Why do objects fall at the same speed in free fall?

[Jay Addy]

This is something I haven't been able to wrap my head around yet.

In physics, I've always been told that gravity is a force that ALWAYS works between "objects" with mass. Now, it seems clear to me that if a feather and a hammer were to be dropped at the same time on Earth (without air resistance doing work on the two objects) they would indeed fall at the same velocity. Here is my problem: if gravity works between all objects, and if "insert planet or object" pulls with a greater force on the heavier object, to compensate for the heavier mass, wouldn't the object also pull on the "insert planet or object"?

Let's say you have a planet with the weight of 6,0 * 10^20 kg. You drop two objects: a feather (with the weight 0,1 kg) and a gigantic ball (with the weight 6,0 * 10^19 kg. Given that the two objects are only affected by gravity, why won't the big ball (with a mass almost equal to the planet's mass) pull the planet closer to itself, thus reaching impact before the feather does?

I love physics, and I just recently started taking classes in high school. Any answers are greatly appreciated.

 

[A.T.]

Let's say you have a planet with the weight of 6,0 * 10^20 kg. You drop two objects: a feather (with the weight 0,1 kg) and a gigantic ball (with the weight 6,0 * 10^19 kg. Given that the two objects are only affected by gravity, why won't the big ball (with a mass almost equal to the planet's mass) pull the planet closer to itself, ...

It will. The equal fall is an approximation for objects of negligible own mass compared to the gravity source.

 

[Jay Addy]

Okay.

Is it correct of me to inform people who believe that: "All objects fall at the same speed in free fall" are wrong?

 

[SchroedingersLion]

Okay.

Is it correct of me to inform people who believe that: "All objects fall at the same speed in free fall" are wrong?

Technically yes. But for objects with a mass far smaller than Earth's mass, there is no significant difference.

 

[sophiecentaur]

Okay.

Is it correct of me to inform people who believe that: "All objects fall at the same speed in free fall" are wrong?

Yes - but only if you explain where the limits to that statement apply. The statement is only true for a massive object (planet etc.) and a much smaller object (battleship) when there is no atmosphere to provide drag.
If you don't include those caveats, people will be back in the pre-Galileo attitude to gravity. If this is a bit of a crusade for you then you have to be well informed to deal with any counter arguments you may get thrown at you.

 

[A.T.]

Is it correct of me to inform people who believe that: "All objects fall at the same speed in free fall" are wrong?

It depends on details, like frame of reference and how you perform the experiment.

 

[sophiecentaur]

It depends on details, like frame of reference and how you perform the experiment.

. . . and people should realize that ALL SCIENCE depends on details. Every so called 'Law' can be limited in its range of applications - some more so than others.

 

[ZapperZ]

This is something I haven't been able to wrap my head around yet.

In physics, I've always been told that gravity is a force that ALWAYS works between "objects" with mass. Now, it seems clear to me that if a feather and a hammer were to be dropped at the same time on Earth (without air resistance doing work on the two objects) they would indeed fall at the same velocity. Here is my problem: if gravity works between all objects, and if "insert planet or object" pulls with a greater force on the heavier object, to compensate for the heavier mass, wouldn't the object also pull on the "insert planet or object"?

Let's say you have a planet with the weight of 6,0 * 10^20 kg. You drop two objects: a feather (with the weight 0,1 kg) and a gigantic ball (with the weight 6,0 * 10^19 kg. Given that the two objects are only affected by gravity, why won't the big ball (with a mass almost equal to the planet's mass) pull the planet closer to itself, thus reaching impact before the feather does?

I love physics, and I just recently started taking classes in high school. Any answers are greatly appreciated.

https://www.physicsforums.com/insights/why-is-acceleration-due-to-gravity-a-constant/

Zz.

 

[Jay Addy]

https://www.physicsforums.com/insights/why-is-acceleration-due-to-gravity-a-constant/

Zz.

Great article. I read all of it :).

To the previous posts: Context is obviously very important ;-).

 

[Dale]

Okay.

Is it correct of me to inform people who believe that: "All objects fall at the same speed in free fall" are wrong?

The “all objects fall at the same speed in free fall” is probably a misquote anyway. Gravity causes acceleration, not speed.

What is correct in Newtonian gravity is that $a_2=G m_1/r_{1,2}^2$ where the subscripts indicate the object. That remains true regardless of $m_2$

 

[Vanadium 50]

To give you an idea on how minor this niggle is, if you drop a 1 kg weight from your left and and a 2 kg weight from your right, the combined inhomogeneity and non-sphericity of the Earth on meter scales is a bigger effect than the one you are worried about.

 

[Jay Addy]

The “all objects fall at the same speed in free fall” is probably a misquote anyway. Gravity causes acceleration, not speed.

What is correct in Newtonian gravity is that $a_2=G m_1/r_{1,2}^2$ where the subscripts indicate the object. That remains true regardless of $m_2$

But if m₂ (the ball in the previous examples) has a mass almost identical to that of the planet the two objects "fall" closer towards, wouldn't m₂ create a greater pull on the planet, thus colliding with it before the feather m₁ will. Therefore, although the accelerations m₁ and m₂ experience are equal, m₂ will reach "impact" sooner by causing the planet to accelerate towards it?

 

[ZapperZ]

But if m₂ (the ball in the previous examples) has a mass almost identical to that of the planet the two objects "fall" closer towards, wouldn't m₂ create a greater pull on the planet, thus colliding with it before the feather m₁ will. Therefore, although the accelerations m₁ and m₂ experience are equal, m₂ will reach "impact" sooner by causing the planet to accelerate towards it?

You need to reread the link I gave.

IF the two masses are comparable, then you have to solve this in the center of mass frame. You can no longer have a situation where one mass is held fixed.

Zz.

 

[Jay Addy]

You need to reread the link I gave.

IF the two masses are comparable, then you have to solve this in the center of mass frame. You can no longer have a situation where one mass is held fixed.

Zz.

Okay. Thank you!

 

[Dale]

But if m₂ (the ball in the previous examples) has a mass almost identical to that of the planet the two objects "fall" closer towards, wouldn't m₂ create a greater pull on the planet, thus colliding with it before the feather m₁ will. Therefore, although the accelerations m₁ and m₂ experience are equal, m₂ will reach "impact" sooner by causing the planet to accelerate towards it?

Sure, but time to impact is not the same as acceleration. The acceleration of $m_2$ does not depend on $m_2$ and the acceleration of $m_1$ does not depend on $m_1$ regardless of the relative size of $m_1$ and $m_2$

 

[David Lewis]

Is it correct of me to inform people who believe that: "All objects fall at the same speed in free fall" are wrong?

No. In testing the theory, it is assumed you will not be dropping planets.

 

[DaveC426913]

The “all objects fall at the same speed in free fall” is probably a misquote anyway. Gravity causes acceleration, not speed.

Pfft.
Gil Grissom from CSI said "Well, terminal velocity is 9.8 metres per second [sic], so falling from a 5 story building would put him... here."

And Gil's never wrong.

 

[mfb]

Sure, but time to impact is not the same as acceleration. The acceleration of $m_2$ does not depend on $m_2$ and the acceleration of $m_1$ does not depend on $m_1$ regardless of the relative size of $m_1$ and $m_2$

The position of the mass $m_1$ after some time will depend on the value of $m_2$. While the initial acceleration is the same, the acceleration later differs. But that is a completely negligible effect for all cases where people say "objects of different mass fall with the same acceleration".

 

[Jay Addy]

The statement: "On earth, all objects fall in the same way" is therefore correct. If it's about the duration. AND that the objects come from the earth.
Dylan

While this might be true, this answer is not really what I was looking for when I posted my question. If you neglect the fact that the two test objects are from earth, would they fall at the same way?
They would initially, but if m₂ is significantly large (close to Earth's mass) then it will pull on Earth and thus shorten the distance between them compared to m₁ (the mass of the feather). At least this is what I've been told by a lot of people, and this ultimately lead to the question at hand: "Why do objects fall at the same speed in free fall?" and if this is not true 100% of the time, then to what extent is it true?
-If the masses of m₁ and m₂ are in no way comparable to the mass of the object they "fall" towards?
-If both objects are from earth?
-If both objects fall on earth?

Some of these questions have been answered in this thread (to a certain degree).

 

[mfb]

Sometimes descriptions are simplified and not everything is explicitly discussed in every sentence. That’s how languages are used. If you don’t like it, use the mathematics.

 

[TurtleMeister]

The center of mass of two bodies is determined by the difference in their inertial mass. If the two bodies have the same inertial mass then this point will be one half the distance between them. Or more specifically, the distance of M₁ to the center of mass is $D\frac{M_{2}}{M_{1}+M_{2}}$ and the distance of M₂ to the center of mass is $D\frac{M_{1}}{M_{1}+M_{2}}$, where D is the distance between the two bodies. So if the two bodies are on a collision course then this is the point where they will meet (barring their physical size by assuming test masses). If the two bodies are in orbit about each other then this is the point they will orbit around.

However, the acceleration, the time to impact, and the period of the orbit, are all determined by another property of mass. And that property is called the active gravitational mass. A measure of this property is called the standard gravitational parameter, usually denoted as $\mu$. The acceleration of two bodies on a collision course with each other is determined by this property, or more specifically $\frac{\mu_{1}+\mu_{2}}{R^2}$, where R is the distance between the two bodies.

Now, even though these two properties are completely different and have different units of measure they are proportionally equivalent. See the Equivalence Principle. So if you know the value of one, you can calculate the value of the other by using their factor of proportionality, which is the universal gravitational constant, or big G.

One of the main sticking points in understanding the universality of free fall, or UFF, is using an improper frame of reference. The UFF is only valid when using the center of mass as the frame of reference. However, notice in the second paragraph, which describes the effect that $\mu$ has on the acceleration, that the frame of reference is not the center of mass. The acceleration in this frame is called the relative acceleration and will not work for the UFF. It is the acceleration as viewed from one body to the other. To get the proper frame of reference we can use the method from paragraph one:

$$A_{1}=\frac{\mu_{1}+\mu_{2}}{R^2}\frac{M_{2}}{M_{1}+M_{2}}=\frac{\mu_{1}+\mu_{2}}{M_{1}+M_{2}}\frac{M_{2}}{R^2}=G\frac{M_{2}}{R^2}$$

$$A_{2}=-\frac{\mu_{1}+\mu_{2}}{R^2}\frac{M_{1}}{M_{1}+M_{2}}=-\frac{\mu_{1}+\mu_{2}}{M_{1}+M_{2}}\frac{M_{1}}{R^2}=-G\frac{M_{1}}{R^2}$$

The UFF is always true regardless of the difference in mass of the bodies.

 

[Jay Addy]

The center of mass of two bodies is determined by the difference in their inertial mass. If the two bodies have the same inertial mass then this point will be one half the distance between them. Or more specifically, the distance of M₁ to the center of mass is $D\frac{M_{2}}{M_{1}+M_{2}}$ and the distance of M₂ to the center of mass is $D\frac{M_{1}}{M_{1}+M_{2}}$, where D is the distance between the two bodies. So if the two bodies are on a collision course then this is the point where they will meet (barring their physical size by assuming test masses). If the two bodies are in orbit about each other then this is the point they will orbit around.

However, the acceleration, the time to impact, and the period of the orbit, are all determined by another property of mass. And that property is called the active gravitational mass. A measure of this property is called the standard gravitational parameter, usually denoted as $\mu$. The acceleration of two bodies on a collision course with each other is determined by this property, or more specifically $\frac{\mu_{1}+\mu_{2}}{R^2}$, where R is the distance between the two bodies.

Now, even though these two properties are completely different and have different units of measure they are proportionally equivalent. See the Equivalence Principle. So if you know the value of one, you can calculate the value of the other by using their factor of proportionality, which is the universal gravitational constant, or big G.

One of the main sticking points in understanding the universality of free fall, or UFF, is using an improper frame of reference. The UFF is only valid when using the center of mass as the frame of reference. However, notice in the second paragraph, which describes the effect that $\mu$ has on the acceleration, that the frame of reference is not the center of mass. The acceleration in this frame is called the relative acceleration and will not work for the UFF. It is the acceleration as viewed from one body to the other. To get the proper frame of reference we can use the method from paragraph one:

$$A_{1}=\frac{\mu_{1}+\mu_{2}}{R^2}\frac{M_{2}}{M_{1}+M_{2}}=\frac{\mu_{1}+\mu_{2}}{M_{1}+M_{2}}\frac{M_{1}}{R^2}=G\frac{M_{1}}{R^2}$$

$$A_{2}=-\frac{\mu_{1}+\mu_{2}}{R^2}\frac{M_{1}}{M_{1}+M_{2}}=-\frac{\mu_{1}+\mu_{2}}{M_{1}+M_{2}}\frac{M_{1}}{R^2}=-G\frac{M_{1}}{R^2}$$

The UFF is always true regardless of the difference in mass of the bodies.

Thank you for breaking down a large amount of information into something as easily readable and understandable as what you just wrote. Well done. You gave me a better understanding of the subject at hand for sure :-).

Jay.

 

[TurtleMeister]

Thank you for breaking down a large amount of information into something as easily readable and understandable as what you just wrote. Well done. You gave me a better understanding of the subject at hand for sure :-).

Jay.

I noticed a typo in one of my equations which has been corrected, but I cannot correct it in your post where you quoted me. I'm glad I was able to help with your understanding of the UFF.

 

[Dale]

The position of the mass m1m1m_1 after some time will depend on the value of m2m2m_2. While the initial acceleration is the same, the acceleration later differs.

Yes, but at no point does the acceleration of $m_1$ in free fall depend on $m_1$.

 

[PeroK]

Yes, but at no point does the acceleration of $m_1$ in free fall depend on $m_1$.

That depends what you mean by "point". If at any time the mass $m_1$ is replaced by a different mass, then that is true. But, as a function of time or position, say, the acceleration of $m_1$ does depend on $m_1$.

 

[Dale]

If at any time the mass m1m1m_1 is replaced by a different mass, then that is true.

Yes, this is what I was implying.

If you write out the equations for the evolution of the system as a whole, that does depend on $m_1$. But at any point in the state space the acceleration of $m_1$ does not depend on $m_1$.

All other things being equal, the acceleration of $m_1$ is independent of $m_1$. In the classical two body problem all other things are not kept equal, and that is where the differences come in.

 

[Brad Jensen]

Actually all objects affected by mutual gravitation accelerate towards each other. When you drop a feather the Earth accelerates towards it. However the effect is so small you can't measure it.

So technically it is untrue that unequal masses accelerate (or 'fall') at the same rate, but the difference in acceleration is not observable in practical demonstrations.

 

[Brad Jensen]

Pfft.
Gil Grissom from CSI said "Well, terminal velocity is 9.8 metres per second [sic], so falling from a 5 story building would put him... here."

And Gil's never wrong.

View attachment 215571

"The Sun is going down! " "No, the horizon is moving up!" - Firesign Theatre

 

[Dale]

Actually all objects affected by mutual gravitation accelerate towards each other.

You have to be careful here. The acceleration of object A is not the same thing as the acceleration of A-B. You are talking about the acceleration of A-B. The acceleration of A, at any moment, is independent of the mass of A. The acceleration of A-B depends on the reduced mass of the system.

 

[mfb]

The acceleration of A-B depends on the reduced mass of the system.

It depends on the sum. a_rel=G(M+m)/R².

 

[Dale]

It depends on the sum. a_rel=G(M+m)/R².

Oops, yes you are right. The reduced mass gives the equivalent force for a 1 body problem.