Raindrops falling

Classical Mechanics Level 3

When a raindrop falls through the atmosphere, it suffers a drag force which increases with its speed. Eventually, the raindrop attains a terminal (constant) velocity.

If a raindrop of surface area A A is suffering a drag force given by F d = 1 2 C D A ρ air v 2 F_d = \tfrac 12 C_D A \rho_\text{air} v^2 , where C D C_D is the drag coefficient, ρ air \rho_{\text{air}} is the density of air, and v v is the speed of the raindrop, then which of the following is the terminal velocity of the raindrop proportional to?


Details and Assumptions:

  • Raindrops are nearly spherical.
A 1 / 2 A^{1/2} A 1 / 3 A^{1/3} A 1 / 4 A^{1/4} A 3 / 2 A^{3/2}

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Arjen Vreugdenhil by Arjen Vreugdenhil , 44, USA

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1 solution

Arjen Vreugdenhil
May 20, 2017

The drag force on the raindrop is F d = 1 2 C D A ρ a i r v 2 . F_d = \tfrac 12 C_D A \rho_{air} v^2. The mass of the raindrop is m = ρ w V = ρ w 4 π 3 r 3 = ρ w 1 6 π A 3 / 2 m = \rho_w V = \rho_w \frac{4\pi}3 r^3 = \rho_w \frac 1{6\sqrt{\pi}} A^{3/2} Thus force of gravity on the raindrop is F g = m g = 1 6 π ρ w g A 3 / 2 . F_g = m g = \frac 1{6\sqrt{\pi}} \rho_w g A^{3/2}. At terminal velocity, the drag force and gravitational force balance: 1 2 C D A ρ a i r v 2 = 1 6 π ρ w g A 3 / 2 . \tfrac 12 C_D A \rho_{air} v^2 = \frac 1{6\sqrt{\pi}} \rho_w g A^{3/2}. v 2 = ρ w g 3 π C D ρ a i r A 1 / 2 . v^2 = \frac {\rho_w g}{3\sqrt{\pi} C_D \rho_{air}} A^{1/2}. v = ρ w g 3 π C D ρ a i r A 1 / 4 . v = \sqrt{\frac {\rho_w g}{3\sqrt{\pi} C_D \rho_{air}}} A^{1/4}. Thus v ∝ A 1 / 4 . v \propto A^{1/4}.

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