Mass doesn not effect speed of dropped objects

[_DJ_british_?]

Pengwuino's post resume what I'm saying.

 

[_DJ_british_?]

I think you're interchanging field and force, which are VERY different thing

 

[ernestpworrel]

Hum, yes it does. if we asssume g(vector) (the g-field) the be constant), the g-force can be expressed as m*g(vector).

A lump of star matter might move slower in the gravitational field of the Earth than a simple ball bearing. That doesn't automatically mean it feels less force though.

 

[JeepinBob]

Thanks guys. the world makes sense once again and I feel just a little bit smarter.

 

[ernestpworrel]

It HAS to feel a greater gravitational force. If you assume they both have the same acceleration, but vastly different masses, they MUST have vastly different accelerations. Since we know more massive and less massive objects fall with the same acceleration, they must feel different forces.

But the lump of star matter exerts a greater gravitational force on the Earth than the ball bearing, so the total gravitational force is greater. That still doesn't mean it feels a greater force though. It's the equivalent of a greater force so the vector fields are different, but the gravitational force of the Earth is constant so there's no real difference in forces.

 

[ernestpworrel]

Thanks guys. the world makes sense once again and I feel just a little bit smarter.

Glad I could help.

 

[mikelepore]

So if I understand correctly, the force of pulling on the heavier object would be greater but the added mass counters the force. that is the same force on a heavier object would cause less acceleration.

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".

 

[_DJ_british_?]

Basically, a 1 kg mass 10m from the Earth surface doesn't feel more or less acceleration than a 100000000000000000000000000000000000 kg mass (in the absence of other force) at the same distance.

 

[ernestpworrel]

Thank you.

 

[_DJ_british_?]

Wow, I am really drunk because I typed the OPPOSITE of what i tried to say. Still, it's basically the same. On a equipolar surface, a is a constant. If a is 9.8 meter per second squared, the ration of F and m is a constant in classical mechanics. If F augment, M must augment and if M diminish, F must diminish. SO the more the mass, the more the force (to have tha same acceleration)

 

[ernestpworrel]

Okay, but that still doesn't mean that an object feels more force just because it has more mass. It means that the force imparted by an object that has a mass of 1(10^23)kg will be greater than the force imparted by an object of 1 kg.

 

[D H]

But the lump of star matter exerts a greater gravitational force on the Earth than the ball bearing, so the total gravitational force is greater.

The magnitude of the gravitational force exerted by that ball bearing on the Earth is exactly equal to that exerted by the Earth on the ball bearing. The magnitude of the gravitational force exerted by the Earth on the Sun is exactly equal to that exerted by the Sun on the Earth.

That still doesn't mean it feels a greater force though.

Yes, it does.

It's the equivalent of a greater force so the vector fields are different, but the gravitational force of the Earth is constant so there's no real difference in forces.

That is nonsense.

Glad I could help.

You have not been helping.

 

[D H]

The OP is asking about the equivalence principle. One consequence of this principle is that gravitational mass (F=Gm₁m₂/r²) and inertial mass (F=ma) are one and the same thing. Given those two equations, it should be obvious that that gravitational acceleration is independent of mass.

The above of course is the Newtonian perspective of gravity. The equivalence principle (in fact, an even stronger version of the equivalence principle) is a cornerstone of general relativity. The equivalence principle now stands as one of the most precisely tested concepts in physics. Some PhysicsWorld articles:

http://physicsworld.com/cws/article/print/21148
http://physicsworld.com/cws/article/news/20870

 

[TurtleMeister]

The magnitude of the gravitational force exerted by that ball bearing on the Earth is exactly equal to that exerted by the Earth on the ball bearing.

Yes, that's what I thought. So am I also correct in assuming that this is wrong?

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? The force is the same for both Earth and the object (just in opposite directions). It's the accelerations that are different (and in opposite directions). That's because the "a" in F=ma is derived from GM/r².

So

F = ma = m * GM/r² = GMm/r².
Likewise,
F' = MA = M * Gm/r² = GMm/r².

F and F' are equal and opposite.

So

a = GM/r²
and
A' = Gm/r²

Obviously "a" (the acceleration of the object) is much larger than A' (the acceleration of the earth) because of the large difference in inertial mass.

 

[Pengwuino]

But the lump of star matter exerts a greater gravitational force on the Earth than the ball bearing, so the total gravitational force is greater. That still doesn't mean it feels a greater force though. It's the equivalent of a greater force so the vector fields are different, but the gravitational force of the Earth is constant so there's no real difference in forces.

Ok, you seem to be attempting to disprove what I say by using the principle of my argument to try to argue something different?

Ok, tell me, if you drop a 1000kg ball from an airplane not too far off the ground right next to a 1kg ball, what will their acceleration be? And what will be the magnitude of the force that accelerates each object?

 

[ernestpworrel]

Okay, look Pengwino. All objects in freefall near the Earth's surface can be treated as having equal masses when considering only the force imparted to them. 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). The magnitude of the force that is communicated to each object is the same. The magnitude of the force delivered by the larger, more massive object is greater, but that means that it will have more momentum on impact and do more damage. It does not mean that the "heavier' object will fall faster or feel more force or anything. It feels the same force as the "lighter" object because the force causing acceleration is the same, the mass of the earth. F=ma describes the amount of work that an object with momentum will do on an object it collides with.

 

[TurtleMeister]

The magnitude of the force that is communicated to each object is the same. The magnitude of the force delivered by the larger, more massive object is greater, but that means that it will have more momentum on impact and do more damage.

What is "communicated to" and "delivered by"? There is only one force in F = GMm/r². It acts on both Earth and the object, just in opposite directions. The force acting between the Earth and the larger object is greater than the force acting between the Earth and the smaller object. But the accelerations of the large object and the small object are the same because the increased force on the larger object is canceled by it's equally larger inertial mass. But you are correct that the larger object will have greater momentum.

To clarify and avoid confusion (if that's possible at this point). This is not the same argument as my previous post. This argument is about the difference of force and acceleration between two objects in free fall. My previous post was about only one object in free fall.

 

[ernestpworrel]

Exactly, and I agree with you. I'm not contesting that the total gravity between the Earth and a 50000 kg object will be greater than that between the Earth and a 5 kg object. All I'm saying is that the same force is imparted to each. Each may contribute a different amount of force to the field, but we aren't free to infer that more massive objects feel greater forces. Quite the contrary, more massive objects deliver greater forces.

 

[D H]

That makes almost no sense, and what little sense it does make contradicts Newton's third law.

 

[ernestpworrel]

Were you speaking to me? I'm not contradicting Newton's third law. The rate to which an object is accelerated varies directly with the mass of the object. Greater acceleration means more momentum and more force delivered upon impact. For example, if you drop a Mack truck from 500 ft it will do significantly more damage than a bicycle.

 

[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.