# Falling Objects With Different Mass

1. Mar 15, 2010

### BL4CKCR4Y0NS

I've heard and read many times about Galileo and his standing up in the Leaning Tower of Piza. How he dropped two objects of different mass and proving that the two objects hit the floor at the same time if released at the same time.

What I never understood ... was WHY this happened. What causes the two objects to hit the floor at the same moment.

Thanks in advance.

2. Mar 15, 2010

### Fightfish

This is because, ignoring air resistance, all objects accelerate downwards with the same acceleration, g. While the gravitational force acting on a more massive object is greater than that acting on a less massive object, the more massive object still accelerates downwards at the same acceleration as the less massive object due to the more massive object possessing a larger inertial mass.

To simplify issues, we approximate the earth's gravitational field to be uniform such that the value of g remains constant. The gravitational force acting on a body of mass m on earth is then $$F_{g} = mg$$. From the simplified version of Newton's Second Law, we have $$F = ma$$. Combining, we see that the acceleration a of the body is simply g - independent of its mass.

3. Mar 15, 2010

### BL4CKCR4Y0NS

Oh yeah .. that makes sense.

Thanks. :)

4. Mar 15, 2010

### HallsofIvy

Saying that "we approximate the earth's gravitational field to be uniform such that the value of g remains constant" may appear to be begging the question.

In more detail, The gravitational force between two bodies is $F= GmM/r^2$ where G is a universal constant, m and M are the masses of the two bodies and r is the distance between their centers. Take the earth to be one body and the object to be dropped to be the other. The radii of two different objects are so small compared to the radius of earth that "r" can be taken as a constant for any object dropped at the surface of the earth. As Fightfish said, the second law is F= ma so for any object we have $GmM/r^2= ma$ and can cancel the two "m"s: $a= GM/r^2$ for any object dropped at the surface of the earth.

Since two objects always have the same acceleration and start with speed 0 ("dropped" not "thrown down"), the always have the same speed at any time and so always drop the same distance in the same time.

5. Mar 15, 2010

### TurtleMeister

The problem I had in understanding the universality of free fall was this: If increasing the mass of the falling object also increases the force, then why is there no effect on acceleration? The truth of the matter is that the relative acceleration does increase. In other words, the time that it will take for the two objects to meet will decrease. However, the universality of free fall does not refer to relative acceleration because the relative acceleration is a non-inertial frame of reference. The frame of reference for the universality of free fall is the center of mass. So if you increase the mass of the falling object you are also shifting the center of mass closer to the falling object. This has the effect of keeping it's acceleration (relative to the center of mass) the same, even though the relative acceleration has increased. So $a=GM/r^2$ and $a_{rel}=G(M+m)/r^2$.

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