I've seen an experiment where a closed book with a piece of paper lying on its cover is dropped a short distance (the piece of paper is smaller than the book; the book's cover is parallel to the Earth's surface). Both book and paper fall together and this is supposed to demonstrate that all objects (regardless of their mass) experience the same acceleration when in a state of free fall. But are there other reasons why the book and paper don't separate during their fall? Would not air pressure keep the paper pressed against the book's cover? I guess I'm thinking about the experiment where you can break a ruler under a newspaper because of air pressure.
[Physics] Falling book and paper
accelerationaerodynamicsdragnewtonian-gravitynewtonian-mechanics
Related Solutions
Suppose the balloon with air (A) weighs $m_a$ and the balloon with concrete (B) weights $m_b$. The force accelerating the balloons downwards is $m_a g$ for A and $m_b g$ for B, where $g$ is the acceleration due to gravity. In the absence of air the acceleration is simply this force divided by the mass, so both balloons accelerate at the same rate of $g$. So far so good.
Now suppose the air resistance is F. We don't need to worry exactly what F is. The force on balloon A is $m_a g - F$, so its acceleration is
$$ a_a = \frac{m_a g - F}{m_a} = g - \frac{F}{m_a} $$
and likewise the acceleration of balloon B is
$$ a_b = \frac{m_b g - F}{m_b} = g - \frac{F}{m_b} $$
So the balloons don't accelerate at the same rate. In fact the difference in the accelerations is simply
$$ \Delta a_{ba} = a_b - a_a = \frac{F}{m_a} - \frac{F}{m_b} $$
Since $m_b \gg m_a$ the difference is positive, i.e. balloon B accelerates at a much greater rate than balloon A.
Response to comment
I think there are a couple of possible sources of confusion. Let me attempt to clarify these, hopefully without making things even more confused!
Firstly the air resistance affects both the acceleration and the terminal velocity. I've only discussed acceleration because terminal velocity can get complicated.
Secondly, and I think this is the main source of confusion, the gravitational force on an object depends on its mass while the air resistance doesn't. However the air resistance depends on the velocity of the object while the gravitational force doesn't. That means the two forces change in different ways when you change the mass and speed of the object.
Incidentally, Briguy37 is quite correct that buoyancy has some effect, but to keep life simple let's assume the object is dense enough for buoyancy to be safely ignored.
Anyhow, for a mass $m$ the gravitational force is $F = mg$ so it's proportional to mass and it doesn't change with speed. Since $a = F/m$ the acceleration due to gravity is the same for all masses.
By contrast the air resistance is (to a good approximation) $F = Av^n$, where $A$ is a constant that depends on the object's size and shape and $n$ varies from 1 at low speeds to 2 at high speeds. In your example the size and shape of the ballons is the same so $F$ just depends on the speed and for any given speed will be the same for the two ballons. The deceleration due to air resistance will depend on the object mass: $a = Av^n/m$, so air resistance slows a heavy object less than it slows a light object.
When you first release the balloons their speed is zero so the air resistance is zero and they would start accelerating at the same rate. On the moon there is no air resistance, so the acceleration is independant of speed and the two objects hit the ground at the same speed.
On the Earth the acceleration is initially the same for both balloons, but as soon as they start moving the air resistance builds up. At a given velocity the force due to air resistance is the same for both ballons because it only depends on the shape and speed. However because acceleration is force divided by mass the deceleration of the two balloons is different. The deceleration of the heavy balloon is much less than the deceleration of the light balloon, and that's why it hits the floor first.
From the way you phrased your question I'm guessing that and answer based on calculus won't be that helpful, but for the record the way we calculate the trajectory of the falling balloons would be to write:
$$ \frac{dv}{dt} = g - \frac{Av^n}{m} $$
This equation turns out to be hard to solve, and we'd normlly solve it numerically. However you can see immediately that the mass of the object appears in the equation, so the change of velocity with time is affected by the mass.
If you place the book on the desk and a piece of paper on it, if there is no wind turbulence it will stay there until the cleaning lady comes.
Why? because the gravitational force keeps it there, and in addition frictional forces, i.e. electromagnetic interactions between the paper molecules and book molecules . This would be true in vacuum too. Atmospheric pressure on top of the paper and "vacuum", i.e. displacement of air between the book and paper contact surface adds to the stability.
When both fall, unless a turbulence lifts a side of the paper and air enters underneath, they will fall together. If falling from a great enough height, turbulence will separate them inevitably. Just needs some air to intervene between the two surfaces.
Best Answer
To an extent, possibly.
This is not the main reason why it is done though; although it does relate to your newspaper-ruler example.
The reason would be to minimize the effects of air drag on the piece of paper. Although all objects influenced the same by gravity; they react differently with the air. A piece of paper would fall a lot slower than the book if they were both exposed to air drag. Instead; the piece of paper avoids the drag by travelling in the streamline of the book.
Realistically, if the piece of paper were to fall slower than the book under the influence of gravity (due to lower mass) then even with the air pressure, it should still slowly separate from the book. (the reason the ruler breaks in the newspaper experiment is because you are trying to act against the pressure very quickly).
Either way; this isn't really a great experiment to demonstrate how different masses react the same to gravitational force. Both the book and the paper have very large areas; therefore the drag force is substantial. To eliminate the drag force they set up the experiment in a way that pressure differences could keep the paper and the book falling with similar speeds, even if gravity was acting unevenly.
Really, this test should be done in a vacuum to eliminate this source of potential error. It's a good sign that you can recognize this error in the experiment. Noticing details like this is important when designing good experiments.