Mass doesn not effect speed of dropped objects

[JeepinBob]

Okay I am no physics pro, but Galileo claimed/proved that two objects of differing mass will fall at the same rate/speed.

As I understand the force of gravity is a function of mass and distance. There is gravitational pull between all objects and this force is determined by the mass of the objects and their distance from each other.

If I understand this then gravitational pull would be greater on an object of greater mass. Now I can see why Galileo's tests showed no difference as the the difference in mass of any two objects he tested would be insignificant compared to the mass of the earth.

What am I missing here?

Thanks for your help and please don't get too technical.

 

[Pengwuino]

You need to be careful about which mass you're talking about. Also, it's the acceleration that is the same, not the speed. Although if both objects are dropped with the same speed, then they will continue to have the same speeds (although both speeds are constantly increasing) through their falls. The force a mass m_1, for example the Earth, exerts on a second mass, m_2, say for example a person, due to gravity is given by F = \frac{Gm_1m_2}{r^2}. However, the force acting on an object is F = ma so for the person, m_1, you can equate the forces, F = m_1a = \frac{Gm_1m_2}{r^2} and you'll see that the acceleration depends only on the mass of the other object and the distance between them.

As you can see, however, this works no matter what the scale of the objects. You can say instead of looking at the person, you look at the force the person exerts on the planet! Using the same equation, you see that it also gives an acceleration independent of it's own mass.

 

[TurtleMeister]

It is true that the acceleration of an object in Earth free-fall is determined solely by the mass of the Earth and not by the mass of the object. It's a fact. However, you should keep in mind that the Earth is also accelerated by the object. Their relative acceleration is the product of their masses as shown in the universal law of gravitation.

edit:
If you increase the mass of the falling object then the acceleration of the Earth will increase, but not the acceleration of the object itself. Obviously, as you've already pointed out, this increase would be undetectable (unless we're talking about an astronomical size object).

 

[JeepinBob]

Turtle, I do understand and agree with what you said. Pengwuino, can you tell me what all the letters represent? What are G r and a? I got F and m1 and m2. I will probably have further questions.

 

[Feldoh]

G is a constant, r is a distance between two masses and a is acceleration.

 

[russ_watters]

Turtle, it is best not to confuse the OP with something that is a secondary issue.

JeepinBob: If you look at the gravitational force equation that Penguino posted, you can see that the force of gravity scales linearly with the mass of your small object (m1). In other words, if you double the mass, you double the force. But then you plug them into f=ma and you find that if you double the force and double the mass, the acceleration stays the same.

 

[Naty1]

Now I can see why Galileo's tests showed no difference as the the difference in mass of any two objects he tested would be insignificant compared to the mass of the earth.

You did not state the relationship correctly: Objects of different masses accelerate at the same rate in a particular gravitational field. Any object will accelerate in a gravitational field at the same rate, regardless of the mass of the object. This has nothing to do with the relative mass of Earth and relatively small masses. All masses accelerate at about g = 9.8 m/s/s at the Earth's surface. If one mass is twice another, it is subject to twice the gravitational force resultring in the same acceleration g.

 

[TurtleMeister]

Turtle, it is best not to confuse the OP with something that is a secondary issue.

Sorry, I wanted to be "not too technical" as the op requested. But I guess I did not understand the question.

JeepinBob, this page has helped me.
http://csep10.phys.utk.edu/astr161/lect/history/Newtongrav.html

 

[Mentallic]

It might also be helpful to add that, while the acceleration due to gravity is equal on both masses, galileo would probably not have seen both masses hit the ground at exactly the same time - no matter how careful he was to drop both masses from the same height and time.

This is because of air resistance. You might think that air resistance is equal on both masses, but this isn't exactly so. Galileo used larger objects for the heavy mass, and smaller objects for the small mass. Their sizes are different. This means that more air is bombarding the larger object (resistance is a force that counteracts the force of the object due to gravity). But contrastingly, the larger object also has a higher mass and thus higher force due to F=ma. Eventually, as the object keeps increasing speed, the air resistance force will keep increasing until the force due to gravity (which is constant as the mass and acceleration of gravity are constant) will be equal to each other. This is when the net force (total sum of forces - gravity and air) is 0 and thus there is no more acceleration. Then the object experiences a constant speed and this is what is meant by terminal velocity.

Now, you'll probably find that even though the larger object experiences more air resistance, the increased mass is still more profound and thus the large object will have a high terminal velocity. So, if you drop the two objects from a very high location at the same time, the larger one will hit the ground first (the higher the tower, the bigger the difference).

 

[JeepinBob]

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.

 

[mgb_phys]

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.

Exactly, an object that is twice as heavy feels twice as much gravitational force, but it also takes twice as much force to accelerate it.

 

[Mentallic]

I'm not sure exactly what you're trying to say simply because of the lack of terminology.

the force of pulling on the heavier object

Do you mean the air resistance force, or gravity force here?

Anyway, I'll try explain it with example. Imagine you have two spheres (I'll refrain from using the term balls ) and each sphere is the same size. However, the only difference is that one sphere is twice as heavy as the other (one sphere could be hollow).
Now, looking at the equation F=ma, the acceleration is gravity and this is the same for all objects, but the only difference here is the mass. The heavier ball will have twice the constant force compared to the other.

Since the two spheres are the same size, the air resistance that acts on them is also equal. As we drop both spheres from a large height, the accelerate (keep speeding up) and as their speed increases, the air resistance force which tries to counter-act the force of the spheres gets larger too. At some point, the air resistance force will become equal to the force of the lighter sphere. This means there is no more force being applied so no more acceleration. The speed of the lighter sphere is now constant and the air resistance force doesn't change anymore.

There is still the force of gravity acting, but the air resistance is also acting so no acceleration.
The twice as heavy sphere still has a larger gravity force so it continues to accelerate until the air resistance also becomes twice as much.
Now, as you can probably visualize, the heavier sphere would hit the ground first because it manages to get to a greater speed and keep that higher speed.


But, the heavier objects aren't always the first to hit the ground. Think of a man in a parachute versus an apple.

 

[Mentallic]

Oh I understand what you meant now after seeing mgb_phys' response.

 

[ernestpworrel]

Okay I am no physics pro, but Galileo claimed/proved that two objects of differing mass will fall at the same rate/speed.

That's right. All objects are accelerated toward the Earth at the same rate, regardless of diameter.

Galileo's tests showed no difference as the the difference in mass of any two objects he tested would be insignificant compared to the mass of the earth.

There would not be any difference even if the masses he used were greater than those of the earth. Regardless of the mass of the object, the acceleration imparted to it by the gravity of the Earth is still the same. The mass is the factor that determines the amount of momentum-changing force that the object experiences. Shape is also a factor when calculating for air resistance. But in freefall all objects can be treated as having equivalent masses. This means that if there were no air to provide resistance, a feather would fall at the same rate as a bowling ball.

 

[ernestpworrel]

Exactly, an object that is twice as heavy feels twice as much gravitational force, but it also takes twice as much force to accelerate it.

Why would an object that is twice as heavy feel twice as much gravitational force? I think what you mean is that a heavy object exerts more gravitational force than a lighter one.

 

[Pengwuino]

Why would an object that is twice as heavy feel twice as much gravitational force? I think what you mean is that a heavy object exerts more gravitational force than a lighter one.

That's how gravity works, it's been explained at least a couple times in this thread. On Earth, an object, A, that is twice as heavy as another object, B, will feel twice the gravitational force from the Earth as object B.

 

[ernestpworrel]

This means there is no more force being applied so no more acceleration. The speed of the lighter sphere is now constant and the air resistance force doesn't change anymore.

There is still the force of gravity acting, but the air resistance is also acting so no acceleration.

There would still be acceleration, wouldn't there? Just no net force?

 

[Pengwuino]

There would still be acceleration, wouldn't there? Just no net force?

There's still two forces involved, but they exactly cancel at terminal velocity, giving a 0 net force. That means there is 0 acceleration: F = ma.

 

[ernestpworrel]

That's how gravity works, it's been explained at least a couple times in this thread. On Earth, an object, A, that is twice as heavy as another object, B, will feel twice the gravitational force from the Earth as object B.

Only when it's on the surface though. There's no heavy or light in freefall conditions.

 

[ernestpworrel]

There's still two forces involved, but they exactly cancel at terminal velocity, giving a 0 net force. That means there is 0 acceleration: F = ma.

So air resistance changes gravitational acceleration?

 

[_DJ_british_?]

Why would an object that is twice as heavy feel twice as much gravitational force? I think what you mean is that a heavy object exerts more gravitational force than a lighter one.

Remember, F=m*dp\dt. You double the mass, you double the force. The gravitational field (ie. the interaction of said mass with other masses) is indepedent of the mass of the other objects. The interaction that another mass receive from your field (your interaction) is proportionnal to its mass. Bad analogy: Imagine a bully and a nerd. They have a certain strenght. The bully hit the nerd. His "hitting force" is independent of the nerd, he would hit with the same strenght on something else. But the pain the nerd receive is dependent of his strenght. If he's slim, he'll get hit hard. Maybe that was a morally bad exemple, but that get the point done. The "hitting force" of a mass is it's gravitational field, it's ability to interact with your mass, the "pain" it does on you is the force, the effect it have on you, namely, an acceleration.

 

[Pengwuino]

Only when it's on the surface though. There's no heavy or light in freefall conditions.

No, it feels the same force anywhere. Having a "weight" is not the same as feeling an acceleration. Imagine you have a ball bearing and you drop it out of an airplane. Let's say you also have a ball bearing sized sample of... neutron star matter, something undeniably dense and massive. Now, if they are indeed feeling the same force, by F = ma, the neutron star matter wouldn't pretty much not even budge compared to the ball bearing since the mass of the neutral star bearing is far far greater.

So air resistance changes gravitational acceleration?

No, it's its own force. It simply acts in the opposite direction of the gravitational force if the object if moving strictly in the direction of the gravitational force. Now, if you shot a ball horizontally, it will feel a horizontal resistance that has nothing to do with the vertical resistance.

 

[ernestpworrel]

But you're agreeing with me, british.

 

[_DJ_british_?]

But you're agreeing with me, british.

Mmh, about what? I only made a particular comment about a particular post!

 

[ernestpworrel]

No, it feels the same force anywhere. Having a "weight" is not the same as feeling an acceleration. Imagine you have a ball bearing and you drop it out of an airplane. Let's say you also have a ball bearing sized sample of... neutron star matter, something undeniably dense and massive. Now, if they are indeed feeling the same force, by F = ma, the neutron star matter wouldn't pretty much not even budge compared to the ball bearing since the mass of the neutral star bearing is far far greater.

But the ball bearing and the lump of neutron star matter would still be influenced by the same amount of gravitational force and would both be accelerated at the same rate. I know the equation F=ma but I'm just saying that that doesn't mean that a heavier object will feel a greater or lesser gravitational force.

 

[_DJ_british_?]

Okay forget my posts, I'm sleepy and little bit drunk

 

[ernestpworrel]

Mmh, about what? I only made a particular comment about a particular post!

Well then, excuse me. I thought you were trying to debate me.

 

[_DJ_british_?]

I know the equation F=ma but I'm just saying that that doesn't mean that a heavier object will feel a greater or lesser gravitational force.

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).

 

[Pengwuino]

But the ball bearing and the lump of neutron star matter would still be influenced by the same amount of gravitational force and would both be accelerated at the same rate. I know the equation F=ma but I'm just saying that that doesn't mean that a heavier object will feel a greater or lesser gravitational force.

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.

 

[_DJ_british_?]

Near the surface of the earth, mases have the same acceleration, which is F\m. That ratio is approximately 9.8 m\s2. Now imagine: If F augment, how m change? And if m change, how F change?

 

[_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.