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

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1. 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
2. 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.

3. 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.

4. Nov 25, 2017

### Staff: Mentor

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

5. 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 $m_1$.

6. Nov 26, 2017

### Staff: Mentor

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.

7. Nov 28, 2017

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.

8. Nov 28, 2017

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

9. Nov 28, 2017

### Staff: Mentor

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.

10. Nov 29, 2017

### Staff: Mentor

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

11. Nov 29, 2017

### Staff: Mentor

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