-mg}{ \lambda } + C$$ $$C = v_0 + \frac{mg}{ \lambda }$$

Making the appropriate substitution

$$v \left( t \right) = \frac{-mg}{ \lambda } + \left( v_0 + \frac{mg}{ \lambda } \right) e^{ -\frac{ \lambda}{m}t }$$

From this equation it is important to note what the velocity of the object tends if it is permitted to fall for a sufficient amount of time. In other words

$$\lim_{t \rightarrow \infty} v \left( t \right) = \frac{-mg}{ \lambda }$$

The velocity of the object tends to a constant. This ultimate ultimate velocity is called the terminal velocity. Also, it is important to note that an object’s terminal velocity is proportional to its mass. This helps to explain why a brick dropped from a great height will hit the ground after a shorter duration than a paper ball dropped from the same height.