Galileo Goes To The Moon

I guess you all have played with bouncy balls. Did you ever notice that no matter how hard you throw, straight towards the ground or at an angle, they never ''bounce up'' higher than their drop height? Do you wonder why? 

Bouncy balls follow a fundamental principle of physics - the conservation of momentum and energy. The exact reason why a bouncy ball never reaches a height greater than its drop height can be understood as follows: When a bouncy ball hits the ground, its potential energy gets converted into kinetic energy (this is true for all objects), a part of which dissipates in the form of sound and heat during the exact moment of collision. As the ball bounces up, it possesses less kinetic energy compared to what it had before. In the most idealistic scene, where the collision with the ground is perfectly elastic, the ball will bounce to its initial drop height and not an inch higher. 

Suppose a ball is supported, by some means, at a height h above the ground. Initially, it has a potential energy mgh, where m is its mass and g is the acceleration due to gravity. As the ball falls, it accelerates under gravity, gaining its speed in proportion to g and h, and strikes the ground with the usual velocity of √2gh. In the process, the ball also gains kinetic energy given by (1/2)mv², where v is the velocity at the instant it hits the ground. Since energy and momentum are the conserved quantities in a perfectly elastic collision, the ball rebounds with the same kinetic energy with which it hit the ground. Understandably, the ball bounces up to its drop height, whereby its kinetic energy transforms entirely into potential energy and the whole process restarts, with the ball bouncing up and down, ad infinitum. 

A word of caution, though. The above process does not imply a perpetual motion machine. An undergrad would effortlessly point out the loophole in there. For the ball to drop, someone/something has to lift it to the height of h. In doing so, that particular someone/something would have to do work prior, extracting energy from the system, and this is where the whole argument for perpetual machines breaks down. The laws of physics abhor perpetual machines. 

If you want to seek out further details of energy/momentum conservation during free fall, look at the following external article - https://farside.ph.utexas.edu/teaching/301/lectures/, or you could also take up any undergrad textbook on elementary mechanics.

If the ball was to bounce higher than its drop height, it would have needed more potential energy than what it is supposed to have at height h, i.e., P.Eₕ > mgh. This does not make any logical sense. Where does this extra energy come from? The answer is nowhere because energy can not be created at will. It can only be transformed from one form to another - mechanical energy can be transformed into heat and sound, but it is impossible to create any form of energy out of nowhere. You can not make a clapping sound without actually clapping. So potential energy at h cannot exceed mgh in the real world. Every time the bouncy ball impacts the surface, it loses some energy in the form of heat and sound, not to mention that it also undergoes translation and rotation due to friction with the ground. As a result, the energy after impact falls short of the energy before. With each successive impact on the ground, the ball keeps on losing energy. Plus, there are factors like air resistance, magnus force, buoyancy, the "bounciness of the ball", material of the ball, and the like. I am not going into such details. In summary, the bouncy ball loses energy while going through an infinite number of bounces (but in a finite time) before coming to dead rest. 

It is nevertheless possible to make a bouncy ball bounce higher than its drop height. In that case, we need to drop a stack of balls instead of a single ball. But there are certain things to consider. We can not make an arbitrary stack of balls. It requires a very specific arrangement of balls of decreasing size, with the largest ball at the bottom, followed by a smaller ball on top, an even smaller ball on top of it, and so on. The schematic diagram below gives a standard picture of such a stack. The conservation of momentum states that if this stack is dropped, provided the stack maintains its orientation (which can be done by aligning them together with a light string passing through the vertical diameter of each ball), then the topmost ball will gain the momentum of all the lower balls and as such, will rebound to a height many orders greater than its drop height. 

 Schematic of a stack of 7 balls.
Not to scale.
Image Credits: Author's computer 

Ladies and gentlemen, I present to you the Galilean Cannon. However, I am pretty sure Galileo had nothing to do with this contraption. When this entire thing is dropped, the bottommost ball hits the ground first and rebounds, transferring its momentum to the one on top of it, which in turn, rebounds and transfers its momentum to the ball on top of it, and on and on, with the topmost ball bouncing to a great height, with the combined momentum of all the balls below. Theoretically speaking, one can even go to the Moon in this way. I will come to that later. 

Let us think of a simple scenario. Suppose a stack of two balls, one larger (also heavier) than the other, is dropped from a height h in such a way that they maintain their orientation throughout and do not deviate from a straight line-trajectory, as shown in the figure below. The vertical dotted line is a light string keeping the ball in a stack. Obviously, we are working under the assumption that the balls undergo perfectly elastic collision. Further, there is no air resistance, and the balls are not otherwise spinning or undergoing any such rotational motion. We intend to find out the speed of the smaller (red) ball and the height to which it rebounds and see if the rebounding height can exceed the drop height.  

Fig. 1 - Mechanics of a stack of two balls.
The black the red arrow indicates the velocity vector as seen by an observer on the ground and on the yellow ball, respectively. 
Image Credits: Author's computer 

As the figure shows, the stack is supported at a height of h from the ground, with the yellow ball starting to drop from height h. The red ball is actually at a height (h + d) where d is the diameter of the yellow ball. But for the mere convenience of calculation, we can ignore d. The separation in fig. 1.2 has been introduced for visual aid, for if they were falling in a vacuum, the separation would not exist. To an observer on the ground, the yellow ball and the red ball falls with velocity v. When the yellow ball hits the ground, it rebounds with the same velocity v. Now suppose if there is an observer on the yellow ball, then with respect to his frame of reference (i.e., of the yellow ball), he will be approaching the ground at the rate of v, and when the yellow ball rebounds, he will witness the red ball approaching him with velocity 2v and the ground receding away at velocity v. Now, when the red ball hits the rebounding yellow ball, the observer on the yellow ball will see the red ball rebounding with a velocity 2v. Understandably, the observer on the ground will see the red ball going up with a velocity 3v.  

The conservation of momentum for the (above) stack of two balls implies: 

expression 1.0

where m₁ is the mass of the heavier (yellow) ball.

           m₂ is the mass of the lighter (red) ball.

           v'₁ is the rebounding velocity of the yellow ball.

           v'₂ is the rebounding velocity of the red ball.

           v₁ is the velocity with which the yellow ball drops from rest at height h.

           v₂ is the velocity with which the red ball drops from rest at height.  

Similarly, the conservation of energy for the above system dictates: 

expression 1.1

Solving the pair of expressions at hand for v'₂, we get

expression 1.2

Since the balls are falling from the same height and under no air resistance, we can take v₁ to be equal to v₂. When the yellow ball rebounds, v'₁ equals v₁, and further, v₁ equals negative v₂, whereby the negative sign indicates that the direction of v₁ (or v'₁) and v₂ are opposite. Therefore we get, 

expression 1.3

or taking the modulus, 

expression 1.4

If m₁ >> m₂, then we can neglect the latter, and the above expression reduces to 

expression 1.5

From the standard expression for the velocity of a freely falling body, i.e., 

expression 1.5

we easily understand that the red ball will rebound to 9 times its original drop height.  

At this point, it is fairly easy to generalize the problem for a stack of n balls. If you are a freshman physics undergrad, you can solve this particular problem as your homework exercise on pp. 177, problem 5.23 of David Morin's Classical Mechanics. 

For a stack of n balls, the rebounding velocity of the nth ball is given by

expression 1.6

The nth ball rebounds to a height of 

expression 1.7

where H is the rebounding height, h is the drop height and l is the diameters of the (n-1) balls. 

In an youtube video, Brian Greene demonstrates a Galilean cannon and it turns out that he wins the Guinness record for the highest Galilean cannon launched. This was done in 2016 prior to the New York Science Festival. Brian's record was later broken by a team from the North Carolina Science Festival when they launched it to a maximum height of 13.5 m (~ 44 feet). 

As Brian says, the most powerful and energetic explosions in the universe, supernovae, essentially occur along the basic principles of the Galilean cannon. When an aged star collapses gravitationally, the outermost atoms fly apart at colossal speeds, ripping away whatever stands in its way.  

Here comes the best part. Recall that I said one could go to the Moon onboard a Galilean cannon. Do you see how?

In expression 1.7, if we take h to be 1 meter, and we want to reach a height of 100 km, i.e., we want to cross the Karman line and reach space, we would need a stack of at least 8 balls. Since the calculation is fairly easy I believe most of my readers can do it in a jiffy. Earth's escape velocity is roundabout 11,200 m/s. Fixing the initial drop height at 1 m, one needs to drop a stack of 12 balls with the condition m₁ >> m₂ >> ... >> m₁₂. However, this way of going to the Moon is totally unrealistic and physically absurd in several ways. Since the mass of each ball out of the 12-ball stack decreases in proportion, the twelfth ball will be too small to fit a person. 

Remembering Einstein's quote, ''Imagination is more important than knowledge'', I would like to imagine a universe where Galileo is trying his best to reach the Moon, sitting comfortably in an armchair on the twelfth ball.