Terminal Velocity of Body under Fall Retarded Proportional to Square of Velocity

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Theorem

Let $B$ be a body falling in a gravitational field.

Let $B$ be falling through a medium which exerts a resisting force of magnitude $k v^2$ upon $B$ which is proportional to the square of the velocity of $B$ relative to the medium.


Then the terminal velocity of $B$ is given by:

$v = \sqrt {\dfrac {g m} k}$


Proof

Let $B$ start from rest.

The differential equation governing the motion of $B$ is given by:

$m \dfrac {\d^2 \mathbf s} {\d t^2} = m \mathbf g - k \paren {\dfrac {\d \mathbf s} {\d t} }^2$

Dividing through by $m$ and setting $c = \dfrac k m$ gives:

$\dfrac {\d^2 \mathbf s} {\d t^2} = \mathbf g - c \paren {\dfrac {\d \mathbf s} {\d t} }^2$

By definition of velocity:

$\dfrac {\d \mathbf v} {\d t} = \mathbf g - c \mathbf v^2$

for some constant $c$.

and so taking magnitudes of the vector quantities:

\(\ds \int \dfrac {\d v} {g - c v^2}\) \(=\) \(\ds \int \rd t\)
\(\ds \leadsto \ \ \) \(\ds \dfrac 1 {2 c} \sqrt {\frac c g} \ln \paren {\frac {\sqrt {\frac c g} + v} {\sqrt {\frac c g} - v} }\) \(=\) \(\ds t + c_1\) Primitive of $\dfrac 1 {a^2 - x^2}$: Logarithm Form


We have $v = 0$ when $t = 0$ which leads to $\ln 1 = 0 + c_1$ and thus $c_1 = 0$.


Hence:

\(\ds \ln \paren {\frac {\sqrt {\frac c g} + v} {\sqrt {\frac c g} - v} }\) \(=\) \(\ds 2 t \sqrt {c g}\)
\(\ds \leadsto \ \ \) \(\ds \frac {\sqrt {\frac c g} + v} {\sqrt {\frac c g} - v}\) \(=\) \(\ds e^{2 t \sqrt {c g} }\)
\(\ds \leadsto \ \ \) \(\ds \sqrt {\frac c g} + v\) \(=\) \(\ds \paren {\sqrt {\frac c g} - v} e^{2 t \sqrt {c g} }\)
\(\ds \leadsto \ \ \) \(\ds v \paren {e^{2 t \sqrt {c g} } + 1}\) \(=\) \(\ds v \paren {e^{2 t \sqrt {c g} } - 1} \sqrt {\frac g c}\)
\(\ds \leadsto \ \ \) \(\ds v\) \(=\) \(\ds \sqrt {\frac g c} \paren {\frac {e^{2 t \sqrt {c g} } - 1} {e^{2 t \sqrt {c g} } + 1} }\)
\(\ds \) \(=\) \(\ds \sqrt {\frac g c} \paren {\frac {1 - e^{-2 t \sqrt {c g} } } {1 + e^{-2 t \sqrt {c g} } } }\)


Since $c > 0$ it follows that $v \to \sqrt {\dfrac g c}$ as $t \to \infty$.

Thus in the limit:

$v = \sqrt {\dfrac g c} = \sqrt {\dfrac {g m} k}$

$\blacksquare$


Sources

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