Mass doesn not effect speed of dropped objects

[TurtleMeister]

Since you agree with my post, I must assume we are just having communication problems. I don't know which of us DH was referring to, but after reading over my post I can see where I could have done a better job. For example, I may appear to be contradicting myself in the third sentence when I use the word "force" twice in a way that could be interpreted as two separate forces. I should have used "magnitude of the force".

After reading over your post several times, I think I understand the idea behind what you are trying to convey, and I agree with it.

 

[ernestpworrel]

Okay. Turtlemeister, you're fine. DH is on the list of people I don't agree with.

 

[D H]

Were you speaking to me?

Yes, I am talking to you. You are posting utter nonsense and in doing so you are not helping. In fact, you are hurting the discussion. Post #46, just to pick one of your egregious posts, is chock-full of nonsense.

All objects in freefall near the Earth's surface can be treated as having equal masses when considering only the force imparted to them.

Nonsense. The force imparted on objects in freefall is proportional to their gravitational mass, and gravitational mass is equal to inertial mass.

They are all accelerated at the same rate. Whether an object's rest mass is 5 kg or 5000 kg, it's going to accelerate toward Earth at the same rate as any other object (ignoring air resistance).

Correct. One out of six is not a very good mark, however.

The magnitude of the force that is communicated to each object is the same.

Nonsense. You are contradicting Newton's law of gravitation here.

It feels the same force as the "lighter" object because the force causing acceleration is the same, the mass of the earth.

Nonsense. You are once again contradicting Newton's law of gravitation.

F=ma describes the amount of work that an object with momentum will do on an object it collides with.

Nonsense. F=ma has nothing to do (directly) with work. It has everything to do with force, mass, and acceleration. The nonsense you are posting contradicts the following:

1. Newton's second law of motion, F=ma. Suppose two objects with measurably distinct inertial masses are observed to undergo the same acceleration. The force on those two objects *must* be different if Newton's second law has any validity.
2. Newton's law of gravitation, F=GMm/r². Now apply this to two different objects at the same distance from the primary object: F₁=GMm₁/r² and ₂=GMm₂/r². Different masses yield different forces. Now apply Newton's second law to determine the accelerations of these different masses: a₁=F₁/m₁=GM/r² and a₂=F₂/m₂=GM/r². Even though the forces acting on the two objects are different, the two objects will undergo the same acceleration. This assumes of course that gravitational mass (the mass quantities in Newton's law of gravitation) and inertial mass (the m in F=ma) are one and the same.
3. The equivalence principle, which (among other things) posits that inertial mass and gravitational mass are indeed one and the same. The equivalence principle is one of the most accurately tested concepts in all of physics.

Now back to the last egregious post.

I'm not contradicting Newton's third law.

Yes, you are.

The rate to which an object is accelerated varies directly with the mass of the object.

Now you are contradicting yourself. In post #46 you explicitly said "They are all accelerated at the same rate." Moreover you are contradicting over 400 years worth of experimental observations.

 

[ernestpworrel]

The force imparted on objects in freefall is proportional to their gravitational mass, and gravitational mass is equal to inertial mass.

Fine. A 500 kg object will hit harder than a 500 g object.

F=ma has nothing to do (directly) with work. It has everything to do with force, mass, and acceleration.

Work equals (blank) times distance.

Suppose two objects with measurably distinct inertial masses are observed to undergo the same acceleration. The force on those two objects *must* be different if Newton's second law has any validity.

What force on those two objects? You mean the force of the Earth's gravitational acceleration? Isn't it a constant force? Doesn't it accelerate all objects equally? Doesn't f=ma describe the momentum transferred in a collision? When an object is in free fall, what (besides the air) is it colliding with?

Different masses yield different forces. Now apply Newton's second law to determine the accelerations of these different masses: a₁=F₁/m₁=GM/r² and a₂=F₂/m₂=GM/r². Even though the forces acting on the two objects are different, the two objects will undergo the same acceleration. This assumes of course that gravitational mass (the mass quantities in Newton's law of gravitation) and inertial mass (the m in F=ma) are one and the same.

Okay, I think I need to clarify that I have been talking about the gravitational acceleration of the Earth only. That is the only force (neglecting air resistance) that an object in free fall will feel. Note that I said an object in free fall, not an object that has fallen. Now I understand what you and many others here have been trying to convince me of. We've apparently been talking about two different things. I've been talking about the only force acting on objects in free fall. You've been talking about that force along with the force that resists change in momentum. The term r² refers to distance. Work is force times distance. So the work done by an object that collides with the Earth is determined by the distance it takes to slow to zero times the distance over which it was accelerated along the y axis. If I looked at it your way, I would agree with you. The reason I wasn't looking at it your way was that someone said that objects that are more massive feel a greater force which is not correct. Example: If the gravity of the Earth were the only force affecting the moon, the moon would collide with the Earth but would still only be accelerated at 9.8 m/s². Now I know that the Earth will be accelerated toward the moon at a lesser rate. But this has nothing to do with the force that the moon will feel until it collides with the earth.

Now you are contradicting yourself. In post #46 you explicitly said "They are all accelerated at the same rate." Moreover you are contradicting over 400 years worth of experimental observations.

You're right here. I should've said the rate to which an object is accelerated on the surface of the Earth varies according to its mass.

 

[D H]

You mean the force of the Earth's gravitational acceleration? Isn't it a constant force? Doesn't it accelerate all objects equally?

Questions answered in order:

• "You mean the force of the Earth's gravitational acceleration?": This does not make sense.
  Gravitation is a force. I was talking about the Earth's gravitational force exerted on a pair of objects of different mass.
• "Isn't it a constant force?": No.
  F=GMm/r². Keeping M (the mass of the Earth) and r constant (the distance from the center of the Earth) but varying m (the mass of the other object) and it should be obvious that the gravitational force is not constant.
• "Doesn't it accelerate all objects equally?": Yes and no.
  For all objects that are at the same distance from the Earth, the answer is yes.
  For all objects period, the answer is no. An object a few meters above the surface of the Earth accelerates at 9.81 m/s² toward the Earth. The gravitational acceleration of Pluto toward the Earth is considerably less than that.

Doesn't f=ma describe the momentum transferred in a collision?

No.

 

[ernestpworrel]

DH, the Earth exerts the same gravitational force in all directions. I know that gravitational acceleration diminishes with increasing distance but that is a function of distance, not mass. It doesn't matter how massive the object is, it will still be accelerated toward the Earth at a rate that approaches 9.8 m/s². Once it reaches that rate of acceleration, then the rate will not increase. Also, I'm sure you know that two objects that are the same distance from the Earth will be accelerated at the same rate, regardless of mass. That is what I meant by constant. I did not mean that all objects, regardless of distance, approach the Earth at 9.8 m/s². Sorry for the ambiguity.

If F=ma does not describe the momentum transferred from one object to another in a collision then what does it describe? Surely it does not describe the gravitational force exerted by an object onto another object since it does not vary with distance.

 

[D H]

If F=ma does not describe the momentum transferred from one object to another in a collision then what does it describe? Surely it does not describe the gravitational force exerted by an object onto another object since it does not vary with distance.

Ernest, Newton's second law of motion, F=ma, is extremely basic physics. That you do not know the meaning of this is very telling. Newton's law of gravitation, F=Gm₁m₂/r², is also extremely basic physics. The gravitational force varies with mass and distance. That you are arguing this is also very telling.

I strongly suggest that you gain an understanding of these very basic concepts before you start than arguing about them. Some reading material:

The Wikipedia articles on Newton's laws of motion and Newton's universal law of gravitation are
http://en.wikipedia.org/wiki/Newton's_laws_of_motion
http://en.wikipedia.org/wiki/Newton's_law_of_universal_gravitation

The Physics Classroom has some good basic tutorials on these concepts; see
http://www.physicsclassroom.com/Class/newtlaws
http://www.physicsclassroom.com/Class/circles/u6l3c.cfm

Google "Newton's laws of motion" and "Newton's law of gravitation" and you will find much, much more about these concepts.

 

[ernestpworrel]

Ernest, Newton's second law of motion, F=ma, is extremely basic physics. That you do not know the meaning of this is very telling. Newton's law of gravitation, F=Gm₁m₂/r², is also extremely basic physics. The gravitational force varies with mass and distance. That you are arguing this is also very telling.

I can't believe that you're accusing me of not knowing these simple things. I can understand your confusion regarding the way I stated Newton's second law because it's not normally stated that way. However, let's look at Newton's first law which says that a body with a certain velocity tends to maintain that velocity unless acted upon by some other force. The object has a certain mass and is accelerating at a certain rate. Multiply these two and you will get the force that it takes to slow the object down to zero velocity. Now Newton's third law says "[w]henever a particle A exerts a force on another particle B, B simultaneously exerts a force on A with the same magnitude in the opposite direction." That means that, assuming no change in environmental conditions, if A were to collide with B then B would transfer a force of the same magnitude as the force transferred to it by A. I'm not trying to say that A has to collide with something for its velocity to be reduced to zero nor did I ever say that. I'm just describing the force that A has when it is in motion in terms of the work it would do on impact. Really, I think it's better to desribe it that way because that helps clarify the point that when a massive object hits a less massive object then it will not instantaneously slow to zero. Rather it will push the less massive object for a distance, until it has had enough work done to it. This is not to say that the other object will not push back. But in short, it takes more time for a 1 kg object to deliver a force of 10 j than it does for a 10 kg object to do 10 j.
As far as the other equation is concerned, I understand that as the equation for gravitational attraction between two bodies. This equation absolutely does not mean that either body will feel that force though. It describes how each body will behave under the influence of the gravity of the other. I don't see how I'm wrong in conveying these obviously correct ideas on this forum.

 

[D H]

[Off-topic discussion of collisions]

The issue of collision dynamics has nothing to do with the topic of this thread, Ernest. This is a hijack on your part.

As far as the other equation is concerned, I understand that as the equation for gravitational attraction between two bodies. This equation absolutely does not mean that either body will feel that force though. It describes how each body will behave under the influence of the gravity of the other. I don't see how I'm wrong in conveying these obviously correct ideas on this forum.

This is nonsense.

Addendum
What exactly do you mean by the word "feel" here, Ernest?

 

[ernestpworrel]

The issue of collision dynamics has nothing to do with the topic of this thread, Ernest. This is a hijack on your part.


This is nonsense.

Addendum
What exactly do you mean by the word "feel" here, Ernest?

What I meant by feel was be influenced by, as in it does not make sense to say that an object in free fall towards the Earth feels a combination of the Earth's gravity and its own because it never accelerates faster than 9.8 m/s² yet when it hits the surface of the Earth it does feel (is influenced by) both forces. I'm curious to know what you think I meant. Now let me ask you something. Do you think that someone who goes around calling everything anyone else says nonsense without ever providing any explanation for their views should be respected just because they have tags like Science Advisor attached to their name? I don't know you from Adam, so I don't know why you think that just because you call something I said nonsense I should instantly snap and say "Oh yeah, it is nonsense. What was I thinking?" I will ignore your accusation of hijacking although I "feel" that it is nonsense.

 

[Doc Al]

What I meant by feel was be influenced by, as in it does not make sense to say that an object in free fall towards the Earth feels a combination of the Earth's gravity and its own because it never accelerates faster than 9.8 m/s²

What do you mean by "feels a combination"? The gravitational force that Earth and object exert on each other is given by Newton's law of gravity. The resulting acceleration is given by Newton's 2nd law. Basic stuff.

yet when it hits the surface of the Earth it does feel (is influenced by) both forces.

The additional forces exerted on the object upon collision with the Earth have nothing to do with gravity.

 

[ernestpworrel]

What do you mean "resulting acceleration" Doc? Are you talking about the net force between the Earth and another planet? What does that have to do with the force imparted by the Earth to the planet? I just thought of a perfect analogy of what I'm getting here. A pair of apples is not always a pair of apples because sometimes an orange is added to it giving you a total of three pieces of fruit. Now I know what you will use as an example to illustrate your point to me so don't bother. It would be nonsense.

 

[ernestpworrel]

Oh, by the way no one has told me what they think f=ma means. Smugly referring me to Wikipedia does no good.

 

[Doc Al]

What do you mean "resulting acceleration" Doc?

I'm talking about the acceleration of a falling object due to the gravitational force between it and the earth.

Are you talking about the net force between the Earth and another planet?

Huh? Why are you dragging in another planet?

What does that have to do with the force imparted by the Earth to the planet?

Huh? Review Newton's 3rd law.

I just thought of a perfect analogy of what I'm getting here. A pair of apples is not always a pair of apples because sometimes an orange is added to it giving you a total of three pieces of fruit. Now I know what you will use as an example to illustrate your point to me so don't bother. It would be nonsense.

Huh? More nonsense.

Oh, by the way no one has told me what they think f=ma means.

What you mean what "they think"? Newton's 2nd law is basic stuff. Why not open a physics book and learn about it? In a nutshell, it relates the net force on an object to its mass and acceleration. For an object in free fall, the only force (and thus the net force) is the Earth's gravitational pull on it.

Why don't you tell us what you think "F = ma" means?

 

[D H]

What I meant by feel was be influenced by, as in it does not make sense to say that an object in free fall towards the Earth feels a combination of the Earth's gravity and its own because it never accelerates faster than 9.8 m/s² yet when it hits the surface of the Earth it does feel (is influenced by) both forces. I'm curious to know what you think I meant.

I didn't have any idea what you meant. That is why I asked.

We observe everything released near the surface of the Earth to accelerate Earthward at 9.81 m/s² because everything we release has a mass that is negligible compared to that of the Earth. To see an object accelerate Earthward at 9.82 m/s², the released object would have to be incredibly massive: about 1/12 of the Moon's mass, or an iron sphere about 1140 km in diameter.

Suppose we did just that: Construct a 1140 km diameter iron sphere, raise it a short distance above the Earth, and let it fall. From the perspective of an inertial observer, the sphere would accelerate Earthward at 9.81 m/s² while the Earth would accelerate toward the sphere at 0.01 m/s².

We can't do that, but nature does do exactly that for us. The ratio of the Earth's mass to that of the Moon is about 81:1. The Earth and the Moon are orbiting each other. (The Earth-Moon barycenter is displaced from the center of the Earth by about 2/3 Earth radii.) Suppose the Moon's mass was negligible compared to that of the Earth. This would make a month be about four hours longer than it actually is.

Ernest, you appear to have a deep misunderstanding of Newton's laws of motion and of Newton's law of gravitation. I gave you some reading material. There is plenty more available on the web and in your local public library. Please take advantage of these resources.

 

[ernestpworrel]

Why don't you tell us what you think "F = ma" means?

Here is what I think. A mass m that is has acceleration a will impart a force F to any object it collides with. In other words, force F of impact is equal to the mass m times the rate of acceleration a.

 

[russ_watters]

That's worded very badly: the first sentence isn't true the way it is worded, but the second basically is. Regardless, F=ma isn't just for impacts, it is for all accelerations - so I think you are confusing yourself by trying to consider collisions.

Anyway, there is no need for all of this running around in circles and trying to figure out what people mean by their usages of certain words. It isn't going anywhere or achieving anything. Newton's equations for gravitational force and for acceleration are listed in post #2 in this thread. If you can't figure out how they work just by looking at them, you'll need to actually start using them.

Here's what I want you to do (and this isn't optional: this thread is not going to continue going around in circles):

1. Use the gravitational force equation to calculate the gravitational force between the Earth and a 1 kg object and a 2 kg object at its surface to 4 decimal places.

2. Use the force, mass and acceleration equation to calculate the acceleration of each object under the force you calculated in step 1 (also to 4 decimal places).

3. Compare the results and tell us what you have learned.

Btw:
The mass of the Earth is 5.9736 x 10^24 kg
G = 6.6730 x 10^-11 m^3/kg-s^2
You now have everything you need to do this problem available in the thread.

 

[ernestpworrel]

What! What force mass? What is that?

 

[Jimmy]

Use the force mass and acceleration equation...

force, mass, and acceleration: F = ma

 

[russ_watters]

force, mass, and acceleration: F = ma

Sorry, yes - I missed a comma in there. It's corrected now.

The equation relates the three and can be used (as can any 3 term equation, typically) to find whichever of the three you don't know, given the other two. Obviously, it needs to be rearranged in order to be used to calculate acceleration...

 

[Jimmy]

I understood exactly what you meant, Russ. I just posted for ernest's sake. I get the impression that he's being deliberately difficult.

 

[russ_watters]

I understood exactly what you meant, Russ. I just posted for ernest's sake. I get the impression that he's being deliberately difficult.

I figured you did - I was confirming and amplifying.

 

[mikelepore]

The way I like to say that is:
F=ma
a=F/m
The ratio (F_big / m_big) is equal to the ratio (F_small / m_small).
Both are the same value of "a".

What is F_big and F_small?

I meant, for example, drop two objects near the surface of the earth.

Drop a 10 kg object (the smaller mass) and a 100,000 kg object (the bigger mass).

For each object, express its acceleration by expressing the ratio of the gravitational force that the Earth exerts on the object to the mass of the object.

10 kg object:
a = F_small / m_small = 98 N / 10 kg = 9.8 m/s^2

100,000 kg object:
a = F_big / m_big = 980,000 N / 100,000 kg = 9.8 m/s^2

The two objects experience the same acceleration.

 

[TurtleMeister]

I meant, for example, drop two objects near the surface of the earth.

I understand. Thanks for the clarification. I was thinking of the force between the Earth and only one object. My fault in misunderstanding.