# Which ball falls faster, the cool one or the hot one?

Suppose we're on the top of the Tower of Pisa (or a larger version of it) with two identical cannonballs. We heat one up (say, to 200 degrees Celsius, or some other high temperature before it starts melting and deforming). Then we drop both. Which one will reach the ground first?

• What do you think, and why? – Carl Witthoft Dec 29 '14 at 19:23
• Do you wish to neglect or include air resistance? – dmckee --- ex-moderator kitten Dec 29 '14 at 19:26
• OK, so now you have a theory (good!). Edit your question to reflect that theory and see if you can't put in some equation (hinthintwikipedia) for the air resistance as a function of ball size, speed, mass, etc. – Carl Witthoft Dec 29 '14 at 19:28
• I would think that the cannonball would be moving too fast for it to be able to appreciably increase the ambient air's temperature. – Kyle Kanos Dec 29 '14 at 19:28
• @KyleKanos: Yeah, and air resistance doesn't matter super duper much for cannonballs until they're going darn fast. – DanielSank Dec 29 '14 at 19:35

I presume that the question is actually "which of these experiences greater acceleration due the difference between the force of gravity and atmospheric forces on the cannon ball" because "which one will reach the ground first" has all kinds of flippant answers waiting.

The hotter cannonball has a few properties that may be relevant:

1. it will be larger: more buoyancy, more drag
2. It might heat the air around it - does that affect drag?

Let's estimate a few quantities. We will assume the cannon ball is made of lead (iron or stone could be used - the answers would not change all that much), and that it's a 15 cm diameter ball - volume about 1.8 liter, mass 20 kg. Coefficient of thermal expansion of lead is about $28\cdot10^{-6}/\mathrm{K}$, so with a 200 °C increase in temperature the the diameter increases by 0.56% and the volume by 1.69%. The volume of the additional displaced air is 22 ml, and the mass of this displaced air $\Delta m$ is about 27 mg.

Let's first estimate the drag. For the above cannon ball, Reynolds number is on the order of $10^5$ and drag coefficient of a sphere is relatively constant around 0.5 and drag force is given by

$$F = \frac12 \rho C_v A v^2$$

where $\rho$ is the density of the fluid, $A$ is the apparent area, $C_v$ is the drag coefficient and $v$ is the velocity. At the bottom of the tower, when the cannon ball has a speed of about 33 m/s, this drag amounts to 12 N - this is actually quite a significant amount! The velocity of an object faling under the influence of gravity and a $v^2$ drag force can be written in terms of the height and terminal velocity:

$$v_t = \sqrt{\frac{2mg}{C_v\rho A}}\approx 191 m/s$$ $$v(y) = v_t\sqrt{1-\exp\left(\frac{-2g(y-y_0)}{v_t^2}\right)}$$

Numerically integrating this expression shows that the drag on the above cannon ball will slow it down by about 0.26 m/s (velocity on impact); the difference in time (vs landing without drag) is about 8 ms, and the distance covered in that time is 0.30 m - quite a lot!

The difference in drag due to the extra size of the hot cannon ball is sufficient to add another 3 mm to that distance, giving the hot ball quite a bit more drag. It will definitely be slower.

But wait - there's more.

Now a sphere moving through a medium experiences an apparent inertia equal to its own mass plus half the mass of the displaced fluid, so the hot cannonball will have an apparent increase in inertia of $\Delta m/2$ and an increase in buoyancy of $\Delta m g$.

The time to fall is given by

$$t = \sqrt{\frac{2h}{g'}}$$

where $g'$ is the effective acceleration, namely net force divided by apparent inertia

$$g' = \frac{mg - V\rho g}{m + V\rho / 2}$$

With a height of 56 m, the approximate time to drop (without taking into account these additional effects) is 3.4 seconds. The fractional difference between the two is roughly given by $0.75 \Delta m / m = 1.4\cdot 10^{-6}$. This makes the difference in time about 4.6µs - tricky to measure, since at that speed (about 34 m/s) it corresponds to one ball being ahead of the other by 0.15 mm. It's possible you could measure that - but it's less than a grain of sand. The tower is already leaning - better make sure you have the two balls exactly level when you drop them; drop them at exactly the same time; and have them land on an exactly flat surface... The buoyancy is therefore definitely a less important factor, but drag is actually surprisingly significant.

Incidentally - exactly how the balls are dropped does play a part. If you keep the tops of the balls level before you drop them, then the bottom of the hot ball is already well ahead (by about 0.6 mm) - more than the effect of buoyancy, but less than the effect of drag.

The question of heating the air and affecting drag is interesting, but hard to estimate. The hot cannonball might heat the air and lower the density - but drag is in essence caused by imparting kinetic energy to a column of air "swept" by the cannon ball, and the amount of turbulence created in the wake of the ball is likely sufficient to mix the air and leave the density essentially unchanged. If the air were actually heated by 200°C, and the pressure remained constant, then the lower density of the air would result in a significant reduction in drag (remember $F\propto A\rho$) so making the hot cannonball faster. But for this to work, the air needs to be heated by several degrees on average (we would want the product $\rho A$ to be lower, and since $A$ increased by about 1.1%, we would need $\rho$ to decrease by more than that: a 4 degrees temperature change would be enough. Getting this kind of heating in the air when the ball is moving at 30 m/s (the drag is greatest just before the ball hits the ground: that's where we need the effect) means that we need to be able to transfer a sufficient amount of heat from the cannonball to the air.

If we want to heat a column of air that is 30 m long by 15 cm diameter by 4 degrees in one second, the heat required is

$$U = \frac{\pi}{4}(0.15)^2*1.2*4 = 2.6 \mathrm{J}$$

Using an h factor of 40 $\mathrm{W/m^2/K}$ (slightly outside the range of values given at http://www.engineeringtoolbox.com/convective-heat-transfer-d_430.html) with a surface area of 0.07 $\mathrm{m^2}$ and a temperature difference of 200 degrees actually allows for a heat transfer of 560 J - much larger than the 2.6 J that would be needed.

So - rather surprisingly, even though the ball is falling rapidly, it is capable of heating the air.

But will it heat the right air to lower the drag, or the wrong air to increase the drag?

Hypothesis 1: by increasing the temperature of the boundary layer, you increase the viscosity. This increases the "apparent size" of the cannonball and increases the drag

Hypothesis 2: the heating of the air causes expansion of the air just in front of the cannonball and results in additional pressure. This slows the ball down

Hypothesis 3: since the cannonball feels warm air "all around it", the net density of the air it interacts with is slightly lower. This lowers the drag and makes the ball drop faster.

I don't have a good enough intuition to know which of these three is correct - I have some colleagues with excellent CFD software and I am going to try to persuade them to run the model for me.

The heating of the air will probably completely dominate the process, but whether this means that the hot cannonball will reach the ground first is in my mind not proven.

• My first thought on seeing the question was what convection would do to the air beneath the hot ball as it fell, rather than expansion of the hot ball. Would warmer air try to rise, slowing the ball more? Would warmer air expand, becoming less dense, slowing the ball less? – Russell Borogove Dec 30 '14 at 0:23
• @RussellBorogove I should think more about those questions... Will update later. Thanks for suggesting it! – Floris Dec 30 '14 at 1:01
• Any idea if the hot ball would heat up the air enough to change the drag coefficient? – DanielSank Dec 30 '14 at 2:17
• I have never heard of a lead cannonball. Cannonballs were made from stone early on, and iron later on. – aaaaaaaaaaaa Dec 30 '14 at 10:02
• @GlenTheUdderboat I am thinking about this some more. The effect of the heating of the surroundings air is interesting and more complex than my current answer suggests. I will update next year... – Floris Jan 1 '15 at 3:04

The heated ball has a larger volume, therefore by Archimedes' law it has higher buoyancy than the other. Perhaps the friction plays a role as well, but at least we already have a principle that says which one will reach the ground first.

• Yeah, but by how much? – Carl Witthoft Dec 29 '14 at 21:47
• Can be easily determined by applying Archimedes law – Phoenix87 Dec 29 '14 at 21:49
• Friction would also tend to make the larger block fall slower, just as buoyancy does. – BMS Dec 29 '14 at 21:58
• @CarlWitthoft you asked "how much". I have provided an estimate. – Floris Dec 29 '14 at 23:05

This is what I think about this problem - As the cannon balls fall, 3 forces act on them:

1.Stokes force (upwards) - 6πηrv 2.Weight (downwards) - mg 3.Buoyant force (upwards) - ρ1Vg

let the variables be - η=viscosity of air, r=radius of sphere, V=volume of sphere, v=velocity of sphere, m=mass of sphere, g=acceleration due to gravity, ρ1=density of air, ρ2=density of solid.

Initially, both spheres moves with decreasing acceleration which is given by-

a=g(1- ρ12) - 9ηv/2ρ2r2

As the stokes force is velocity dependent, it increases gradually and when the sum of Stokes force and Buoyant force equals weight of sphere, the sphere moves with terminal velocity and acceleration is zero

vT = 2r2g(ρ2 - ρ1)/9η

Now suppose if due to temperature of hot ball, even if its volume increases by very small amount, there will not be much effect on "a" and "v" .

We can approximate that both will reach at same time.

Maybe the shape of hot ball may change, as it is heated till it starts deforming, and it may acquire shape of a raindrop.

• We do have MathJax here that makes your equations look better. You can see the notation page in help center for details, if you aren't familiar with it. – Kyle Kanos Mar 11 '19 at 12:27