How do these factors affect the free fall of objects?

Question

Take an example of two balls of different masses being dropped from about 250 meters from the ground.

How do the following factors affect free fall:

1. Air resistance.
2. Absence of air resistance.
3. Conduction of experiment on the moon.

For factor 2, I believe that they should fall at an equal rate, one of the reasons being that the force on heavier ball is greater than the second's however it has more inertia balancing it out.

Answer 1

Here is an extremely simple explanation:

1. Force = Mass x Acceleration
2. Force / Mass = Acceleration
3. (Mass x Acceleration due to Gravity) / Mass = Acceleration
4. Acceleration due to Gravity = Acceleration

For further intuition, consider this:

1. The greater the mass, the greater the inertia.
2. The greater the inertia, the greater the difficulty to accelerate the mass.
3. The greater the mass, the greater the weight (i.e. gravitational force acting on the mass).

So, looking at this you can understand that the increased difficulty of acceleration is nullified by the increased gravitational force. They sort of 'cancel' each other out, leaving only acceleration due to gravity behind.

To answer your question:

1. Air resistance for each object would depend on the surface area of the side facing down to the ground (see aerodynamics), density and shape.
2. If we neglect air resistance, both objects would accelerate equally (as just proven).
3. The time taken to reach the ground would be approximately six times the time squared (using the ever-helpful SUVAT equations).

Answer 2

If only the forces of gravity are present, all objects fall at the same rate. This is what one calls equivalence principle. In classical mechanics it shows up in the force law for two particles of gravitating mass $m_G$ and $M_G$, where $M_G$ shall denote the earth's mass. $$ m_i \cdot \vec{a} = -G \cdot \frac{m_G \cdot M_G}{|\vec{r} - \vec{r} '|^2 } \cdot \frac{\vec{r} - \vec{r} '}{|\vec{r} - \vec{r} '|} $$ If the inertial mass $m_i$ is equal to the gravitating mass $m_G$ (this is what we observer!), the accelerations is independent of the particle's mass, because they cancel in the equation. Thus, all object fall at the same rate.

This is however only half of the story. Because usually one has to take care of different forces, too. For example if friction (due to an atmosphere) is present. These forces orginate from electromagnetic forces on tiny scales and are modeled to depend on parameters like the size and shape of an object and its surface texture. Also, there is no such thing like the equivalance principle for EM-forces. If an object is heavier, it will "resist" friction more easy if the parameters like size, shape and surface texture are equal, leaving an heavier object falling at a higher rate in an atmosphere than the lighter one. Furthermore, friction is most often modeled to the depend on the objects velocity. For this reason, there is a finite, maximal velocity for any object falling in an atmosphere. This velocity depends on the objects density.

On the moon, however, there is no atmosphere (at least in very good approximation), thus objects on the moon will always fall at the same rate, no matter of their properties!

Answer 3

Without air resistance all objects are accelerated with gravity $$g = 9.81 \frac{m}{s^2}$$ Only the air causes a "slower" acceleration. This effect depends from the density and shape.