Does the terminal velocity of a sphere in free drop increase with increasing diameter?

Question

I was trying today to calculate the terminal velocity of a hailstones with increasing diameter and mass. I was trying to figure out if larger hailstones will have a higher impact velocity or lower.

I performed the calculation (I was a bit surprised by the end result), which I am presenting as a potential answer. I would be very happy to see if there are any improvements to this answer (if there are additional factors that I need to consider)

Answer 1

The terminal velocity $V_t$ will be reached when the drag coefficient is equal to the force of gravity:

$$F_{drag} = mg$$ $$C_D\frac{1}{2}\cdot \rho_{air} \cdot A \cdot V_{t}^2= m\cdot g$$

where:

• $C_D$: is the drag coefficient (for a sphere is 0.5)
• $\rho_{air}$ is density of the liquid the sphere is passing through (if air then 1.225 kg/m3)
• $A$: cross-sectional area of the hailstone assuming its a sphere $\frac{\pi d^2}{4} =\pi r^2 $
• $V_{t}$: the terminal velocity of the sphere
• $m$: the mass of the sphere (assuming its a sphere the volume is $\frac{4}{3}\pi r^3$, and the density of the sphere is $\rho_{sphere}$)
• $g$: the acceleration of gravity

Therefore: $$C_D\frac{1}{2} \rho_{air} \cdot \left(\pi r^2\right)\cdot V_{t}^2= \frac{4}{3}\pi r^3 \rho_{sphere}\cdot g$$

$$ V_{t}^2= \frac{4\cdot 2 \cdot\pi r^3 \rho_{sphere}\cdot g}{3C_D \rho_{air} \cdot \left(\pi r^2\right)}$$ $$ V_{t}= \sqrt{\frac{8}{3} \cdot \frac{g}{C_D } \cdot \frac{\rho_{sphere}}{\rho_{air} }\cdot r }$$

which indicates that the terminal velocity of a sphere increases proportionally to the square root of the diameter (radius).

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if the buoyancy is considered then:

$$ V_{t}= \sqrt{\frac{8}{3} \cdot \frac{g}{C_D } \cdot \frac{\rho_{sphere} - \rho_{air }}{\rho_{air} }\cdot r }$$