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

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1. Nov 24, 2017

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.

2. Nov 24, 2017

### A.T.

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

3. Nov 24, 2017

Okay.

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

4. Nov 24, 2017

### SchroedingersLion

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

5. Nov 24, 2017

### sophiecentaur

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.

6. Nov 24, 2017

### A.T.

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

7. Nov 24, 2017

### sophiecentaur

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

8. Nov 24, 2017

### ZapperZ

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

Zz.

9. Nov 24, 2017

Great article. I read all of it :).

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

10. Nov 24, 2017

### Staff: Mentor

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 $m_2$

11. Nov 24, 2017

Staff Emeritus
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.

12. Nov 24, 2017

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?

13. Nov 24, 2017

### ZapperZ

Staff Emeritus

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.

14. Nov 24, 2017

Okay. Thank you!

15. Nov 24, 2017

### Staff: Mentor

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 $m_2$

16. Nov 24, 2017

### David Lewis

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

17. Nov 24, 2017

### DaveC426913

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.

18. Nov 25, 2017

### Staff: Mentor

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".

19. Nov 25, 2017

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).

Last edited: Nov 25, 2017
20. Nov 25, 2017

### Staff: Mentor

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.

21. Nov 25, 2017

### 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.

Last edited: Nov 25, 2017
22. Nov 25, 2017

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.

23. Nov 25, 2017

### TurtleMeister

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.

24. Nov 25, 2017

### Staff: Mentor

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

25. Nov 26, 2017

### PeroK

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 $m_1$.