If an object with more mass experiences a greater gravitational force, why don't more massive objects fall faster? [duplicate] The Next CEO of Stack OverflowDon't heavier objects actually fall faster because they exert their own gravity?Why do two bodies of different masses fall at the same rate (in the absence of air resistance)?Don't heavier objects actually fall faster because they exert their own gravity?Why do two bodies of different masses fall at the same rate (in the absence of air resistance)?Why do objects with different masses fall at the same rate?Free falling objectsWhy do objects fall at the same acceleration?Was Aristotle Actually Correct About Gravitation?A contradiction statement to $F=ma$Galileo proved wrong?If inertial mass were equal to for example half of gravitational mass why would things not still fall at same rate?How does Newtonian gravity violate law of inertia?
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If an object with more mass experiences a greater gravitational force, why don't more massive objects fall faster? [duplicate]
The Next CEO of Stack OverflowDon't heavier objects actually fall faster because they exert their own gravity?Why do two bodies of different masses fall at the same rate (in the absence of air resistance)?Don't heavier objects actually fall faster because they exert their own gravity?Why do two bodies of different masses fall at the same rate (in the absence of air resistance)?Why do objects with different masses fall at the same rate?Free falling objectsWhy do objects fall at the same acceleration?Was Aristotle Actually Correct About Gravitation?A contradiction statement to $F=ma$Galileo proved wrong?If inertial mass were equal to for example half of gravitational mass why would things not still fall at same rate?How does Newtonian gravity violate law of inertia?
$begingroup$
This question already has an answer here:
Don't heavier objects actually fall faster because they exert their own gravity?
11 answers
According to Sir Isaac Newton, the gravity equation runs like this:
$$ F = fracGm_1m_2r^2 $$
where $F$ is the gravitational force, $G$ the gravitational constant, $m_1$ and $m_2$ are the respective masses of two bodies being pulled together by said force, and $r$ is the distance between them.
Now Galileo asserted that regardless of the difference in mass of two objects falling onto Earth, their acceleration rate will always be the same, i.e. 9.8 meters per second squared.
Question:
Is this actually true?
It is clear from the above equation that the larger the object's mass (either of the $m$'s), the greater the force ($F$).
In other words, should we leave the distance ($r$) the same but increase the mass of either object, the force ($F$) would increase proportionately, resulting in a greater acceleration rate. Which was what Aristotle and everyone else after him thought until Galileo decided Aristotle was wrong and proved it by throwing two different objects from the tower of Pisa. They hit the ground at the same time: yes. So far as the naked eye could judge, anyway.
To clarify:
Compared to the Earth's mass, the masses of the two objects would have been so small that any difference in their respective acceleration rates would have been too tiny to detect, i.e. NEGLIGIBLE.
In other words, to all, or most, PRACTICAL (i.e. earth-based) purposes, Galileo was right.
But was he TECHNICALLY right?
Einstein's theory states that gravity and inertia are the same exact force. This strikes me as perfectly reasonable. However, it is asserted that this somehow confirms Galileo's conclusion. I don't see how.
Please explain.
newtonian-mechanics mass acceleration free-fall equivalence-principle
$endgroup$
marked as duplicate by John Rennie, ZeroTheHero, knzhou, Qmechanic♦ Mar 24 at 15:25
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
|
show 7 more comments
$begingroup$
This question already has an answer here:
Don't heavier objects actually fall faster because they exert their own gravity?
11 answers
According to Sir Isaac Newton, the gravity equation runs like this:
$$ F = fracGm_1m_2r^2 $$
where $F$ is the gravitational force, $G$ the gravitational constant, $m_1$ and $m_2$ are the respective masses of two bodies being pulled together by said force, and $r$ is the distance between them.
Now Galileo asserted that regardless of the difference in mass of two objects falling onto Earth, their acceleration rate will always be the same, i.e. 9.8 meters per second squared.
Question:
Is this actually true?
It is clear from the above equation that the larger the object's mass (either of the $m$'s), the greater the force ($F$).
In other words, should we leave the distance ($r$) the same but increase the mass of either object, the force ($F$) would increase proportionately, resulting in a greater acceleration rate. Which was what Aristotle and everyone else after him thought until Galileo decided Aristotle was wrong and proved it by throwing two different objects from the tower of Pisa. They hit the ground at the same time: yes. So far as the naked eye could judge, anyway.
To clarify:
Compared to the Earth's mass, the masses of the two objects would have been so small that any difference in their respective acceleration rates would have been too tiny to detect, i.e. NEGLIGIBLE.
In other words, to all, or most, PRACTICAL (i.e. earth-based) purposes, Galileo was right.
But was he TECHNICALLY right?
Einstein's theory states that gravity and inertia are the same exact force. This strikes me as perfectly reasonable. However, it is asserted that this somehow confirms Galileo's conclusion. I don't see how.
Please explain.
newtonian-mechanics mass acceleration free-fall equivalence-principle
$endgroup$
marked as duplicate by John Rennie, ZeroTheHero, knzhou, Qmechanic♦ Mar 24 at 15:25
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
2
$begingroup$
What is your argument against Galileo's conclusion?
$endgroup$
– my2cts
Mar 24 at 9:38
2
$begingroup$
$F=ma$, so the gravitational acceleration is independent of the mass of the falling body. However, see physics.stackexchange.com/q/3534
$endgroup$
– PM 2Ring
Mar 24 at 9:49
1
$begingroup$
Inertia isn't a force
$endgroup$
– Aaron Stevens
Mar 24 at 10:24
1
$begingroup$
@Ricky The difference is due to how much the Earth accelerates upwards to meet the falling object, in the centre of mass frame of (the Earth + object), the object's acceleration towards the centre of mass is still independent of its mass.
$endgroup$
– PM 2Ring
Mar 24 at 10:32
1
$begingroup$
Possible duplicate of Why do two bodies of different masses fall at the same rate (in the absence of air resistance)?
$endgroup$
– knzhou
Mar 24 at 13:13
|
show 7 more comments
$begingroup$
This question already has an answer here:
Don't heavier objects actually fall faster because they exert their own gravity?
11 answers
According to Sir Isaac Newton, the gravity equation runs like this:
$$ F = fracGm_1m_2r^2 $$
where $F$ is the gravitational force, $G$ the gravitational constant, $m_1$ and $m_2$ are the respective masses of two bodies being pulled together by said force, and $r$ is the distance between them.
Now Galileo asserted that regardless of the difference in mass of two objects falling onto Earth, their acceleration rate will always be the same, i.e. 9.8 meters per second squared.
Question:
Is this actually true?
It is clear from the above equation that the larger the object's mass (either of the $m$'s), the greater the force ($F$).
In other words, should we leave the distance ($r$) the same but increase the mass of either object, the force ($F$) would increase proportionately, resulting in a greater acceleration rate. Which was what Aristotle and everyone else after him thought until Galileo decided Aristotle was wrong and proved it by throwing two different objects from the tower of Pisa. They hit the ground at the same time: yes. So far as the naked eye could judge, anyway.
To clarify:
Compared to the Earth's mass, the masses of the two objects would have been so small that any difference in their respective acceleration rates would have been too tiny to detect, i.e. NEGLIGIBLE.
In other words, to all, or most, PRACTICAL (i.e. earth-based) purposes, Galileo was right.
But was he TECHNICALLY right?
Einstein's theory states that gravity and inertia are the same exact force. This strikes me as perfectly reasonable. However, it is asserted that this somehow confirms Galileo's conclusion. I don't see how.
Please explain.
newtonian-mechanics mass acceleration free-fall equivalence-principle
$endgroup$
This question already has an answer here:
Don't heavier objects actually fall faster because they exert their own gravity?
11 answers
According to Sir Isaac Newton, the gravity equation runs like this:
$$ F = fracGm_1m_2r^2 $$
where $F$ is the gravitational force, $G$ the gravitational constant, $m_1$ and $m_2$ are the respective masses of two bodies being pulled together by said force, and $r$ is the distance between them.
Now Galileo asserted that regardless of the difference in mass of two objects falling onto Earth, their acceleration rate will always be the same, i.e. 9.8 meters per second squared.
Question:
Is this actually true?
It is clear from the above equation that the larger the object's mass (either of the $m$'s), the greater the force ($F$).
In other words, should we leave the distance ($r$) the same but increase the mass of either object, the force ($F$) would increase proportionately, resulting in a greater acceleration rate. Which was what Aristotle and everyone else after him thought until Galileo decided Aristotle was wrong and proved it by throwing two different objects from the tower of Pisa. They hit the ground at the same time: yes. So far as the naked eye could judge, anyway.
To clarify:
Compared to the Earth's mass, the masses of the two objects would have been so small that any difference in their respective acceleration rates would have been too tiny to detect, i.e. NEGLIGIBLE.
In other words, to all, or most, PRACTICAL (i.e. earth-based) purposes, Galileo was right.
But was he TECHNICALLY right?
Einstein's theory states that gravity and inertia are the same exact force. This strikes me as perfectly reasonable. However, it is asserted that this somehow confirms Galileo's conclusion. I don't see how.
Please explain.
This question already has an answer here:
Don't heavier objects actually fall faster because they exert their own gravity?
11 answers
newtonian-mechanics mass acceleration free-fall equivalence-principle
newtonian-mechanics mass acceleration free-fall equivalence-principle
edited Mar 24 at 15:25
Qmechanic♦
107k121991231
107k121991231
asked Mar 24 at 9:30
RickyRicky
349311
349311
marked as duplicate by John Rennie, ZeroTheHero, knzhou, Qmechanic♦ Mar 24 at 15:25
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by John Rennie, ZeroTheHero, knzhou, Qmechanic♦ Mar 24 at 15:25
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
2
$begingroup$
What is your argument against Galileo's conclusion?
$endgroup$
– my2cts
Mar 24 at 9:38
2
$begingroup$
$F=ma$, so the gravitational acceleration is independent of the mass of the falling body. However, see physics.stackexchange.com/q/3534
$endgroup$
– PM 2Ring
Mar 24 at 9:49
1
$begingroup$
Inertia isn't a force
$endgroup$
– Aaron Stevens
Mar 24 at 10:24
1
$begingroup$
@Ricky The difference is due to how much the Earth accelerates upwards to meet the falling object, in the centre of mass frame of (the Earth + object), the object's acceleration towards the centre of mass is still independent of its mass.
$endgroup$
– PM 2Ring
Mar 24 at 10:32
1
$begingroup$
Possible duplicate of Why do two bodies of different masses fall at the same rate (in the absence of air resistance)?
$endgroup$
– knzhou
Mar 24 at 13:13
|
show 7 more comments
2
$begingroup$
What is your argument against Galileo's conclusion?
$endgroup$
– my2cts
Mar 24 at 9:38
2
$begingroup$
$F=ma$, so the gravitational acceleration is independent of the mass of the falling body. However, see physics.stackexchange.com/q/3534
$endgroup$
– PM 2Ring
Mar 24 at 9:49
1
$begingroup$
Inertia isn't a force
$endgroup$
– Aaron Stevens
Mar 24 at 10:24
1
$begingroup$
@Ricky The difference is due to how much the Earth accelerates upwards to meet the falling object, in the centre of mass frame of (the Earth + object), the object's acceleration towards the centre of mass is still independent of its mass.
$endgroup$
– PM 2Ring
Mar 24 at 10:32
1
$begingroup$
Possible duplicate of Why do two bodies of different masses fall at the same rate (in the absence of air resistance)?
$endgroup$
– knzhou
Mar 24 at 13:13
2
2
$begingroup$
What is your argument against Galileo's conclusion?
$endgroup$
– my2cts
Mar 24 at 9:38
$begingroup$
What is your argument against Galileo's conclusion?
$endgroup$
– my2cts
Mar 24 at 9:38
2
2
$begingroup$
$F=ma$, so the gravitational acceleration is independent of the mass of the falling body. However, see physics.stackexchange.com/q/3534
$endgroup$
– PM 2Ring
Mar 24 at 9:49
$begingroup$
$F=ma$, so the gravitational acceleration is independent of the mass of the falling body. However, see physics.stackexchange.com/q/3534
$endgroup$
– PM 2Ring
Mar 24 at 9:49
1
1
$begingroup$
Inertia isn't a force
$endgroup$
– Aaron Stevens
Mar 24 at 10:24
$begingroup$
Inertia isn't a force
$endgroup$
– Aaron Stevens
Mar 24 at 10:24
1
1
$begingroup$
@Ricky The difference is due to how much the Earth accelerates upwards to meet the falling object, in the centre of mass frame of (the Earth + object), the object's acceleration towards the centre of mass is still independent of its mass.
$endgroup$
– PM 2Ring
Mar 24 at 10:32
$begingroup$
@Ricky The difference is due to how much the Earth accelerates upwards to meet the falling object, in the centre of mass frame of (the Earth + object), the object's acceleration towards the centre of mass is still independent of its mass.
$endgroup$
– PM 2Ring
Mar 24 at 10:32
1
1
$begingroup$
Possible duplicate of Why do two bodies of different masses fall at the same rate (in the absence of air resistance)?
$endgroup$
– knzhou
Mar 24 at 13:13
$begingroup$
Possible duplicate of Why do two bodies of different masses fall at the same rate (in the absence of air resistance)?
$endgroup$
– knzhou
Mar 24 at 13:13
|
show 7 more comments
6 Answers
6
active
oldest
votes
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Others have addressed the mathematics; here is an intuitive explanation. (I'm not sure if this will help the asker, given that their question shows a good grasp of the maths, but it may help others.)
Let's say my identical twin and I are pushing two different boulders along the ground. Because we're identical, we provide the same force. But my boulder is much heavier than his. The result is that my boulder moves more slowly.
So I find someone much more muscled to push my boulder for me. Muscly McMuscles applies a greater force, enough to get the boulder moving exactly as fast as my twin's.
Yes, bigger objects have a greater force acting on them.
No, this doesn't make them fall faster.
The greater force and the greater mass exactly balance each other out, so the big and small objects move just as fast (more precisely, they accelerate the same).
$endgroup$
1
$begingroup$
Again an answer which ignores the problem of observing the phenomenon from a non-inertial frame. It's a pity that the OP was not able to grasp the point.
$endgroup$
– GiorgioP
Mar 24 at 21:13
$begingroup$
@If thinking that makes you happy, then ... etc ...
$endgroup$
– Ricky
Mar 25 at 0:13
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@GiorgioP: I answered the question as summarised in the title: doesn't greater force produce greater acceleration? I'd love to see an answer that delves into the maths of the non-inertial frame (it's been far too long since I did any of that stuff myself!), but even your own answer and comments stop at saying "there's a difference but it's unobservably small".
$endgroup$
– Tim Pederick
Mar 25 at 10:12
add a comment |
$begingroup$
Now Galileo asserted that regardless of the difference in mass of two objects falling onto Earth, their acceleration rate will always be the same, i.e. 9.8 meters per second squared. Question: Is this actually true?
Yes, and it is fairly straightforward to derive. For an object in free fall near the earth’s surface.
$$F=fracGMmr^2$$
$$ma=fracGMmr^2$$
$$a=fracGMr^2$$
$$a=g$$
Where $F$ is the gravitational force, $G$ is the universal gravitational constant, $M$ is the mass of the earth, $m$ is the mass of the object, $r$ is the radius of the earth, and $a$ is the object’s acceleration.
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add a comment |
$begingroup$
The correct replies to this question have already been written elsewhere in SE.Physics (better if ignoring the negative score answers).
However, it is probably useful to stress the basic issue underlying the answer to this question: if we describe motions in non-inertial frames, accelerations depend on the frame. Therefore, the fact that different masses are acclerated in the same way in inertial frames, does not imply that the same is true in non-inertial frames. And observing free fall of bodies on the Earth, means that we are using a non-inertial frame whose acceleration depends on the force between Earth and falling body. It is only the huge ratio between the mass of Earth and that of (reasonable size) falling bodies which makes practically unobservable the difference of accelerations.
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". It is only the huge ratio between the mass of Earth and that of (reasonable size) falling bodies which makes practically unobservable the difference of accelerations." That's what I said. "Practically unobservable" does not mean that it doesn't exist.
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– Ricky
Mar 24 at 20:36
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@Ricky yeah, but I would not insist on claiming that , in absence of air drag, a iron ball would fall faster than a feather, if observed from Earth. From the experimental point of view unobservable differences are equivalent to no difference at all.
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– GiorgioP
Mar 24 at 21:04
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@Ricky Unless the falling body has a mass that's a significant fraction of Earth's, the acceleration it induces on the Earth is so tiny that not only is it immeasurable, it will also be swamped by various other effects, mainly air resistance, but also including the asymmetric distribution of mass in the Earth, seismic disturbances, tidal forces from the Sun, Moon, and the other planets (primarily Jupiter), other people dropping things at other locations, et cetera. If you insist on including the Earth's acceleration in your model you should also include those other things.
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– PM 2Ring
Mar 25 at 8:22
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@PM2Ring: I absolutely agree with you on all points here.
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– Ricky
Mar 25 at 16:11
add a comment |
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This might sound confusing i agree. A body of greater mass does experience a greater force compared to a lighter object , but then the acceleration is the ratio of Force experienced over the mass which equals gravitational constant times Mass of earth( in this case) over distance squared , which is independent of the mas of the object that is falling . You can also understand it this way : a greater force is greater is required to maintain a given acceleration for a more massive body than the force that is required to maintain the same accelaration for a less massive body.In both cases force varies but accelaration remains same
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add a comment |
$begingroup$
The gravitational force is proportional to the masses of the two interacting objects. Since force is mass times acceleration, instantaneous gravitational acceleration is independent of mass. This is a approximation. The solution to the full two body problem does depend on both masses. See: https://en.m.wikipedia.org/wiki/Gravitational_two-body_problem.
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add a comment |
$begingroup$
To put more words into @Dale answer, the greater the gravitational mass,the greater the force due to gravity per Newton's law of gravity. But the greater the mass the greater its inertial mass as well, i.e., its resistance to a change in velocity, and therefore a greater force is required to accelerate the mass per Newton's second law, $F=ma$. Gravitational mass equals inertial mass, thus the acceleration is the same for all masses in the gravitational field.
Hope this helps.
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add a comment |
6 Answers
6
active
oldest
votes
6 Answers
6
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Others have addressed the mathematics; here is an intuitive explanation. (I'm not sure if this will help the asker, given that their question shows a good grasp of the maths, but it may help others.)
Let's say my identical twin and I are pushing two different boulders along the ground. Because we're identical, we provide the same force. But my boulder is much heavier than his. The result is that my boulder moves more slowly.
So I find someone much more muscled to push my boulder for me. Muscly McMuscles applies a greater force, enough to get the boulder moving exactly as fast as my twin's.
Yes, bigger objects have a greater force acting on them.
No, this doesn't make them fall faster.
The greater force and the greater mass exactly balance each other out, so the big and small objects move just as fast (more precisely, they accelerate the same).
$endgroup$
1
$begingroup$
Again an answer which ignores the problem of observing the phenomenon from a non-inertial frame. It's a pity that the OP was not able to grasp the point.
$endgroup$
– GiorgioP
Mar 24 at 21:13
$begingroup$
@If thinking that makes you happy, then ... etc ...
$endgroup$
– Ricky
Mar 25 at 0:13
$begingroup$
@GiorgioP: I answered the question as summarised in the title: doesn't greater force produce greater acceleration? I'd love to see an answer that delves into the maths of the non-inertial frame (it's been far too long since I did any of that stuff myself!), but even your own answer and comments stop at saying "there's a difference but it's unobservably small".
$endgroup$
– Tim Pederick
Mar 25 at 10:12
add a comment |
$begingroup$
Others have addressed the mathematics; here is an intuitive explanation. (I'm not sure if this will help the asker, given that their question shows a good grasp of the maths, but it may help others.)
Let's say my identical twin and I are pushing two different boulders along the ground. Because we're identical, we provide the same force. But my boulder is much heavier than his. The result is that my boulder moves more slowly.
So I find someone much more muscled to push my boulder for me. Muscly McMuscles applies a greater force, enough to get the boulder moving exactly as fast as my twin's.
Yes, bigger objects have a greater force acting on them.
No, this doesn't make them fall faster.
The greater force and the greater mass exactly balance each other out, so the big and small objects move just as fast (more precisely, they accelerate the same).
$endgroup$
1
$begingroup$
Again an answer which ignores the problem of observing the phenomenon from a non-inertial frame. It's a pity that the OP was not able to grasp the point.
$endgroup$
– GiorgioP
Mar 24 at 21:13
$begingroup$
@If thinking that makes you happy, then ... etc ...
$endgroup$
– Ricky
Mar 25 at 0:13
$begingroup$
@GiorgioP: I answered the question as summarised in the title: doesn't greater force produce greater acceleration? I'd love to see an answer that delves into the maths of the non-inertial frame (it's been far too long since I did any of that stuff myself!), but even your own answer and comments stop at saying "there's a difference but it's unobservably small".
$endgroup$
– Tim Pederick
Mar 25 at 10:12
add a comment |
$begingroup$
Others have addressed the mathematics; here is an intuitive explanation. (I'm not sure if this will help the asker, given that their question shows a good grasp of the maths, but it may help others.)
Let's say my identical twin and I are pushing two different boulders along the ground. Because we're identical, we provide the same force. But my boulder is much heavier than his. The result is that my boulder moves more slowly.
So I find someone much more muscled to push my boulder for me. Muscly McMuscles applies a greater force, enough to get the boulder moving exactly as fast as my twin's.
Yes, bigger objects have a greater force acting on them.
No, this doesn't make them fall faster.
The greater force and the greater mass exactly balance each other out, so the big and small objects move just as fast (more precisely, they accelerate the same).
$endgroup$
Others have addressed the mathematics; here is an intuitive explanation. (I'm not sure if this will help the asker, given that their question shows a good grasp of the maths, but it may help others.)
Let's say my identical twin and I are pushing two different boulders along the ground. Because we're identical, we provide the same force. But my boulder is much heavier than his. The result is that my boulder moves more slowly.
So I find someone much more muscled to push my boulder for me. Muscly McMuscles applies a greater force, enough to get the boulder moving exactly as fast as my twin's.
Yes, bigger objects have a greater force acting on them.
No, this doesn't make them fall faster.
The greater force and the greater mass exactly balance each other out, so the big and small objects move just as fast (more precisely, they accelerate the same).
edited Mar 24 at 14:58
answered Mar 24 at 14:13
Tim PederickTim Pederick
42448
42448
1
$begingroup$
Again an answer which ignores the problem of observing the phenomenon from a non-inertial frame. It's a pity that the OP was not able to grasp the point.
$endgroup$
– GiorgioP
Mar 24 at 21:13
$begingroup$
@If thinking that makes you happy, then ... etc ...
$endgroup$
– Ricky
Mar 25 at 0:13
$begingroup$
@GiorgioP: I answered the question as summarised in the title: doesn't greater force produce greater acceleration? I'd love to see an answer that delves into the maths of the non-inertial frame (it's been far too long since I did any of that stuff myself!), but even your own answer and comments stop at saying "there's a difference but it's unobservably small".
$endgroup$
– Tim Pederick
Mar 25 at 10:12
add a comment |
1
$begingroup$
Again an answer which ignores the problem of observing the phenomenon from a non-inertial frame. It's a pity that the OP was not able to grasp the point.
$endgroup$
– GiorgioP
Mar 24 at 21:13
$begingroup$
@If thinking that makes you happy, then ... etc ...
$endgroup$
– Ricky
Mar 25 at 0:13
$begingroup$
@GiorgioP: I answered the question as summarised in the title: doesn't greater force produce greater acceleration? I'd love to see an answer that delves into the maths of the non-inertial frame (it's been far too long since I did any of that stuff myself!), but even your own answer and comments stop at saying "there's a difference but it's unobservably small".
$endgroup$
– Tim Pederick
Mar 25 at 10:12
1
1
$begingroup$
Again an answer which ignores the problem of observing the phenomenon from a non-inertial frame. It's a pity that the OP was not able to grasp the point.
$endgroup$
– GiorgioP
Mar 24 at 21:13
$begingroup$
Again an answer which ignores the problem of observing the phenomenon from a non-inertial frame. It's a pity that the OP was not able to grasp the point.
$endgroup$
– GiorgioP
Mar 24 at 21:13
$begingroup$
@If thinking that makes you happy, then ... etc ...
$endgroup$
– Ricky
Mar 25 at 0:13
$begingroup$
@If thinking that makes you happy, then ... etc ...
$endgroup$
– Ricky
Mar 25 at 0:13
$begingroup$
@GiorgioP: I answered the question as summarised in the title: doesn't greater force produce greater acceleration? I'd love to see an answer that delves into the maths of the non-inertial frame (it's been far too long since I did any of that stuff myself!), but even your own answer and comments stop at saying "there's a difference but it's unobservably small".
$endgroup$
– Tim Pederick
Mar 25 at 10:12
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@GiorgioP: I answered the question as summarised in the title: doesn't greater force produce greater acceleration? I'd love to see an answer that delves into the maths of the non-inertial frame (it's been far too long since I did any of that stuff myself!), but even your own answer and comments stop at saying "there's a difference but it's unobservably small".
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– Tim Pederick
Mar 25 at 10:12
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Now Galileo asserted that regardless of the difference in mass of two objects falling onto Earth, their acceleration rate will always be the same, i.e. 9.8 meters per second squared. Question: Is this actually true?
Yes, and it is fairly straightforward to derive. For an object in free fall near the earth’s surface.
$$F=fracGMmr^2$$
$$ma=fracGMmr^2$$
$$a=fracGMr^2$$
$$a=g$$
Where $F$ is the gravitational force, $G$ is the universal gravitational constant, $M$ is the mass of the earth, $m$ is the mass of the object, $r$ is the radius of the earth, and $a$ is the object’s acceleration.
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add a comment |
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Now Galileo asserted that regardless of the difference in mass of two objects falling onto Earth, their acceleration rate will always be the same, i.e. 9.8 meters per second squared. Question: Is this actually true?
Yes, and it is fairly straightforward to derive. For an object in free fall near the earth’s surface.
$$F=fracGMmr^2$$
$$ma=fracGMmr^2$$
$$a=fracGMr^2$$
$$a=g$$
Where $F$ is the gravitational force, $G$ is the universal gravitational constant, $M$ is the mass of the earth, $m$ is the mass of the object, $r$ is the radius of the earth, and $a$ is the object’s acceleration.
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add a comment |
$begingroup$
Now Galileo asserted that regardless of the difference in mass of two objects falling onto Earth, their acceleration rate will always be the same, i.e. 9.8 meters per second squared. Question: Is this actually true?
Yes, and it is fairly straightforward to derive. For an object in free fall near the earth’s surface.
$$F=fracGMmr^2$$
$$ma=fracGMmr^2$$
$$a=fracGMr^2$$
$$a=g$$
Where $F$ is the gravitational force, $G$ is the universal gravitational constant, $M$ is the mass of the earth, $m$ is the mass of the object, $r$ is the radius of the earth, and $a$ is the object’s acceleration.
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Now Galileo asserted that regardless of the difference in mass of two objects falling onto Earth, their acceleration rate will always be the same, i.e. 9.8 meters per second squared. Question: Is this actually true?
Yes, and it is fairly straightforward to derive. For an object in free fall near the earth’s surface.
$$F=fracGMmr^2$$
$$ma=fracGMmr^2$$
$$a=fracGMr^2$$
$$a=g$$
Where $F$ is the gravitational force, $G$ is the universal gravitational constant, $M$ is the mass of the earth, $m$ is the mass of the object, $r$ is the radius of the earth, and $a$ is the object’s acceleration.
answered Mar 24 at 11:22
DaleDale
6,5671829
6,5671829
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add a comment |
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The correct replies to this question have already been written elsewhere in SE.Physics (better if ignoring the negative score answers).
However, it is probably useful to stress the basic issue underlying the answer to this question: if we describe motions in non-inertial frames, accelerations depend on the frame. Therefore, the fact that different masses are acclerated in the same way in inertial frames, does not imply that the same is true in non-inertial frames. And observing free fall of bodies on the Earth, means that we are using a non-inertial frame whose acceleration depends on the force between Earth and falling body. It is only the huge ratio between the mass of Earth and that of (reasonable size) falling bodies which makes practically unobservable the difference of accelerations.
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". It is only the huge ratio between the mass of Earth and that of (reasonable size) falling bodies which makes practically unobservable the difference of accelerations." That's what I said. "Practically unobservable" does not mean that it doesn't exist.
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– Ricky
Mar 24 at 20:36
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@Ricky yeah, but I would not insist on claiming that , in absence of air drag, a iron ball would fall faster than a feather, if observed from Earth. From the experimental point of view unobservable differences are equivalent to no difference at all.
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– GiorgioP
Mar 24 at 21:04
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@Ricky Unless the falling body has a mass that's a significant fraction of Earth's, the acceleration it induces on the Earth is so tiny that not only is it immeasurable, it will also be swamped by various other effects, mainly air resistance, but also including the asymmetric distribution of mass in the Earth, seismic disturbances, tidal forces from the Sun, Moon, and the other planets (primarily Jupiter), other people dropping things at other locations, et cetera. If you insist on including the Earth's acceleration in your model you should also include those other things.
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– PM 2Ring
Mar 25 at 8:22
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@PM2Ring: I absolutely agree with you on all points here.
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– Ricky
Mar 25 at 16:11
add a comment |
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The correct replies to this question have already been written elsewhere in SE.Physics (better if ignoring the negative score answers).
However, it is probably useful to stress the basic issue underlying the answer to this question: if we describe motions in non-inertial frames, accelerations depend on the frame. Therefore, the fact that different masses are acclerated in the same way in inertial frames, does not imply that the same is true in non-inertial frames. And observing free fall of bodies on the Earth, means that we are using a non-inertial frame whose acceleration depends on the force between Earth and falling body. It is only the huge ratio between the mass of Earth and that of (reasonable size) falling bodies which makes practically unobservable the difference of accelerations.
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". It is only the huge ratio between the mass of Earth and that of (reasonable size) falling bodies which makes practically unobservable the difference of accelerations." That's what I said. "Practically unobservable" does not mean that it doesn't exist.
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– Ricky
Mar 24 at 20:36
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@Ricky yeah, but I would not insist on claiming that , in absence of air drag, a iron ball would fall faster than a feather, if observed from Earth. From the experimental point of view unobservable differences are equivalent to no difference at all.
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– GiorgioP
Mar 24 at 21:04
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@Ricky Unless the falling body has a mass that's a significant fraction of Earth's, the acceleration it induces on the Earth is so tiny that not only is it immeasurable, it will also be swamped by various other effects, mainly air resistance, but also including the asymmetric distribution of mass in the Earth, seismic disturbances, tidal forces from the Sun, Moon, and the other planets (primarily Jupiter), other people dropping things at other locations, et cetera. If you insist on including the Earth's acceleration in your model you should also include those other things.
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– PM 2Ring
Mar 25 at 8:22
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@PM2Ring: I absolutely agree with you on all points here.
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– Ricky
Mar 25 at 16:11
add a comment |
$begingroup$
The correct replies to this question have already been written elsewhere in SE.Physics (better if ignoring the negative score answers).
However, it is probably useful to stress the basic issue underlying the answer to this question: if we describe motions in non-inertial frames, accelerations depend on the frame. Therefore, the fact that different masses are acclerated in the same way in inertial frames, does not imply that the same is true in non-inertial frames. And observing free fall of bodies on the Earth, means that we are using a non-inertial frame whose acceleration depends on the force between Earth and falling body. It is only the huge ratio between the mass of Earth and that of (reasonable size) falling bodies which makes practically unobservable the difference of accelerations.
$endgroup$
The correct replies to this question have already been written elsewhere in SE.Physics (better if ignoring the negative score answers).
However, it is probably useful to stress the basic issue underlying the answer to this question: if we describe motions in non-inertial frames, accelerations depend on the frame. Therefore, the fact that different masses are acclerated in the same way in inertial frames, does not imply that the same is true in non-inertial frames. And observing free fall of bodies on the Earth, means that we are using a non-inertial frame whose acceleration depends on the force between Earth and falling body. It is only the huge ratio between the mass of Earth and that of (reasonable size) falling bodies which makes practically unobservable the difference of accelerations.
answered Mar 24 at 11:52
GiorgioPGiorgioP
4,2501628
4,2501628
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". It is only the huge ratio between the mass of Earth and that of (reasonable size) falling bodies which makes practically unobservable the difference of accelerations." That's what I said. "Practically unobservable" does not mean that it doesn't exist.
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– Ricky
Mar 24 at 20:36
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@Ricky yeah, but I would not insist on claiming that , in absence of air drag, a iron ball would fall faster than a feather, if observed from Earth. From the experimental point of view unobservable differences are equivalent to no difference at all.
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– GiorgioP
Mar 24 at 21:04
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@Ricky Unless the falling body has a mass that's a significant fraction of Earth's, the acceleration it induces on the Earth is so tiny that not only is it immeasurable, it will also be swamped by various other effects, mainly air resistance, but also including the asymmetric distribution of mass in the Earth, seismic disturbances, tidal forces from the Sun, Moon, and the other planets (primarily Jupiter), other people dropping things at other locations, et cetera. If you insist on including the Earth's acceleration in your model you should also include those other things.
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– PM 2Ring
Mar 25 at 8:22
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@PM2Ring: I absolutely agree with you on all points here.
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– Ricky
Mar 25 at 16:11
add a comment |
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". It is only the huge ratio between the mass of Earth and that of (reasonable size) falling bodies which makes practically unobservable the difference of accelerations." That's what I said. "Practically unobservable" does not mean that it doesn't exist.
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– Ricky
Mar 24 at 20:36
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@Ricky yeah, but I would not insist on claiming that , in absence of air drag, a iron ball would fall faster than a feather, if observed from Earth. From the experimental point of view unobservable differences are equivalent to no difference at all.
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– GiorgioP
Mar 24 at 21:04
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@Ricky Unless the falling body has a mass that's a significant fraction of Earth's, the acceleration it induces on the Earth is so tiny that not only is it immeasurable, it will also be swamped by various other effects, mainly air resistance, but also including the asymmetric distribution of mass in the Earth, seismic disturbances, tidal forces from the Sun, Moon, and the other planets (primarily Jupiter), other people dropping things at other locations, et cetera. If you insist on including the Earth's acceleration in your model you should also include those other things.
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– PM 2Ring
Mar 25 at 8:22
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@PM2Ring: I absolutely agree with you on all points here.
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– Ricky
Mar 25 at 16:11
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". It is only the huge ratio between the mass of Earth and that of (reasonable size) falling bodies which makes practically unobservable the difference of accelerations." That's what I said. "Practically unobservable" does not mean that it doesn't exist.
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– Ricky
Mar 24 at 20:36
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". It is only the huge ratio between the mass of Earth and that of (reasonable size) falling bodies which makes practically unobservable the difference of accelerations." That's what I said. "Practically unobservable" does not mean that it doesn't exist.
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– Ricky
Mar 24 at 20:36
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@Ricky yeah, but I would not insist on claiming that , in absence of air drag, a iron ball would fall faster than a feather, if observed from Earth. From the experimental point of view unobservable differences are equivalent to no difference at all.
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– GiorgioP
Mar 24 at 21:04
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@Ricky yeah, but I would not insist on claiming that , in absence of air drag, a iron ball would fall faster than a feather, if observed from Earth. From the experimental point of view unobservable differences are equivalent to no difference at all.
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– GiorgioP
Mar 24 at 21:04
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@Ricky Unless the falling body has a mass that's a significant fraction of Earth's, the acceleration it induces on the Earth is so tiny that not only is it immeasurable, it will also be swamped by various other effects, mainly air resistance, but also including the asymmetric distribution of mass in the Earth, seismic disturbances, tidal forces from the Sun, Moon, and the other planets (primarily Jupiter), other people dropping things at other locations, et cetera. If you insist on including the Earth's acceleration in your model you should also include those other things.
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– PM 2Ring
Mar 25 at 8:22
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@Ricky Unless the falling body has a mass that's a significant fraction of Earth's, the acceleration it induces on the Earth is so tiny that not only is it immeasurable, it will also be swamped by various other effects, mainly air resistance, but also including the asymmetric distribution of mass in the Earth, seismic disturbances, tidal forces from the Sun, Moon, and the other planets (primarily Jupiter), other people dropping things at other locations, et cetera. If you insist on including the Earth's acceleration in your model you should also include those other things.
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– PM 2Ring
Mar 25 at 8:22
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@PM2Ring: I absolutely agree with you on all points here.
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– Ricky
Mar 25 at 16:11
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@PM2Ring: I absolutely agree with you on all points here.
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– Ricky
Mar 25 at 16:11
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This might sound confusing i agree. A body of greater mass does experience a greater force compared to a lighter object , but then the acceleration is the ratio of Force experienced over the mass which equals gravitational constant times Mass of earth( in this case) over distance squared , which is independent of the mas of the object that is falling . You can also understand it this way : a greater force is greater is required to maintain a given acceleration for a more massive body than the force that is required to maintain the same accelaration for a less massive body.In both cases force varies but accelaration remains same
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add a comment |
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This might sound confusing i agree. A body of greater mass does experience a greater force compared to a lighter object , but then the acceleration is the ratio of Force experienced over the mass which equals gravitational constant times Mass of earth( in this case) over distance squared , which is independent of the mas of the object that is falling . You can also understand it this way : a greater force is greater is required to maintain a given acceleration for a more massive body than the force that is required to maintain the same accelaration for a less massive body.In both cases force varies but accelaration remains same
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add a comment |
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This might sound confusing i agree. A body of greater mass does experience a greater force compared to a lighter object , but then the acceleration is the ratio of Force experienced over the mass which equals gravitational constant times Mass of earth( in this case) over distance squared , which is independent of the mas of the object that is falling . You can also understand it this way : a greater force is greater is required to maintain a given acceleration for a more massive body than the force that is required to maintain the same accelaration for a less massive body.In both cases force varies but accelaration remains same
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This might sound confusing i agree. A body of greater mass does experience a greater force compared to a lighter object , but then the acceleration is the ratio of Force experienced over the mass which equals gravitational constant times Mass of earth( in this case) over distance squared , which is independent of the mas of the object that is falling . You can also understand it this way : a greater force is greater is required to maintain a given acceleration for a more massive body than the force that is required to maintain the same accelaration for a less massive body.In both cases force varies but accelaration remains same
answered Mar 24 at 13:45
Maxwell's GhostMaxwell's Ghost
132
132
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The gravitational force is proportional to the masses of the two interacting objects. Since force is mass times acceleration, instantaneous gravitational acceleration is independent of mass. This is a approximation. The solution to the full two body problem does depend on both masses. See: https://en.m.wikipedia.org/wiki/Gravitational_two-body_problem.
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add a comment |
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The gravitational force is proportional to the masses of the two interacting objects. Since force is mass times acceleration, instantaneous gravitational acceleration is independent of mass. This is a approximation. The solution to the full two body problem does depend on both masses. See: https://en.m.wikipedia.org/wiki/Gravitational_two-body_problem.
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add a comment |
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The gravitational force is proportional to the masses of the two interacting objects. Since force is mass times acceleration, instantaneous gravitational acceleration is independent of mass. This is a approximation. The solution to the full two body problem does depend on both masses. See: https://en.m.wikipedia.org/wiki/Gravitational_two-body_problem.
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The gravitational force is proportional to the masses of the two interacting objects. Since force is mass times acceleration, instantaneous gravitational acceleration is independent of mass. This is a approximation. The solution to the full two body problem does depend on both masses. See: https://en.m.wikipedia.org/wiki/Gravitational_two-body_problem.
answered Mar 24 at 14:35
my2ctsmy2cts
5,7522719
5,7522719
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To put more words into @Dale answer, the greater the gravitational mass,the greater the force due to gravity per Newton's law of gravity. But the greater the mass the greater its inertial mass as well, i.e., its resistance to a change in velocity, and therefore a greater force is required to accelerate the mass per Newton's second law, $F=ma$. Gravitational mass equals inertial mass, thus the acceleration is the same for all masses in the gravitational field.
Hope this helps.
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To put more words into @Dale answer, the greater the gravitational mass,the greater the force due to gravity per Newton's law of gravity. But the greater the mass the greater its inertial mass as well, i.e., its resistance to a change in velocity, and therefore a greater force is required to accelerate the mass per Newton's second law, $F=ma$. Gravitational mass equals inertial mass, thus the acceleration is the same for all masses in the gravitational field.
Hope this helps.
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To put more words into @Dale answer, the greater the gravitational mass,the greater the force due to gravity per Newton's law of gravity. But the greater the mass the greater its inertial mass as well, i.e., its resistance to a change in velocity, and therefore a greater force is required to accelerate the mass per Newton's second law, $F=ma$. Gravitational mass equals inertial mass, thus the acceleration is the same for all masses in the gravitational field.
Hope this helps.
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To put more words into @Dale answer, the greater the gravitational mass,the greater the force due to gravity per Newton's law of gravity. But the greater the mass the greater its inertial mass as well, i.e., its resistance to a change in velocity, and therefore a greater force is required to accelerate the mass per Newton's second law, $F=ma$. Gravitational mass equals inertial mass, thus the acceleration is the same for all masses in the gravitational field.
Hope this helps.
answered Mar 24 at 15:15
Bob DBob D
4,3932318
4,3932318
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2
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What is your argument against Galileo's conclusion?
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– my2cts
Mar 24 at 9:38
2
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$F=ma$, so the gravitational acceleration is independent of the mass of the falling body. However, see physics.stackexchange.com/q/3534
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– PM 2Ring
Mar 24 at 9:49
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Inertia isn't a force
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– Aaron Stevens
Mar 24 at 10:24
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@Ricky The difference is due to how much the Earth accelerates upwards to meet the falling object, in the centre of mass frame of (the Earth + object), the object's acceleration towards the centre of mass is still independent of its mass.
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– PM 2Ring
Mar 24 at 10:32
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Possible duplicate of Why do two bodies of different masses fall at the same rate (in the absence of air resistance)?
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– knzhou
Mar 24 at 13:13