# G acceleration

1. Jan 8, 2013

### scientifico

Hello, comparing the formula of gravitational attraction with F = m*a you get that the smaller mass disappear.
I don't think this is physically correct, do you ?

thank you

2. Jan 8, 2013

### ZapperZ

Staff Emeritus

Zz.

Last edited by a moderator: May 6, 2017
3. Jan 8, 2013

### Staff: Mentor

Why do you think it's not correct?

4. Jan 8, 2013

### VantagePoint72

To briefly summarize ZapperZ's FAQ entry (since it mostly just derives what you seem to have already realized): what you have noticed is the fact that, ignoring air resistance, all objects fall at the same rate if dropped from the same height. Galileo noticed this a long time ago. It's not only physically correct—it's the defining feature of gravity.

5. Jan 8, 2013

### ZapperZ

Staff Emeritus
I think you missed the point of the "mostly just derives" part. The derivation shows one very important assumption, making it valid for a particular situation, i.e the case for two bodies in which one mass is significantly smaller than the other, i.e. m << M. I wouldn't call this "the defining feature of gravity" since the situation applies only to a specific case.

For m comparable to M, this doesn't work.

Zz.

6. Jan 8, 2013

### A.T.

What doesn't work for m comparable to M? Sure, you have to consider the acceleration of the other body, if you want to integrate over time. But that is not what the OP asks about.

The instantaneous acceleration of an body in a gravitational field is independent of its mass, regardless what the mass ratios are. That's what the OP derived, but doubts. Whether the gravitational field changes over time, because the other object moves too, is a different issue.

7. Jan 8, 2013

### ZapperZ

Staff Emeritus
I wasn't addressing the OP in that last post. I was addressing the statement that all things falling at the same rate is the defining feature of gravity. This is false in the most general case.

For what the OP is asking, the reason is shown in the derivation in the FAQ.

Zz.

8. Jan 8, 2013

### VantagePoint72

This is absurdly pedantic. Do you disagree that the equivalence principle is what distinguishes gravity from other forces? Given the manner in which the geometrization of gravity follows from it, I can't imagine how any one would disagree with that characterization. The weak equivalence principle, at least, and the notion that all bodies fall at the same rate are the same thing. The fact that matter also generates gravity doesn't change the fact that way it responds to it is unique compared to every other force. I think A.T. addressed this point well.

More to the point: the OP said, "I've done this calculation, and it leads to conclusion X which I can't believe is true." Your response was to link to a question that is, "Why is X true?" that shows it's true (at least under the right circumstances) by repeating the same sort of calculation the OP did. That is not very helpful. The OP wouldn't ask, "Why is the conclusion true?" because he doesn't think it's true! All you've done is start from the belief that it's true and repeated what he already knew. Whatever further caveats and restrictions you want to add don't change that fundamental fact. The correct response to OP's question is just: yes, actually, it is essentially true. If they then want to delve into why it's true—why gravitational and inertial mass seem to be equivalent—then that can be delved into as a follow up, as can the caveats that show it's actually just a limiting behaviour.

9. Jan 8, 2013

### A.T.

"Falling at the same rate" is indeed fuzzy. I prefer saying "accelerating the same".

10. Jan 8, 2013

### ZapperZ

Staff Emeritus
I don't understand what this is about.

Why don't you try to work this out yourself. You are on a planet of mass M. Planet A has mass M as well while planet B has mass M/2. Do you really think that from your point of view on your planet, both planets will "fall" at the IDENTICAL rate?

Try it. And don't forget that your planet is also falling towards the center of mass of the 2-planet system.

Zz.

11. Jan 8, 2013

### scientifico

If I have two bodies on the earth m1 and m2 both with m << M, is their gravitational acceleration EXACTLY the same or the heavier body has a very slightly higher acceleration ?

12. Jan 8, 2013

### A.T.

I assume you mean two scenarios with two bodies each: M & m1 and M & m2 ?

m << M doesn't matter for the instantaneous acceleration, which is the same for m1 and m2 at the same distance from M, no matter how great M is.

But the total time until collision will be shorter for the heavier m, if M is not much greater than the m's.

13. Jan 8, 2013

### BobG

What frame of reference are you using?

To an external viewer, both small objects accelerate at the same rate. The only difference is that the acceleration of the large mass is just slightly greater with the heavier small mass.

If your frame of reference is the large mass, then the acceleration of the small objects and the large object have to be added together. A viewer on the large mass will see the heavier small object accelerate at a slightly greater rate than the smaller small object.

14. Jan 8, 2013

### Staff: Mentor

I think it is the defining feature in Newtonian gravity. Given a point mass M at rest at the origin then the acceleration of a second point mass m at some distance r depends only on M and r, and not on m. It doesn't matter at all if m<<M or if m≈M or even if M<<m.

The only way it is not true is if you use a (Newtonian) non inertial frame to measure the rate of falling. I think that must be what you are doing, but I don't know why.

15. Jan 9, 2013

### Low-Q

I have learned that a hammer and a feather (in vacuum) will not have the same acceleration. Appearently it has because these objects are so small compared to the earth, that the earth mass will dominate almost totally. The hammer will hit the ground a split second before the feather. Maybe this was off topic, but might help I hope.

Vidar

16. Jan 9, 2013

### Staff: Mentor

I wouldn't put it that way. Their accelerations (measured from an inertial frame) will be the same, but the earth's acceleration will be different.
If you calculate the acceleration of the earth due to the hammer or a feather you'll find it to be a ludicrously small correction. Way way way lost in the noise and beyond the scope of this simple gravity model for the earth. See: https://www.physicsforums.com/showpost.php?p=343562&postcount=16

17. Jan 9, 2013

### scientifico

so heavier body hit the earth before the lighter one because its stronger gravitational field gets the earth a bit closer to it ?

isn't acceleration directly proportional to force a = F/m, and gravitational formula says force increase dependently on both bodies masses.... F = (G*M*m)/d^2
said that I dont understand physically why acceleration is independent from m

18. Jan 9, 2013

### BruceW

yup.

"The instantaneous acceleration is independent from its mass." Is a statement that, (as others have said), could be taken as fundamental to the physics of gravity. So really, we can simply say it is a postulate. (which means it is an assumption that a theory is based on, and so as long as our theory keeps producing good results, we don't reject our postulate).

But I am guessing you are looking for a good analogy, or intuitive explanation, which will help to drive it home? hmm... by analogy, it is similar to coulomb's law, replacing charge with mass, so then it just so happens that mass gets cancelled out of the equation. By intuition... the gravitational force gets stronger with stronger mass, but then the inertia increases as well, so gravity has more to pull on, so the two effects cancel out, so that increasing the mass doesn't increase the acceleration (or decrease it). I can't think of any other ways to explain it.

19. Jan 9, 2013

### Low-Q

The relative acceleration between those objects will increase. Put the earth surface 1 meter from the surface of a planet 100 times heavier with same size. How would you expect the relative acceleration to be now?

Vidar

20. Jan 9, 2013

### bahamagreen

"The instantaneous acceleration is independent from its mass."

Does this require treating it as a point mass?
I'm thinking that spherical body "A" is placed at rest wrt body "B"...
The center of A lies on the surface of a sphere "b" of equal radii from B...
That surface "b" curves through A and we can distinguish the part of A's volume that is inside b and outside b...

Half of A's volume is outside "b" when "b" approaches infinity,
Majority of A's volume is outside b for finite radii from B,
this proportion increases with A approaching B.

How is the "point" mass A's actual changing distribution of acceleration treated here?

21. Jan 9, 2013

### BruceW

Sorry, I should have said, I was using the particular situation in which the gravitational field is external (static), and then thinking about what a test particle would do in that gravitational field.

22. Jan 9, 2013

### BruceW

In Newtonian gravity, as long as the mass A is a rigid spherical object (with spherically symmetric mass distribution), then it can be treated as if it were a point mass. (Which is nice). So this means that mass B will feel the same gravitational field as if the mass A really were all concentrated at the centre of mass A. Also, mass A will undergo acceleration as if it were a point mass, concentrated at its centre. Of course, there are different rules if one of the objects is inside the other.

23. Jan 9, 2013

### bahamagreen

I understand the spherical mass being treated as a point mass, but what I'm asking is this:

For the spherical mass "A" and a body "B", look at the proportion of A's mass that is inside a shell defined so it's center is at B's center of mass and its radius is defined by the distance between the center of mass of B to the center of mass of A... that proportion of A's mass inside that shell changes with distance.

When the distance is very large that proportion approaches 1/2
When the distance is decreased the proportion of A's mass inside the shell decreases.
If we allow B to be a true point mass so that A's surface may approach all the way to contact with B's center of mass, then the shell radius becomes the radius of A itself.

The volume of intersection of the shell and A becomes

(pi*5r^3)/12

So the proportion of A inside the shell is

((pi*5r^3)/12)/(4/3(pi*r^3)) = 5/16

How does the inverse square relation account for this (I'm sure it does)?

In other words, how does B's tidal effect of gravitational inverse square gradient across the volume of A match the changing distribution of A's mass that falls within and outside the sphere of radius Bcom to Acom?

24. Jan 10, 2013

### BruceW

I don't see any significance. Sorry, maybe there is, but I can't see it.

25. Jan 10, 2013

### bahamagreen

The significance is that whereas the inverse square relation applies to true point masses, I'm not certain that it applies exactly to actual extended masses... the physical distribution of mass in an object stays fixed as the influence on that distribution varies with distance - I'm trying to see how the inverse square relation would match this or not.

I'm sure that this has been examined at some time in the history of gravitational development and that the answer is probably "yes, it matches perfectly", I just have not seen that demonstrated...