How does Newton's third law apply to objects falling at different rates?

[Brin]

I am so very confused by this third law business.

It all started with me thinking about whether it is true that heavy and light objects fall at the same rate (I don't mean approximately).

I figured that they don't. While the Earth attracts, say, an elephant and a small rubber ball, at the same rate, each object, respectively, attracts the Earth in a different amount. Thus the 'gap' between the two objects closes quicker between the Earth and the elephant, than between a rubber ball and an elephant. The elephant hits sooner even if the difference in time is ridiculously small.

Edit: I am thinking of rigid elephants and rubber balls acting at large distances.

BUT, Newton's third law says that the Earth attracts the elephant, and that the elephant has an equal but opposite attraction, right? So my misunderstanding then is, how can the elephant exert 2 forces simultaneously? One force being by its own mass, via gravity, and then another one in equal but opposite response to the Earth's gravity.

I know this is simple, but geez, my brain is all looped up. I think I am missing something.

 

[Tac-Tics]

With your elephant ball scenario, you have three objects. The key to working with these kinds of systems is that forces are additive. The force acting on the Earth is the force of the elephant's gravity PLUS the force of the rubber ball. Similarly, the force acting on the elephant is the force of the Earth's gravity plus the gravity of the rubber ball.

Due to its gravity, the elephant exerts one force ON EACH other massive object. It exerts a force on the ball and another on the earth.

Newton's third law is equivalent to the conservation of momentum. I think this is a simpler way of thinking about the law.

 

[Brin]

Thanks for your response Tic Tacs,
Do you confirm that heavier objects falling equal distance to smaller objects hit at a different times?

 

[Gnosis]

It’s not the mass of the miniscule free-falling objects that are producing any distinguishable gravitational attraction. While an elephant has considerably more mass than a ball, both the ball and elephant produce negligible gravitational attractions especially in comparison to the enormous mass of Earth (approximately 6 x 10^24 kg with a volume of roughly 265 billion cubic miles). Even with the Earth’s almost unimaginable enormous mass, it still only manages to produce a gravitational field strength of a mere 9.8 Newtons per kg to produce a rate of acceleration of only 9.8 m/s^2. This is why after one full second of free-fall, an object has only achieved a distance of 4.9 meters (just a hair over 16 feet) and a velocity of only 9.8 m/s (which is 32.152 ft/s, or a speed of 21.92 MPH). Earth has an enormous mass and yet, it only exerts a few pounds of force per kg. On such a grand scale, the gravitational force produced by the miniscule mass of an elephant or a ball is of no consequence. The Earth is still 2 x 10^21 times greater in mass than a 3,000 kg elephant, so even the elephant’s mass is negligible.

All objects (and without exception in a vacuum) free-fall at precisely the same rate, as free-fall tests have demonstrated time and again. Even in a dense atmosphere such as Earth’s, objects that vary greatly in their mass will still free-fall at precisely the same rate when released from lower heights since velocities achieved are still too low to produce enough air resistance to alter the object’s acceleration.

Over the years, I’ve performed countless free-fall tests from a variety of heights (mostly from .3048 to 4.9 meters, 1 to 16 feet) with objects that differ in their mass by thousands of times and every time, both objects released simultaneously from precisely the same height will make contact at precisely the same instant. One initial impact sound can be heard even though two objects made contact. If one object were to actually free-fall faster, there would be a delay that would cause two initial impact sounds to be heard (as when purposely releasing one object just a tad after the other).

 

[Tac-Tics]

The time it takes for an object to hit the ground is determined solely by it's acceleration (assuming the objects begin at rest, and at the same hight, yadda yadda).

Now the results depend on the exact set up. Let's do an ideal situation. Suppose the ball and the elephant are "practically" on top of each other. This means we're going to ignore the gravitation between the elephant and ball. Additionally, all the forces will be in one dimension, and we don't have to worry about x and y components.

The masses are m_earth, m_elephant, and m_ball, and similarly for other quantities.

Say the Earth is a distance r from the other two. The force acting on the elephant is G(m_earth m_elephant)/r^2. The force acting on the ball is G(m_earth m_ball)/r^2. The force on the Earth is G(m_earth m_ball)/r^2 + G(m_earth m_elephant)/r^2. But it's in the opposite direction of the other two forces, so we'll say it's actually the opposite of that value. So in our total system, we have F_earth, F_elephant, and F_ball:

F_ball = G(m_earth m_ball)/r^2
F_elephant G(m_earth m_elephant)/r^2
F_earth = -G(m_earth m_ball)/r^2 - G(m_earth m_elephant)/r^2

So, how are things accelerating? Use F/m = a:
a_ball = F_ball / m_ball = G(m_earth)/r^2
a_elephant = F_elephant / m_elephant = G(m_earth)/r^2
a_earth = F_earth / m_earth = -G(m_ball + m_elephant)/r^2

Note that the acceleration on the ball is the same as that on the elephant! The acceleration is only based on the mass of the other object. This is one of the distinguishing features of gravity, and is something not seen in any other fundamental force.

 

[claygod]

The force on the Earth is G(m_earth m_ball)/r^2 + G(m_earth m_elephant)/r^2.

.
That's not right.

F_ball = G(m_earth m_ball)/r^2
F_elephant G(m_earth m_elephant)/r^2
F_earth = -G(m_earth m_ball)/r^2 - G(m_earth m_elephant)/r^2 This sounds
like the gravitation attraction on the Earth due to the elephant and the ball is the force
of the elephant minus the force of the ball.

That's not right either. If we have an elephant
and a ball lifting the Earth up, the force should be additive. But you add it and still not right
cause then you would have more than twice the gravitational force of the Earth due to
the elephant and the ball.

So, how are things accelerating? Use F/m = a:
a_ball = F_ball / m_ball = G(m_earth)/r^2
a_elephant = F_elephant / m_elephant = G(m_earth)/r^2
a_earth = F_earth / m_earth = -G(m_ball + m_elephant)/r^2

If the acceleration of the Earth depends on both the mass of the ball and the elephant,
the acceleration on the ball and elephant should be different since they have different mass.
In other words, different masses freefall differently. For free fall rates to be a constant,
gravitational force should have no mass in the equations.

 

[Brin]

Whew, you guys have been fantastic on the replies.

Tac-Tic, I look at your equations, and you are only calculating the gravitational force the Earth exerts on the ball, and not the gravitational force of the ball on the earth. Note, just to make it simpler, let's say we drop the objects at two different times and we somehow time them to some amazing precision.

The definition of "falling" is kind of odd for what I am talking about, what I mean is "as the two objects move closer to each other." Our objects, in this case, are the Earth, an Elephant, and a Ball. They could just as easily be the Moon, Mars, and Earth, however.

Just by borrowing your equations Tac-tic:
I.e.
F_earth = G(m_earth m_ball)/r^2 = m_earth*a_earth_ball
a_earth_ball = G(m_ball)/r^2

This is not the same as the acceleration created by the elephant:
F_earth = G(m_earth m_elephant)/r^2 = m_earth*a_earth_elephant
a_earth_elephant = G(m_elephant)/r^2

i.e.

G(m_ball)/r^2 < G(m_elephant/r^2)
a_earth_ball < a_earth_elephant

This, to me, suggests that the distance between the two objects (i.e. the elephant and the ball) is closing quicker when the object is heavier - like two walls closing in.

Maybe this is more obvious if we dropped, say, the moon, and maybe Mars towards the earth? Clearly the larger of the two would collide with the Earth first, right? Because there is a greater pull between them? Earth wouldn't just stand still in such an event, am I right? We know that the pull from the moon is strong enough to pull Earth towards it (enough to 'wobble' it) so by this same logic, surely heavier objects hit the surface of the Earth right?

Edit: Claygod, you created your post while I was writing mine up - the same equations made me uneasy, so I simplified it a little bit more by saying that we are to conduct two different trials with our three objects.

 

[Tac-Tics]

.
That's not right either. If we have an elephant
and a ball lifting the Earth up, the force should be additive. But you add it and still not right
cause then you would have more than twice the gravitational force of the Earth due to
the elephant and the ball.

I don't see that I made a mistake. The forces, as I have calculated them, ARE addative. You can pull out the constants G and m_earth, and nothing is doubling.

For free fall rates to be a constant,
gravitational force should have no mass in the equations.

The acceleration of a mass by gravity is a function only of the other object's mass.

 

[atyy]

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

Looks like a=m1+m2, so the elephant will hit first.

 

[russ_watters]

Thanks for your response Tic Tacs,
Do you confirm that heavier objects falling equal distance to smaller objects hit at a different times?

Let me put it more simply than the others:

Yes.

But as you describe the scenario, there are two completely separate accelerations (the elephant accelerates one way, the Earth accelerates the other). The acceleration of the elephant and the ball are, in fact, both exactly g with respect to a stationary reference frame (the system's barycenter), but that does not imply that they will hit the ground at the same time. You can use the Earth as a reference frame, but you must accept that it is not a stationary reference frame, it is an accelerating one.

 

[LennoxLewis]

What you must understand is that gravity is not a "force" like me pushing a door. That is clearly a force from me, to the door. Is gravitational force the Earth pulling the elephant closer? Or is the elephant pulling the Earth closer?

The answer is neither. Gravitation is an interaction that goes both ways.

 

[A.T.]

What you must understand is that gravity is not a "force" like me pushing a door. That is clearly a force from me, to the door. Is gravitational force the Earth pulling the elephant closer? Or is the elephant pulling the Earth closer?

The answer is neither. Gravitation is an interaction that goes both ways.

I don't get this one. If you push a door, the door also pushes you with the same force: action = reaction. This interaction also goes both ways.

 

[LennoxLewis]

I don't get this one. If you push a door, the door also pushes you with the same force: action = reaction. This interaction also goes both ways.

The door doesn't push back with the same force unless it's not moving. Remember those videos of a karate dude smashing a set of bricks? That doesn't hurt the hands much at all because the bricks break and don't exert a force back. However, if you fail to break the bricks, they WILL exert the same force and you'll badly hurt your hand.

But that is beyond what i meant. I meant that gravity between two massive objects is an interaction, i.e. it goes both ways and it's from one to the other only, like the example of pushing a door, a sled, or whatever moving object.

 

[Rybose]

The door doesn't push back with the same force unless it's not moving. Remember those videos of a karate dude smashing a set of bricks? That doesn't hurt the hands much at all because the bricks break and don't exert a force back. However, if you fail to break the bricks, they WILL exert the same force and you'll badly hurt your hand.
QUOTE]

I'm not sure if I agree with this last statement. Newton's 3rd law states if particle A exerts a force on particle B then particle B exerts an equal and opposite force on particle A. Period.

I think the examples LL is citing are complicated real-life scenarios where the dynamics of the process may lead to deceptive observations.