Gravitational Attraction: Does Smaller Mass Disappear?

[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

 

[ZapperZ]

Please see this FAQ entry:

https://www.physicsforums.com/showthread.php?t=511172

Zz.

 

[Doc Al]

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 ?

Why do you think it's not correct?

 

[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.

 

[ZapperZ]

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.

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.

 

[A.T.]

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

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.

 

[ZapperZ]

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.

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.

 

[VantagePoint72]

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.

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 anyone 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.

 

[A.T.]

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.

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

 

[ZapperZ]

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 anyone 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.

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.

 

[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 ?

 

[A.T.]

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 ?

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.

 

[BobG]

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 ?

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.

 

[Dale]

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.

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.

 

[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

 

[Doc Al]

I have learned that a hammer and a feather (in vacuum) will not have the same acceleration.

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.

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.

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

 

[scientifico]

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

The instantaneous acceleration of an body in a gravitational field is independent of its mass, regardless what the mass ratios are.

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 don't understand physically why acceleration is independent from m

 

[BruceW]

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

yup.

...said that I don't understand physically why acceleration is independent from m

"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 canceled 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.

 

[Low-Q]

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.

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

 

[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?

 

[BruceW]

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

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.

 

[BruceW]

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

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.

 

[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?

 

[BruceW]

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

 

[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...

 

[Doc Al]

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.

Are you asking whether an extended mass can be represented as a point mass (for points external to the mass)?

 

[bahamagreen]

I'm not sure what you mean by "(for points external to the mass)"...

If you mean wrt another external mass' gravitation, then yes.

The simplification of using a point mass for calculation assumes that the point mass will accelerate identically to the center of mass of an actual extended version of the body, or that this has been mathematically demonstrated.

The simplification is so that the calculations far all the distributed location of mass in an extended body can be avoided. At first glance I'm not seeing the justification for this.

For example, with changing distance between the mass sphere A and point body B, the distribution and proportion of A's mass within the radial distance to B is changing.
I'm not seeing how this change is accounted for... basically I'm asking if the inverse square relation applies exactly for point masses, but has a small error factor when employed for extended masses, especially at close proximity.

Maybe I'm not being clear because I have been using the example of a sphere. Take the example of a long rod...
In a gravity free environment we measure and mark the center of mass of the rod.
Now we put the rod in proximity to another object and place the orientation of the rod to be aligned radially to the other object.
I can see how the measured center of mass would correspond to the point mass location applying the inverse square relation if the rod was a line (a "mass line") without width.

It seems to me that as the rod's width is increased there will be a problem aligning the rod's center of mass with the location of the point mass acting gravitationally because the volume of the rod begins to present unbalanced proportions inside and outside the sphere of constant radius from the other object.
I'm not seeing that discrepancy match the difference in acceleration across the length of the rod...

Here is a specific example to see what I mean:

Let the rod have a width. Call the end of the rod nearest to other body "N".
That end N is a circle area on the end of the rod. For the distance from the other body to N, only the point in the center of the circle area at N is being influenced at the magnitude of the acceleration for that distance. The place on the rod at which the circle's area's circumference lies on the sphere about the other body occurs a little further up the rod. Call that circle's center M.
Now repeat at the far end of the rod approaching the center. Call the center of the circle at the end of the rod "n" and the center of the circumference of the circle that is a section of the sphere about the other body when the sphere surface is at "n", call that point "m".

Because the radius of curvature is different for the sphere centered about the other body for the two ends of the rod with width, |N-M| > |n-m|
There seems to be a net displacement of the gravitational point mass location wrt the static center of mass. The longer the rod, the wider the rod, the more this presents itself. I was using spheres originally because they are the "fattest" volume in all directions. Does it help see what I'm pointing to using the rod as a simpler example?

 

[BruceW]

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.

Oh, yes for sure we cannot apply the inverse square law to the centre of masses of two objects with general mass distributions. (To get the total force on the object). (Although this would give us a first-order approximation).

But luckily, for any two spherically-symmetric mass distributions, we can apply the inverse square law to their centres of masses, and we will get the exact same answer as if we had done the whole calculation exactly.

What I mean by 'exact' calculation, is that if we have some rigid body, in some external gravitational field $\vec{g}$ then
$$\int \rho \vec{g} dV$$
Gives us the explicit formula to get force on a rigid body body due to an external gravitational field acting on that rigid body. I make the distinction 'external' gravitational field because there is also the gravitational field on parts of the rigid body due to other parts (of the same body), but since gravitational fields add up linearly, and we assume that the rigid body without the external gravitational field is not accelerating, then we can ignore any 'self-gravity'.

Edit: wikipedia has a page on 'shell theorem', which seems to be related to this kind of stuff.

 

[bahamagreen]

Thanks BruceW, exactly what I was looking for!