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scientifico

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I don't think this is physically correct, do you ?

thank you

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- Thread starter scientifico
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In summary, the conversation discusses the formula for gravitational attraction and its comparison with F = m*a, which leads to the conclusion that smaller masses seem to disappear. However, this is not physically correct, as explained in a FAQ entry that discusses the equivalence principle and the fact that all objects accelerate at the same rate when dropped from the same height. The conversation also touches on the issue of different masses and their effects on gravitational fields, but this is not the main topic of discussion. Overall, the conversation highlights the importance of considering all factors when analyzing the effects of gravity.

- #1

scientifico

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I don't think this is physically correct, do you ?

thank you

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Please see this FAQ entry:

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

Zz.

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

Zz.

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- #3

Doc Al

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Why do you think it's not correct?scientifico said: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 ?

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VantagePoint72

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LastOneStanding said:

I think you missed the point of the "mostly just derives" part. The derivation shows one very important

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

Zz.

- #6

A.T.

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

The

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A.T. said: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

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

Zz.

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VantagePoint72

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ZapperZ said:I wasn't addressing the OP in that last post. I was addressing the statement that all things falling at the same rate is thedefining featureof 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

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

- #9

A.T.

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"Falling at the same rate" is indeed fuzzy. I prefer saying "accelerating the same".ZapperZ said:I was addressing the statement that all things falling at the same rate is thedefining featureof gravity. This is false in the most general case.

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LastOneStanding said: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 matteralsogenerates gravity doesn't change the fact that way itrespondsto 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 intowhyit'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.

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scientifico

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A.T.

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

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.

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BobG

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scientifico said:

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.

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Dale

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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.ZapperZ said:I wasn't addressing the OP in that last post. I was addressing the statement that all things falling at the same rate is thedefining featureof gravity. This is false in the most general case.

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.

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Low-Q

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Vidar

- #16

Doc Al

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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.Low-Q said:I have learned that a hammer and a feather (in vacuum) will not have the same acceleration.

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=16Appearently 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.

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scientifico

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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^2The instantaneous acceleration of an body in a gravitational field is independent of its mass, regardless what the mass ratios are.

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

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BruceW

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yup.scientifico said: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 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).scientifico said:...said that I don't understand physically why acceleration is independent from m

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.

- #19

Low-Q

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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?BruceW said: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.

Vidar

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bahamagreen

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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

BruceW

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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.Low-Q said: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

- #22

BruceW

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

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bahamagreen

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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?

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BruceW

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

- #25

bahamagreen

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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...

- #26

Doc Al

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

- #27

bahamagreen

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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?

- #28

BruceW

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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).bahamagreen said: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.

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 [itex]\vec{g}[/itex] then

[tex]\int \rho \vec{g} dV [/tex]

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.

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bahamagreen

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Thanks BruceW, exactly what I was looking for!

Gravitational attraction is a natural phenomenon in which objects with mass are drawn towards each other. It is one of the four fundamental forces of nature and is responsible for the motion of planets, stars, and galaxies.

Gravitational attraction is caused by the curvature of space-time around objects with mass. The more massive an object is, the more it curves space-time, causing other objects to be pulled towards it.

No, smaller mass does not disappear due to gravitational attraction. The force of gravity is determined by the mass and distance between two objects. Smaller objects may be pulled towards larger objects, but they do not disappear.

Gravitational attraction cannot be shielded or canceled. It is a fundamental force of nature that cannot be blocked or eliminated. However, its effects can be counteracted by other forces, such as centrifugal force or electromagnetism.

The mass of an object directly affects the strength of its gravitational attraction. The more massive an object is, the stronger its gravitational pull will be. This is why larger objects, such as planets and stars, have a stronger gravitational pull than smaller objects like rocks or humans.

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