public:misconceptions_on_galileo_s_experiment_at_pisa_tower

I think a lot of people have misconceptions on the implication of Galileo's experiment at Pisa Tower. Wikipedia says that Galileo “is said to have dropped two spheres of different masses from the Leaning Tower of Pisa to demonstrate that their time of descent was independent of their mass” [(https://en.wikipedia.org/wiki/Galileo%27s_Leaning_Tower_of_Pisa_experiment)]. This experiment was even repeated on the surface of the Moon.

Understanding Galileo's experiment in this way raises a lot of interesting questions. If two objects have the shape and different mass, would they have the same terminal velocity? Would they accelerate at an equal rate? Does that also mean two bikes with different mass would arrive at the bottom of a hill at the same time?

Last Friday (19/07/19) my research group went out for a dinner, at one point we talked about skydiving. It is kind of surprising to hear some of them mumbling that everyone would have the same terminal velocity. It was really surprising, because two of my labmates have physical science undergraduate background.

I once had a debate with a physicist during a bike ride in York, on whether our mass differences affect our speed at going downhill. Now looking back, starting that debate was rather embarrassing, although that conversation did make me think, so it was a useful conversation.

In this article, we discuss some of the physics behind these questions.

Another part of the Wikipedia says “Galileo proposed that a falling body would fall with a uniform acceleration, as long as the resistance of the medium through which it was falling remained negligible, or in the limiting case of its falling through a vacuum” [(https://en.wikipedia.org/wiki/Galileo_Galilei#Falling_bodies)]. I believe this is better way to understand Galileo's experiment.

The acceleration of an object in vacuum under the influence of gravity can be derived using Newton's second law [(https://en.wikipedia.org/wiki/Newton%27s_laws_of_motion)] and Newton's law of universal gravitation [(https://en.wikipedia.org/wiki/Newton%27s_law_of_universal_gravitation)].

Newton's second law is basically:

$$ F = ma $$

The formula for Newton's law of universal gravitation is:

$$ F = G \frac{m_1 m_2}{r^2}, $$

where $F$ is the gravitational force acting between two objects, $m_1$ and $m_2$ are the masses of the objects, $r$ is the distance between the centers of their masses, and $G$ is the gravitational constant. The gravitational force acting between two objects is also known as *weight*.

By equating the two equations above together, and setting $m = m_2$, we get:

$$ m_2 a = G \frac{m_1 m_2}{r^2}. $$

Let's also assume that $m_1$ is the mass of planet Earth, and $m_2$ is the mass of the object we are interested in.

It is clear that $m_2$, the mass of the object we are interested in, gets cancelled out, leaving us with:

$$ a = G \frac{m_1}{r^2}.$$

The acceleration due to gravity only depends on the mass of the planet $m_1$ and distance between the centre of masses $r$. In practice, on planet Earth at sea level, the acceleration due to gravity is about $9.81ms^{-2}$ [(https://en.wikipedia.org/wiki/Gravity_of_Earth)].

After an object reaches terminal velocity, the object no longer accelerates further. Based on Newton's second law of motion, net force acting on the object has to be zero. The force acting on a falling object are buoyancy [(https://en.wikipedia.org/wiki/Buoyancy)], drags [(https://en.wikipedia.org/wiki/Drag_(physics))] and weight [(https://en.wikipedia.org/wiki/Weight)]. Buoyancy is normally negligible in the atmosphere. So let's look at weight and drag.

The equation for weight $F_g$ is:

$$ F_g = m g, $$

where:

- $m$ is mass, and
- $g$ is local acceleration due to gravity.

The equation for drag $F_D$ is:

$$F_D\, =\, \tfrac12\, \rho\, u^2\, C_D\, A$$

where:

- $\rho$ is the density of the fluid,
- $u$ is the flow velocity relative to the object,
- $A$ is the reference area, and
- $C_D$ is the drag coefficient

At terminal velocity, if we ignore buoyancy, the weight and and the drag are balanced. It is clear that weight only depends on mass. If we assume that the object doesn't change it shape while falling, and the air doesn't change its density, then drag only depends on the object's velocity.

In fact, the formula for terminal velocity $V_t$ is:

$$ V_t= \sqrt{\frac{2mg}{\rho A C_D }}, $$

its derivation can be found here [([(https://en.wikipedia.org/wiki/Terminal_velocity#Derivation_for_terminal_velocity)].

Quite often, the falling object's velocity does not depend on the weight. This is because most of the time the objects don't fall far enough to reach a velocity at which drag matters.

It is important to know that if the object travels at terminal velocity, then its time of descent will depend on its mass. For example, feathers and leaves reach their terminal velocity pretty quickly.

In the previous section, we determined whether mass affects terminal velocity by analysing into the force acting on the falling object. We adopt the same approach in this section.

When a bike rolls downhill, mainly three forces act on the bike - weight, rolling resistance and drag. Assuming that there is no rolling resistance [(https://en.wikipedia.org/wiki/Rolling_resistance)], which implies that the tyre is perfectly rigid and smooth, and the ground is perfectly rigid and smooth, then it is a matter between weight and air resistance. It is clear that weight matters in this scenario.

Let us look at rolling resistance. The rolling resistance of a tyre can be calculated as the following:

$$\ F = C_{rr} N, $$

where:

- $F$ is the rolling resistance force,
- $C_{rr}$ is the dimensionless rolling resistance coefficient or coefficient of rolling friction, and
- $N$ is the normal force, the force perpendicular to the surface on which the wheel is rolling.

The normal force depends on the weight, so it is clear that rolling resistance is affected by weight. Although interesting in this case, the heavier you are, the more rolling resistance you get.

The point of the analysis here is not to derive an equation for calculating the velocity of a bike going downhill at a time point, it is to show that your weight matters when you go downhill.

Galileo's experiment should be understood as a demonstration that “a falling body would fall with a uniform acceleration, as long as the resistance of the medium through which it was falling remained negligible, or in the limiting case of its falling through a vacuum”. In other problems, a falling object having reached terminal velocity, and a bike rolling downhill, a lot of other force components are involved. Galileo's experiment did not take those force components into account, so its results cannot be generalised into those scenarios.

I think these misconception might be showing a bigger problem in science education. I think a lot of people (including me) dogmatically remember the results and conclusion of experiments done by people in the past, rather than believing in their own empirical observation, then come up with an explanation.

Some “clever” students (including me) fudged our experimental results when we were in secondary school / sixth form. I think in these situations, the students demonstrate more trust toward the authorities of science, rather than their own experimental results. I do not think these are the correct mentality / attitude.

I think one motto to remember is the Royal Society's motto, which is the Latin phrase “Nullius in verba”. It can be translated as “on the word of no one” or “take nobody's word for it”. Quite often, it is more important to come up with an explanation of what you observe, rather than nullify your observation because of some theories.

public/misconceptions_on_galileo_s_experiment_at_pisa_tower.txt · Last modified: 2019/07/24 08:37 by fangfufu

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