# Free fall of two spheres made of same materials, different masses, with air resistance

Two people falling with the same parachute will gain different speeds if their masses are different. The upward air drag will needed to be bigger for a heavier person, since gravitational force is bigger in his/hers case.
In parachute case, the shape is almost the same. But what about two iron balls, one with weight 100kg, and other 1kg. The densities of two balls are the same, but the area of a bigger ball is larger. Does this increase in area of bigger ball compensates for bigger mass, do the two balls still fall with different speeds?

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See this article on the NASA site for a detailed discussion of the drag on a sphere. The drag is given by:

$$D = k\space A \space v^2 = k \space \pi r^2 \space v^2$$

where $A$ is the area of the sphere, $r$ is the radius of the sphere and $v$ is the velocity. The constant $k$ includes air density and other factors that don't depend on the size of the sphere.

The downwards force is proportional to the mass of the sphere:

$$F = mg = \rho \frac{4}{3} \pi r^3 g$$

So the drag is proportional to the area ($r^2$), but the mass of your sphere is proportional to its volume ($r^3$). That means the bigger heavier sphere will fall faster. To work out the terminal velocity just set the drag equal to the downward force and this gives:

$$v_t = \sqrt {\frac{4}{3} \frac{r \rho g}{k}}$$

So the terminal velocity will scale as $\sqrt{r}$.

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The mass of the ball scales with the cube of the radius. The frontal area however, scales with the square of the radius $\rightarrow$ mass wins.

Put another way: when the ball grows larger, more and more mass will be "behind" existing frontal area, and less and less mass will create new frontal area.

So: provided their densities are always the equal, more massive balls will always be less affected by air drag than less massive balls.

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