# Why don't heavy objects fall more SLOWLY?

1. Dec 10, 2009

### chudd88

This is a simple question, and I'm sure it has a trivial answer, but this thought occured to me just now.

How is it that gravity is able to accelerate things at the same rate regardless of its mass, while one definition of mass is the measure of resistance to acceleration? If two objects are 100 miles from Earth, I would expect that the Earth's gravity would be able to accelerate the lighter object more quickly, while the heavier object would resist the acceleration and fall more slowly. Why is this not the case?

2. Dec 10, 2009

### TurtleMeister

It is true that the more massive object will resist acceleration more than the less massive object. However, the more massive object has a greater force acting on it. The two are perfectly balanced so that the effect is that all objects will accelerate at the same rate regardless of their mass.

I'm sure this question has been asked many times at PF, so you may be able to use the search function to learn more. Or, please feel free to ask more questions here if you like.

3. Dec 10, 2009

### Staff: Mentor

You may notice that the equation for acceleration due to a force and the equation for weight are just two different forms of the same equation....

f=ma
a=f/m

"a" is the acceleration due to gravity.

4. Dec 11, 2009

### arildno

Because the two objects do not experience the same force.

Gravity is a puzzling force; it is proportional in magnitude to the mass of the object ACTED UPON.
(Very few other forces have this strange property)

Thus, the more massive object is acted upon with a greater force than the lighter object, the net result being that both experience the same acceleration.

5. Dec 11, 2009

### Saw

But this rule (principle of equivalence) is established judging… from which reference frame?

I see two possibilities:

(a) From the perspective of an inertial frame not participating in the interaction. For example, the centre of mass of the system composed by bodies with masses M and m, which are gravitationally interacting.

(b) From the perspective of a frame participating in the interaction and hence an accelerated one.

If we judge (measure) from (a), the inertial frame, the centre of mass (CM) of the system, it seems to me that the principle of equivalence should not apply: the more massive object M would cover the shorter distance to the CM with lesser acceleration and the less massive should traverse its own longer path with more acceleration, so that both meet precisely at the said CM. This would be in line with what the OP suggested and seems to be the natural intuition following the example of collisions (i.e., when the relevant force is contact force).

Given this, the principle of equivalence would apply only with regard to or as measured from (b), i.e., from the reference frame of one of the attracted objects. In the frame of M, for example, the acceleration of m1 would be the sum of m1's acceleration as measured in (a) plus the acceleration of M itself as measured also in (a), in the opposite direction. If we now choose m2, more massive than m1, the first component would be lesser (m2 resists more) but the second would be greater (M is accelerated more). One thing would offset the other and thus, as judged from M’s reference frame, all objects would fall towards it with the same acceleration, regardless their respective masses. The same would apply to the judgment from m1, m2…

Did I guess it right?

Note: If you read this post before, I've edited it to make it clearer. Thanks for any comment.

Last edited: Dec 11, 2009
6. Dec 11, 2009

### Louis Wu

Saw

Its a 3 body Problem .. no CM. CM situs are not i think what the op is talking about.

Huge mass.. the effector

One small mass [sm] and one large mass[lm]..Both > to effector
Why does lm accelerate same dv.
" Why does a feather fall as fast as a lead balloon"

Louis
________________________________
PS i have no formal 'education', everything i say i taught myself.
So pls be patient. I tend to incorporate real life into the crap jargon so many are fond of spouting. Albert tried this and i does not work. I persevere

7. Dec 11, 2009

### D H

Staff Emeritus
Any frame of reference.

It would be silly for the equivalence principle to say something that contraindicates well-explained phenomena. Experiment one: Place an Earth-sized mass and a pea-sized mass at rest with respect to one another but separated by 384,400 km (the Earth-Moon distance). Measure the time to collision, which will be several days. Experiment two: Replace the pea-sized mass with a neutron star. These objects will collide on a much smaller time scale.

Fortunately, the equivalence principle does not say that the time to collision will be the same. It instead says that the initial accelerations of the pea and the neutron star toward the Earth will be exactly the same.

8. Dec 11, 2009

### Saw

Well, yes, that sounds logical.

Yes… I suppose I had not grasped what it really says. But you’ll maybe concede that there’s something confusing in all this. Take for example the formulation in Wikipedia:

And also, a little above in the page:

Reading this, wouldn’t you be led to think that the pie and the neutron star, if "dropped" from the same place", should hit the Earth ground at the same time? Of course, for clarity, we should be assuming that each object is dropped at a different moment, in a different experiment, because otherwise what we have (maybe that’s what Louis was pointing at) is a 3-body problem. My understanding was thus that the principle requires that in each experiment (done from the same place but in a different moment) the fall time is identical. But then if the principle does not entail so, what does it state?

So… only the initial accelerations would be equivalent? And why and when should they start differing? As the fall progresses, the distance is reduced and the acceleration augments…, but that is independent of the mass of the object acted upon…

9. Dec 11, 2009

### arildno

The acceleration of the Earth due to the neutron star's pull would be WAY stronger, so the Earth would swiftly accelerate towards it, decreasing the distance swiftly.

10. Dec 11, 2009

### mikhailpavel

the heavier object falling faster than the lighter in the earth's surface is due to the air resistance acting on the objects not due to the gravity as far as i know.

11. Dec 11, 2009

### arildno

Not if the heavy object is a neutron star as in the given example.

Even in vacuum, a (n extremely) heavier object will, RELATIVE to the Earth surface, accelerate faster than a lighter object, due to its stronger attractive pull on the Earth.

12. Dec 11, 2009

### trust143_raj

what about the black holes? am confused?

13. Dec 11, 2009

### Saw

Yes, I understand that. Certainly, since there is an interaction (both sides suffer effects = acceleration towards each other), it's not indifferent that the Earth faces a pie or a neutron star: if it is the latter, the Earth itself is accelerated more. My problem is how to reconcile that with the equivalence principle, in the formulation quoted above from Wikipedia. In the classical Galileo's (real or thought) experiment two lead balls of different sizes dropped from the Tower of Pisa reach the ground (neglecting air resistance) at the same time. It is said that an austronaut dropped a feather and a hammer on the Moon (where there's no atmosphere) and the two fell at the same rate. What changes if you substitute the neutron star for one of the leaden balls or the hammer? Unless... the difference is that the masses of the balls, feather and hammer are all ridiculous in comparison with the mass of the Earth and so they are all equivalent to a good approximation, for practical purposes... But I have not read that interpretation anywhere, it looks even less orthodox and it would convert the equivalence principle into a mere pragmatic rule of thumb for use under certain conditions...

14. Dec 11, 2009

### SystemTheory

Newton's Law of Universal Gravitation:

$$F = G\frac{m_{1}m_{2}}{r^{2}}$$

where each body has its own acceleration a = F/m toward the other body:

$$a_{1} = G\frac{m_{2}}{r^{2}}$$

$$a_{2} = G\frac{m_{1}}{r^{2}}$$

so then the time of acceleration should be established by taking an integral of a differential equation after applying some convenient frame of reference. It is not clear to me that the times will differ as a function of mass.

15. Dec 11, 2009

### Saw

Yes, that was the next step for discussion. It's not only the formulation in words of the equivalence principle, it's Newton's Laws what seems to require that the neutron star reaches the Earth ground at the same time as the pie...

16. Dec 11, 2009

### D H

Staff Emeritus
Correct. Note that the acceleration of object 1 is independent of the mass of object 1, and the acceleration of object 2 is independent of the mass of object 2.

The acceleration of objects 1 and 2 toward each other is

$$a_{\text{rel}} = \frac{G(m_1+m_2)}{r^2}$$

and this obviously depends on the masses of both objects.

If one of those masses is many orders of magnitude larger than the other, the acceleration for all practical purposes depends only on the mass of the larger object. Suppose that in a vacuum you lift a 1 gram pea 4.9 meters high and release it. It will hit the Earth one second later. Now do the same with a 10 metric ton stone. It too will hit the Earth one second later. Because the Earth's acceleration toward the stone is 10 million times that toward the pea, there will be a very slight difference in the timing. How much? About 1.7×10-21]/sup] seconds. This is of course immeasurably small.

The Moon's mass is about 1/81 that of the Earth. This is enough to affect the Moon's orbit about the Earth. Suppose the Moon's mass was a couple of orders of magnitude smaller than it is. This reduced mass Moon would take about 20 minutes longer to orbit the Earth than does our big honkin' Moon.

17. Dec 11, 2009

### TurtleMeister

I think the way this is worded may have caused some confusion. It is true that the accelerations of the pea and the neutron star will be the same. However, for clarification, it should be pointed out that the accelerations are in relation to the barycenter and not to either one of the objects. Or in other words, the frame of reference is the barycenter.
The simple test that you refer to assumes that the masses of the objects dropped are so small compared to the mass of the earth that the earths acceleration can be ignored. If the objects dropped are of significant mass in relation to the earth, then of course you cannot ignore the acceleration of the earth. Also, the frame of reference for this test is the earth. So the acceleration will be the relative acceleration of the earth and the object toward each other.

Last edited: Dec 11, 2009
18. Dec 11, 2009

### TurtleMeister

I didn't see this post before I made my previous post. The first equations by SystemTheory are the ones relevant to the equivalence principle. It assumes the reference frame to be the barycenter. The second equation by DH for relative acceleration is the one used for the simple test quoted by "Saw". This one assumes that one object (such as the earth) is of much greater mass than the other object.

19. Dec 11, 2009

### Saw

Then if I understand you all correctly:

(a) If the acceleration is measured from the frame of one of the objects participating in the gravitational interaction (i.e., an accelerated frame, say the Earth in our example)..., in other words, if we refer to the relative acceleration between the two interacting objects, the equivalence principle does not strictly speaking apply: it does only under certain circumstances (if one mass is negligible wrt the other), to a good approximation, for most practical purposes.

Accordingly here the formula is:

$$a_{\text{rel}} = \frac{G(M+m)}{r^2}$$

(b) If the acceleration is measured from an inertial frame (like the center of mass of the system formed by the two bodies or barycentre)…, in other words, if we refer to the individual accelerations towards the origin of such frame, then:

those individual accelerations are different for each body:

the more massive body accelerates at the rate of

$$a_{M} = G\frac{m}{r^{2}}$$

which is less than the acceleration rate of the less massive body

$$a_{m} = G\frac{M}{r^{2}}$$

and the equivalence principle applies in the sense that if you put any other body in the side of, for example, m (the neutron star in the place of the pea), the acceleration of the star towards the barycentre will be identical to that of the pea, since in both cases it's a function of the mass of M (apart from G and r).

Did I get it right now?

20. Dec 11, 2009

### TurtleMeister

Yes. At least to the best of my understanding. I agree with everything in your post.