**Galileo: Throwing cannonballs off The Leaning Tower of Pisa **

This post is inspired by the excellent book by Robert Banks – Towing Icebergs. This book would make a great investment if you want some novel ideas for a maths investigation.

Galileo Galilei was an Italian mathematician and astronomer who (reputedly) conducted experiments from the top of the Tower of Pisa. He dropped various objects from in order to measure how long it took for them to reach the bottom, coming to the remarkable conclusion that the objects’ weight did not affect the speed at which it fell. But does that really mean that a feather and a cannonball would fall at the same speed? Well, yes – as long as they were dropped in a vacuum. Let’s have a look at how we can prove that.

**Newton’s Laws:**

For an object falling through the air we have:

p_{s}gV – p_{a}gV – F_{D} = p_{s}Va

p_{s} = The density of the falling object

p_{a} = The density of the air it’s falling in

F_{D} = The drag force

g = The gravitational force

V = The volume of the falling object

a = The acceleration of the falling object

To understand where this equation comes from we note that Newton second law (Force = mass x acceleration) is

F = ma

The LHS of our equation (p_{s}gV – p_{a}gV – F_{D}) represents the forces acting on the object and the RHS (p_{s}Va) represents mass x acceleration.

**Time to simplify things**

Things look a little complicated at the moment – luckily we can make our lives easier through a little simplification. p_{a} will be many magnitudes smaller than than p_{s} – as the density of air is much smaller than the density of objects like cannonballs. Therefore we ignore this part of the equation, giving an approximate equation:

p_{s}gV – F_{D} ≈ p_{s}Va

Next, we can note that in a vacuum F_{D} (the drag force) will be 0 – as there is no air resistance. Therefore this can also be ignored to get:

p_{s}gV ≈ p_{s}Va

g ≈ a

But we have a = dU/dt where U = velocity, therefore,

g ≈ a

g ≈ dU/dt

g dt ≈ dU

and integrating both sides will give:

gt ≈ U

Therefore the velocity (U) of the falling object in a vacuum is only dependent on time and the gravitational force. In other words it is independent of the object’s mass. Amazing!

This might be difficult to believe – as it is quite unintuitive. So if you’re not convinced you can watch the video below in which Brian Cox tests this out in the world’s largest vacuum chamber.

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War maths – how modeling projectiles plays an essential part in waging wars.

## 3 comments

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October 28, 2015 at 10:33 am

Matthew WrightI believe this was also demonstrated on the Moon by one of the later Apollo missions where the astronaut dropped a hammer and a feather.

October 29, 2015 at 5:49 am

Dan PearcyVery nice – love the simplicity of the derivation but the power in the final result. Thanks for posting.

October 29, 2015 at 6:56 am

Ibmathsresources.comThanks for the comments – I have added the video about the moon experiment.

Yes, I was quite surprised that it was so simple to show! I might try and look at some other basic Newtonian laws and what we can derive from them. Next up, parachute jumps!