# Question about objects falling at the same speed

1. Feb 5, 2016

### livethere

Alright so I've learned that in a vacuum two objects, irrespective of mass, fall at the same speed. This however doesn't make since to me, because I've also learned that on the moon objects fall at a different speed then on Earth due to the gravitational pull.

This seems to imply that the gravitational pull upon objects has an affect at which speed the object is falling at.

2. Feb 5, 2016

### A.T.

When placed in the same gravitational field.

3. Feb 5, 2016

### CWatters

The Universal Gravitation Equation is:

F = GMm/R2

where M and m are the two masses and R is the distance between them. M is usually the mass of a planet and m the mass of a small object.

Newton says F=ma where a is the acceleration. Substitute to give..

ma = GMm/R2

m cancels leaving

a = GM/R2

So the acceleration of m is independent of the mass of m. However it's not independent of the mass M of the planet.

4. Feb 5, 2016

### livethere

Hey thanks for the responses. I still have a few questions however.
In regarding the smaller mass object, what occurs if the two masses are the same?

I also interpreted your result as saying that no matter how close the smaller m is to the mass of the big M, it doesn't matter as long as it's below the big M.

5. Feb 5, 2016

### CWatters

No difference.

No I didn't mean to imply that. It's just convention that m is used for the mass of the smaller object and M for the bigger. If you look on Wikipedia you will see they use m1 and m2 instead.

https://en.wikipedia.org/wiki/Gravity

6. Feb 5, 2016

### sophiecentaur

I don't see what you're getting at, here. Have you changed direction? The thread title implies a mass 'falling' on Earth (or a planet). The local g force is given by GM/R2. M is the large mass of the planet.
If you're considering two (vaguely comparable) masses then the force equation (on either one of the masses) is GMm/R2 but R corresponds to the distance from the centre of mass of the pair and 'falling' is not quite the word for the result of the attraction.

7. Feb 5, 2016

### jbriggs444

R is still the distance between the centers of masses of the pair elements in such a case. It is not their individual distances from the system's combined center of mass.

If it were otherwise, Newton's third law would be up for a violation.

8. Feb 5, 2016

### Janus

Staff Emeritus
I'm guessing here that you mean what happens if M and m have equal masses.
If you look at the equations given by CWatters, it is important to note that they are dealing with the acceleration of m due to the gravitational attraction between M and m. However, it is important to note that M is subject to that same attraction. Thus while the the acceleration of m towards M is found by GM/R^2, the acceleration of M towards m is Gm/R^2. (n other words, while m falls toward M, M falls towards m) If M is many many times larger than m the acceleration of M is going to so small that we can ignore it in most cases. For example, if I dropped a 1 kg mass from a height of 1 meter above the surface of the Earth, it would fall towards the Earth with an acceleration of 9.80115 m/s/s. The Earth, on the other hand, would "fall" towards the 1 kg mass with an acceleration of 1.6404e-24 m/s/s. This means that when they meet, the 1kg mass will have traveled all but the tiniest fraction of the 1 meter between it and the ground and the Earth will have moved by an all but immeasurable distance. If we increase the dropped mass by 1000 times to 1000 kg, it will still accelerate at 9.80115 m/s/s and the Earth will accelerate towards it at 1.6404e-21 m/s/s or at a rate 1000 times greater than it did towards the 1 kg mass. But a thousand times almost nothing is still almost nothing. You would have to increase the dropped mass by a lot more before there would be a noticeable difference.

If m and M are equal in value, then they will fall towards each other will equal accelerations and collide at a point halfway between their starting positions. The point towards the two masses fall is called the barycenter of the masses. When M and m are equal, it is halfway between them. The larger M is compared to m, the closer it is to the center of M.

9. Feb 5, 2016

### livethere

I believe that I was being vague in my question. Sorry about that.
Although I do understand your responses and I appreciate you taking the time to get back to me. :)
My question is as follows.

Let's say that a dust particle and a sun are approaching a black hole at the middle of our galaxy.
I understand that the sun would ever so slightly bring the black hole closer to it through its gravity, however that's not my question.
My intuitive thinking tells me that due to the black hole having a greater gravitational affect on the sun than on the dust particle that the sun might approach the black hole in a faster manner.

10. Feb 5, 2016

### sophiecentaur

You have made this a three body problem now. Plus, you can throw in some relativity. You need to redifine your question.

11. Feb 5, 2016

### gmax137

What about this situation do you think is different from the cases described in the previous posts (1 kg mass vs 1000 kg mass falling towards earth)?

12. Feb 5, 2016

### livethere

That post was addressing the gravitational affects regarding two objects.
My question deals with 3 objects.

In my example stated above, we have two objects being attracted to a main attractive object.
However I'm wondering if the greater mass of those two objects would be attracted more and would therefore be moved faster than the object with lesser mass towards the main attractive object.

13. Feb 5, 2016

### mfig

Post #3 showed very simply that the rate of acceleration by gravitational force is independent of the mass of the object being accelerated. That is the answer to your question. Call the main attractive object $M$, and your two smaller objects $m_1$ and $m_2$, then run through the derivation shown in post 3 again. You should end up with the acceleration for both $m_1$ and $m_2$ being equal to $\frac{GM}{R^2}$. If they have the same acceleration, they fall with the same speed when dropped from rest. Period.

Perhaps what is confusing you is that the more massive object would feel a larger force. That is true! The more massive object would feel a larger force if placed in the same gravitational field as the smaller object. But that larger force is accelerating a more massive object, compared to the force accelerating the less massive object, and therefore gets less "bang for its buck" in the acceleration department.

The ratios work out in such a way that the acceleration each experiences is the same.