Alright, today, my teacher was explaining to use the "Freefall Motion Equation," which basically explains why objects of different sizes fall at the same speed when dropped. Anyways, he was saying the equation doesn't take into account air resistance. So, I asked him if the equation then, was actually theoretically, since their is air on Earth, and he never really gave me an answer. Can anyone enlighten me on this? Unless this can be done in outer space, isn't the equation more of like a guideline, since air will never make it ideal?

NCEA Level 2 Physics which I'm doing this year is the same, none of the theory takes air resistance into account. Yes, I think it is more of a guideline.

Yes, you are actually correct. The Freefall Motion Equation is based on a theoretical "vacuum."

If a feather and a steel ball fall from the same height, at the same time, they will not hit the ground at the same time. This is due to differences in magnitude, such as weight, along with external values such as wind current, and direction.

One way in which this situation is possible would be to even the magnitudes, either by creating a vacuum in which wind/air/weight is a non-factor, or by leveling the weight values. If you were to have a ton of feathers and a ton of steel balls, as they weigh exactly the same, they would hit the ground at the same time. However, if we are talking specifically about 1 feather and 1 steel ball, then the greatest example would be to demonstrate this inside a vacuum.

You may also be interested in checking out Newton's 2nd Law of Motion which also deals with the acceleration of two objects dropped at the same time, and why they would land together at the same time. I would read up on Newton's 1st Law though, as a precurser, since the 2nd law is more of an extension of the 1st.

In physics, a vacuum is the absence of matter in a volume of space. A perfect vacuum would be an ideal state that does not really exist, but is best described by outer space. (I think you seem to get this part already.)

So to basically answer your question: Yes, the equation is more or less a guideline of how it should work, but is unproven and unable to exist in reality.

Physics tends to be difficult, but I find it fun in that way that no matter how absurd the idea is, in physics there is a way to validate it, even if you are unable to prove it in a real-life scenario, but in theory there is always a way.

Good luck!

-Priest-

Correct me if I'm wrong, but isn't the question more about air resistance and the terminal velocity one reaches because of it if you were, say, jumping from an airplane?

terminal velocity is a by-product of air resistance, you have no terminal velocity in space (e.g. a vacuum).

In physics, you create model situations. Models are simplifications of reality. They're useful for exploring the simple laws that govern the way things behave, without letting real-life complications get in the way.

Imagine throwing a tennis ball into the air. If you think of the tennis ball as a small, smooth particle travelling in a vacuum, you've created a model. You'd probably also assume that the pull of gravity doesn't change over its short journey. This model isn't "true": every one of those assumptions is false. However, by doing this you create a problem which is easily solved. With a pen and a piece of paper, you could work out how high the ball will go or how long it will spend in the air.

If you factored in all of the complications, you'd need a supercomputer to work for hours to figure it out. After all that time, you'd arrive at an answer that isn't much different from the one you did yourself.

The result you mention isn't false. The maths is completely sound, but it's not a law which fits all circumstances. It fits the particular model which neglects air resistance. For many circumstances, including a cannon ball falling, that's a very good approximation. For others, like a feather, you start to see the holes in the model.

Okay, cool.

Another thing you must remember is that it is not always a feather vs rock situation. You might have two identical balls but with different densities ( hence different weights ) and the "Equation" works even with air resistance considered.

That's close but not quite right.

The formulas depend on a vacuum because of hydrodynamics as well. If you have objects of equivelant mass but disproportionate surface area, the larger (surface area wise) of the two will generally suffer more from the effects of air resistance (drag). The more voluminous of the two will also have a greater amount of buoyancy. Higher Mass objects have higher potential energy as well, so they "resist" air resistance better.

To see the effects of drag (which is only found out of vacuums) you only need to drop a 1 gram paper clip and a 1 gram $1 bill through the air.

Remember that air is a fluid...it has weight and it acts like a fluid, even though its density is very, very low. Thus, the vacuum is required to reduce the effects of drag and buoyancy, among other things. In the case of buoyancy, it's Archimedes' principle at work.

To see an example of buoyancy, drop that same paperclip and dollar bill into a pail of water. While friction is something of an "analog" stat (it's infinitely variable), bouyancy is more of a "digital" stat...either it will float or it won't. I suspect buoyancy would affect drag (since you're decreasing the effective weight of an object in comparison to its surface area), but I don't know for sure and it's lunch time, so I don't want to go look, since I know I'll forget about eating and get lost in this stuff all day.

Buoyancy is related to the concept of mass vs. volume. One liter of air at 15 PSI weighs about 1.2 grams (or just a tad more than a dollar bill). Imagine all the volume in a one-litre bottle!

In theory, If you could somehow get a 1 ton object to displace more than 1 ton of air, it will float.

This is how we made concrete ships in WWII

Make sense now?

But the other example of equal sized spheres with different densities is accurate. They will fall and land at the same rate.

In a vacuum, yes.

They should fall evenly in an atmosphere too. Drag is to do with surface area rather than weight.

It has been said that weight is a factor in determining how fast items will fall, even in a vaccum. This is simply not true - one could drop a feather and, I don't know, an ocean liner in a vaccum and if they started at the same height then they will hit the ground at the same time.

No..anywhere. A vaccuum only eliminates drag. It does not affect gravity. Since both spheres have the same surface area, the air has no adverse effect.

drag has to do with surface area AND weight. The higher the surface area/weight ratio, the more drag.

Which is why we say in a vacuum.

Am I not understanding what you're saying? It sounds to me like you're arguing that objects will fall at the same rate out of a vacuum too. That's silly.

Providing they are essentially the same shape (surface area), regarless of the density, that is true:

http://en.wikipedia.org/wiki/Drag_equation

:charles:

What I am saying is two spheres of EQUAL size and surface texture of DIFFERENT weights will still fall at the same rate since the only factor not taken into account by the theory is that of drag.
Qrop a 12 and 16 pound bowling ball. There surface structure is similar enough to prove the point. They will land at the same time even in your backyard.

Drag is not defined at all by weight (or mass more accurately). Look at any formula for the coefficient of drag.
think about it....it is about wind resistance and the only factor in wind resistance is surface area, air density and speed.
http://www.grc.nasa.gov/WWW/K-12/airplane/dragco.html

Yeah, you're completely right. :">

I was thinking about this in the car on the way home. Drag doesn't care about mass, but that fact actually makes them fall at different rates.


m = mass
a = acceleration
g = acceleration due to gravity (9.8 metres per second per second)
D = drag force, independent of mass.


Force downwards = Force due to gravity - Drag force

F = m a = m g - D

Divide all terms by mass to cancel.

F/m = a = g - D/m

acceleration = acceleration due to gravity (fixed) - ( Drag (fixed) / mass )

So the acceleration is higher if the mass is higher. In an atmosphere, the denser ball will hit the ground first.

Sorry for the confusion, but at least I admitted it.

You're almost there...
Don't confuse force with velocity. They are not the same thing. Neither is momentum. Certainly the heavier ball has more momentum and had more potential energy. However the gravity bewteen the two objects is related to mass since objects of grater mass have a greater interneal gravitational force.
But the gravity between the two objects is not in question, but the gravity of the third object (earth).
The earth gravitational force is 9.8 m/s₂

I didn't say that force, velocity and momentum were the same thing. I said that the resultant force is related to the acceleration of the balls towards the earth. Are we answering the same question?

By the way, 9.8 metres per second per second is not a force, it's the acceleration that an object experiences when free-falling towards the Earth.

THis sentence is what I was trying to allude to.
The denser ball hits the ground at the same time as the less dense ball cause the rate of accelration is equal and independant of mass.
Yes I used the wrong term..of course the rate of acceleration due to gravity is 9.8 m/s/s

Yeah, that's what I thought originally. But, after I thought about it, I realized that hogleg was right, and this is only true in an vacuum. Have a look at the equation I posted to see why.

The equation is incorrect. You are measuring force and not velocity. It is not the same thing.
v=d/t
Like I said...take both bowling balls and drop them. They will land at the same time. At any interval in the fall they will be in the same place.
The more dense ball has more potential energy (m*h) but the energy here is only represented in momentum (tendency to remain in motion) and impact (release of potential energy)and not time or velocity.
The only reason a feather argument is ever used is because we all see feathers fall slowly and some cannot comprehend the air slows it down and not the mass.

Read this and let me know where I'm going wrong.

Force divided by mass gives acceleration. If an object accelerates faster, it will have a higher velocity at a given time and will eventually hit the Earth sooner.

We start with forces because that's what we're faced with: the force of gravity pulling the balls down and resistance forces slowing the fall.

We say that the net (or resultant) force in the downward direction is the gravitational force minus the force due to the air resistance, as it operates in the other direction.

(1) Resultant force = gravitational force - resistance force.

Now force equals mass times acceleration.

(2) Resultant force = mass * resulting acceleration of body.

Plugging (2) into (1) gives

(3) mass * resulting acceleration = gravitational force - resistance force

-----------

Now we know that the gravitational force is the mass of the falling body multiplied by 9.8 m/s/s, the acceleration due to gravity on Earth.

(4) gravitational force = mass * 9.8

Plugging (4) into (3) gives

(5) mass * resulting acceleration = mass * 9.81 - resistance forces

---------

We are also agreed that drag doesn't depend on mass. There is no mass term within the resistance force term. Let's divide the whole equation by mass.

(6) resulting acceleration = 9.81 - (resistance forces / mass)


You can see that now, we've stopped talking about forces. We're now talking about the acceleration of the body towards the Earth. If the object has a larger mass, which it does if is the same shape and size but a higher density, then its acceleration is higher. If it has a smaller mass, it's acceleration is smaller.

Therefore, in an atmosphere, the more massive ball accelerates faster. Hogleg was right and the rest of us (including me) were wrong. The classic result of things landing together only applies in a vacuum. In a vacuum, the green terms are all zero and they do indeed fall together.

You are confusing linear movement, like a car moving across a plane, with a gravitational pull. If lighter objects fell faster, or slower, those astronauts in the Space Shuttle would be either hurled through the windshield or through the rear of the craft as their mass is vastly different than the shuttle.
The force of the earths pull changes exactly as the mass is different.
Just drop the dang bowling balls!!!

Gravitational pull is constant. It's that blue 9.8 in the final equation. Drag itself is the same, but as it has no mass dependence you get a drag/mass ratio when you cancel.

I'll tell you what: I'll drop bowling balls if you drop a ping-pong ball and a golf ball through water and tell me which hits the bottom first. Water is just thick air, after all.

Water is not thick air. Ping pong balls float.
Gravitational acceleration is constant but the attraction is not.
Although heavier objects are attracted more readily to the center of the Earth because of their weight, they also have a greater resistance to change in movement due to their greater mass. These effects perfectly counter act one another, and the objects fall at the same speed.

Look, I agree with you on the gravitational attraction. In a vacuum, the objects would hit the ground together. Nobody is disputing that m a = m g means that a = g because mass cancels out completely.

The trouble is, in an atmosphere there is another term. For drag, m a = drag. Drag has no mass component, so whilst it is the same for both balls, you can't cancel the mass term. You're left with a = drag / mass.

So you can completely cancel and account for the downwards force, but the upwards force still has a mass component.

This is basic maths. Point to the flaw.

Because drag due to air resistance is only dependant on surface area. This is a universally accepted law.

EXACTLY!!!

Any force can be written as F = m a.

So we have two forces, gravity and drag.

Gravity has a mass term. Drag doesn't as you've just said.

So F = m a = (m * 9.8) - drag

Notice there is no "m" in drag. We're agreed up to here.

Divide by "m"

F/m = a = 9.8 - (drag/m)

It's this m here ^ that makes all the difference.

Think of it like this: the mass of a body is a measure of its resistance to an applied force. The ONLY force that this doesn't apply to is gravity. As the balls have the same surface area, they receive the same amount of drag. But as they have differenct masses, they react to it differently. The heavier ball can "resist" more of the drag force and it therefore is less slowed by it.

It is precisely because gravitational force is affected by mass and drag isn't that this works. I know it's counter-intuitive, but if you follow the maths, that's what happens.

When Galileo dropped masses from towers, the masses actually fell at different rates. If you had a sensitive enough stopwatch, you could use the time difference to calculate air resistance for the objects. The difference would be very small because the atmosphere is a *fairly* good approximation of a vacuum. That's why there is no point in me dropping bowling balls: there is no way I could tell without specialist equipment.

YEs but the heavier ball resists the pull in equal amounts to the increase in the affect of gravity on it. Since heavier objects are more affected by gravity this cancels the inertia of the heavier mass to resist it.
All that a vaccuum does is remove the air from the equation, but having identical surface area the effect is the same.

My friend, I am totally agreed with you on gravity. The reason that works is that gravitational attraction increases with mass, and the ball's resistance to it increases with mass too. They cancel each other out.

Agreed. Let's never talk about gravity again.

Drag forces do not increase with mass, as you pointed out above. The balls' resistance to them does increase with mass as normal, however. Therefore they both receive the same upwards force (as they have the same surface area) but resist it differently. Drag is a linear force just like pushing a car along a road.

Do you see the distinction? We've been agreeing about gravity for two pages, but you need to realize that the drag force is fundamentally different to gravity because there is no mass in its equation.

Summary:

Gravity:

Both balls receive different forces and resist it differently. The resistance increases with the attraction, so they are both affected the same way.

Drag:

Both balls receive the same force (same surface area) and resist it differently. Therefore the net result is different for each ball.

Hope this helps. If only gravity is involved (like in a vacuum), they will fall together. If drag is involved too, they won't.

OMG...
You cannot compare frictional forces like pushing a car because the resistance is across a different plane.
Is pushing a heavy car harder? Yes...because it's mass cause two forces. Inertia and STATIC friction.
Stop thinking of feathers and you will get it.
If you drop those two cars from a plane they will hit the ground at about the same time.
Now..take this into consideration.

An astronaut is in orbit is not truly in a zero-grav environment but in a constant stae of free-fall. If he/she opens their hand an releases a gold ball, they are both hurtling towards earth in an environement at teh same time. Yet the ball falls AT THE SAME rate as the astronaut and as the shuttle.

First of all...holy cow...all the equations right here!

TMF, I'm not sure what you're arguing there, but space is a vacuum. It seems to me what you're forgetting is that gravity isn't the only force at work. The atmosphere is a fluid, and so using examples like dropping things through water is entirely valid. Think of air as really, really thin water. If this wasn't the case, we wouldn't have a terminal velocity, and we wouldn't have objects that never reach terminal velocity.

Thats the whole point of everyone's argument, actually. These rules work in a vacuum, not in an atmosphere (or lake), for exactly the reasons you're pointing out. The static friction applies more or less to air molecules sliding over bowling balls, just in a vastly lesser way than rubber tires skidding on concrete. When you calculate the speed of a falling object, and you want to be exact, sooner or later you have to calculate the force of the air pushing back (the resistance), which doesn't happen in a vacuum.

When it comes to inertia and friction, the lighter car is both easier to start pushing and easier to slow down. Maybe it's a different plane of resistance, maybe not, but I can tell you this: Firing a heavier bullet from a gun will result in a lower but more consistent velocity over the path of the projectile, because a heavier bullet is affected more by inertia, both in acceleration and decelleration. A lighter bullet will hit a higher speed and decelerate faster, becoming extreme at longer ranges. It gets so extreme that at 100 yards, a shotgun blast of 3/4 oz. lead shot in an average size (we'll say number 6, and if you could keep the pattern tight enough) will leave the barrel of the gun at better than 1,100 feet per second and won't make it through a cardboard box by the time it's 300 feet away. A 3/4 oz lead slug leaving the barrel at the same velocity would literally tear you apart at 100 yards.

Were the laws we're discussing true outside of a vacuum, I'd have to assume that the bullets, if subjected to the same amount of energy (or at least the amount of energy required to launch them at identical velocities), would accellerate and decellerate at the same rate. Since they do not, I have to assume that air resistance is a quantifiable and significant force.

At least, thats the point I thought I was trying to make.

The inside of the shuttle is not a vaccuum, however. If it was they would require their full gear.
If you fire a bullet exactly parallel to the earth and drop one of the same size at the same time, they would both hit the ground at the exact same time.
You cannot argue that water is thick air because it is not. Bouyancy depends on material and air content.
This is all Newton and Galileo's stuff, not mine. Argue with them...lol.
The reason that a vaccuum is needed to prove the theory is because of air resistance and nothing else.
I can't believe you guys finished high school physics without knowing these principles.

Actually it was archimedes. I never took High School physics, only college.

I am kinda dumb though. But I can lift heavy things, so it all works out in the end.

Allthough I would have to say that the astronaut example is more akin to einstein's explanation of general relativity (only he used a train, not a space shuttle). But then again, I'm not so smart.

I do like all the pretty colors in the posts, though!