What if inertial mass did NOT = grav. mass?

[Farsight]

Thanks Jorrie.

I release a test particle of mass m_p a given distance away from a black hole of mass M and measure their initial closing acceleration as a.

If I now substitute the test particle with a second black whole of mass M, is the initial closing acceleration of the two black holes a or 2a or something else?

Can anybody else offer an answer to simple question?

 

[clj4]

= (you) 2 g / z^2
= (me) 2 g / (2 + eps/R)^2

where's the difference ?

Do you understand the difference between a general solution and a particular case? Apparently not.

 

[vanesch]

Thanks Jorrie.

I release a test particle of mass m_p a given distance away from a black hole of mass M and measure their initial closing acceleration as a.

If I now substitute the test particle with a second black whole of mass M, is the initial closing acceleration of the two black holes a or 2a or something else?

Can anybody else offer an answer to simple question?

Far outside of the event horizon of a Schwarzschild black hole of mass M, one cannot make the distinction between such a black hole, or any other spherically distributed, non-rotating mass. If we are far enough away to be able to apply the Newtonian approximation, then there's no difference between the Newtonian case of having two equal masses accelerating towards each other or black holes doing the same, so the answer is 2a.

There might be some nitpicking on how exactly to define the distance between two black holes, but in the Newtonian approximation, the distance between them is so much greater than their event horizon, that this doesn't matter.

Now, if we cannot use the Newtonian approximation, I don't know the answer. I'm not even sure that any closed form solution exists for two black holes.

 

[vanesch]

Do you understand the difference between a general solution and a particular case? Apparently not.

When they take on the same derivation, and the same end result ? No, I don't, indeed.

Your solution is apparently more general than mine. So there must exist cases that are treated by yours, and not mine. Hence my question, which you obstinately refuse to answer:

Can you say in what particular case your solution applies, and not mine ?

 

[clj4]

When they take on the same derivation, and the same end result ? No, I don't, indeed.

Your solution is apparently more general than mine. So there must exist cases that are treated by yours, and not mine. Hence my question, which you obstinately refuse to answer:

Do you understand the difference between a general derivation and one based on a particular case (example)? Apparently (still) not.

 

[vanesch]

Do you understand the difference between a general derivation and one based on a particular case (example)? Apparently (still) not.

And what more general case do you have in mind ?
That's the essence of my question:

Can you say in what particular case your solution applies, and not mine ?

to which you are apparently not able to provide an answer...

 

[clj4]

And what more general case do you have in mind ?
That's the essence of my question:

Looks like you can't tell the difference between a general, fully symbolic derivation and one based on a partcular case.
You also seem unable to grasp the fact that your explanation starts from physically impossible initial conditions as well.
Too bad.

 

[vanesch]

Looks like you can't tell the difference between a general, fully symbolic derivation and one based on a partcular case.
You also seem unable to grasp the fact that your explanation starts from physically impossible initial conditions as well.
Too bad.

Ok, I'll answer my own question in your place then, because you will not be able to write it down apparently.

I wanted you to trick into saying something like: "well, my technique works also for other things than spheres: I use the completely general potential integration, while you use a theorem that only works for spherical distributions". I think that this is what you think I'm not seeing: that you use the "totally general" integration over the two bodies (actually, you didn't: you only gave a link to a calculation that calculated the potential for a test particle of a sphere, but you never calculated how a sphere was going to react to such a potential - something I had to outline).

So your "completely general" technique would then of course also apply to, say, two cones, right ? Now, does your formula 2 g / z^2 work for two cones ? What's g ? And what's z ? Remember that z was a relative distance between the two centers, divided by the radius of the spheres.
And what points are now being considered ?

You see, from the moment you deviate from two identical spheres, the problem changes entirely. First of all, your normalization of z/R becomes rather strange, because which R are we going to take now ? Next, if the masses of the spheres are different, then you cannot just say "2 a" because the symmetry doesn't work anymore. And if the bodies are anything else but spheres, the relative accelerations of two points on their surface do not only depend on the linear acceleration of the centers of gravity, but also on the rotational motion of the two solid bodies, and which exert a torque on one another. It makes a difference if you are considering the tips of the two cones, or the center of the base, or any other point on its surface. You will now also have to calculate the torques, which you didn't in your "general" calculation.
Also, for a non-spherical body, as we are supposed to compare to the surface acceleration g, this g is now depending on the point on the surface one considers.

So your "general" calculation also only applies to two identical spheres, after all, if:
- you normalize the distance onto R
- you use a factor of 2 out of symmetry
- you use "g" in an indiscriminate way.
- you only use the linear acceleration of the centers of gravity and not also the rotational
degrees of freedom

 

[Garth]

Question:" What if inertial mass did NOT = grav. mass?"

Answer: We set inertial mass = gravitational mass.

If one varied with respect to the other then the value of Newton's Gravitational 'constant' would be different.

Garth

 

[clj4]

Ok, I'll answer my own question in your place then, because you will not be able to write it down apparently.

I wanted you to trick into saying something like: "well, my technique works also for other things than spheres: I use the completely general potential integration, while you use a theorem that only works for spherical distributions". I think that this is what you think I'm not seeing: that you use the "totally general" integration over the two bodies (actually, you didn't: you only gave a link to a calculation that calculated the potential for a test particle of a sphere, but you never calculated how a sphere was going to react to such a potential - something I had to outline).

So your "completely general" technique would then of course also apply to, say, two cones, right ? Now, does your formula 2 g / z^2 work for two cones ? What's g ? And what's z ? Remember that z was a relative distance between the two centers, divided by the radius of the spheres.
And what points are now being considered ?

You see, from the moment you deviate from two identical spheres, the problem changes entirely. First of all, your normalization of z/R becomes rather strange, because which R are we going to take now ? Next, if the masses of the spheres are different, then you cannot just say "2 a" because the symmetry doesn't work anymore. And if the bodies are anything else but spheres, the relative accelerations of two points on their surface do not only depend on the linear acceleration of the centers of gravity, but also on the rotational motion of the two solid bodies, and which exert a torque on one another. It makes a difference if you are considering the tips of the two cones, or the center of the base, or any other point on its surface. You will now also have to calculate the torques, which you didn't in your "general" calculation.
Also, for a non-spherical body, as we are supposed to compare to the surface acceleration g, this g is now depending on the point on the surface one considers.

So your "general" calculation also only applies to two identical spheres, after all, if:
- you normalize the distance onto R
- you use a factor of 2 out of symmetry
- you use "g" in an indiscriminate way.
- you only use the linear acceleration of the centers of gravity and not also the rotational
degrees of freedom

You can't accept the facts, do you?
Ok, since you asked for it, try using your method to solve the following situation:

-you have a hollow sphere of interior radius r and exterior radius R
-somewhere inside the first sphere there is a second ball of radius a
-find out the sphere2 acceleration in its motion towards the sphere1 center

 

[vanesch]

You can't accept the facts, do you?
Ok, since you asked for it, try using your method to solve the following situation:

-you have a hollow sphere of interior radius r and exterior radius R
-isomewhere inside the first sphere there is a second ball of radius a
-find out the sphere2 acceleration in its motion towards the sphere1 center

Answer: there's no acceleration at all.

I do that, using Gauss' theorem, and the theorem that I can replace, gravitationally, a sphere by a point.

If you don't know how to do so, I'll tell you in detail

And now you go and integrate away

 

[clj4]

Answer: there's no acceleration at all.

I do that, using Gauss' theorem, and the theorem that I can replace, gravitationally, a sphere by a point.

If you don't know how to do so, I'll tell you in detail

And now you go and integrate away

Really? The second sphere is not in the center of the first one.

 

[vanesch]

Question:" What if inertial mass did NOT = grav. mass?"

Answer: We set inertial mass = gravitational mass.

If one varied with respect to the other then the value of Newton's Gravitational 'constant' would be different.

Garth

Yes, but that would mean, that if the ratio of inertial mass / gravitational mass was different for different objects, that we would need a different Newton's gravitational constant for different objects.
So it is only when this ratio is the same for all objects that we can do this adaptation. It's what I said in post #21 in this thread.

 

[vanesch]

Really? The second sphere is not in the center of the first one.

Doesn't matter. But just integrate away and show me again where I made an approximation

 

[clj4]

Doesn't matter. But just integrate away and show me again where I made an approximation

So far , you haven't produced any calculations for the problem I gave you, just words and some silly little faces.I am asking you a second time: are you sure that what you are saying is correct? That is, sphere2 , according to you, does not move from its initial position though it is placed away from the center of sphere1? In other words, no matter where sphere2 gets placed inside sphere1 it remains in equilibrum, is this what you are saying? Can you prove this mathematically, not with prose?

Since you challenged the symbolic (non-hack) method, I challenged you back to show your calculations thru your method.
Please show them. You know very well that the potential method works under any circumstances, don't you?

BTW, the potential method uses differentiation (as in variational mechanics) , not integration. I hope that you can tell the difference.

 

[vanesch]

I am asking you a second time: are you sure that what you are saying is correct? That is, the second sphere , according to you, does not move from its initial position though it is placed away from the center of sphere1

Since you challenged the symbolic (non-hack) method, I challenged you back to show your calculations thru the hacky method.
Please show them. You know very well that the potential method works under any circumstances.

Ok, accepted

First of all, it is another theorem if you like, that inside a hollow spherical shell, the potential is constant. This can be shown easily by using Gauss' theorem: By symmetry, the field vector must be radial and only function of r. Its flux integral over a centered sphere is thus equal to its magnitude times the surface of the sphere. And (Gauss), this flux integral is equal to the total mass inside, which is zero. So its magnitude is 0. Hence, there is no field, nowhere, inside a hollow spherical body. The sphericity is needed to be able to require the symmetry of the field vector.

Now, if this is too sophisticated a reasoning, you can also bluntly work it out, using your own favorite page:
http://scienceworld.wolfram.com/physics/SphericalShellGravitationalPotential.html

Look at equation 29, and you will see that, for R < a (inside the hollow sphere), the potential is not a function of R (nor of theta or phi for that matter), so is constant.

It's another theorem to remember: inside a hollow spherical body or mass distribution, there is no field (and hence the potential is constant)

The small sphere bathing in a constant potential, it doesn't feel any force.
If you want to verify that, you can integrate over it , but I can also use my favorite theorem where I replace that smaller sphere by a point mass, which is in a constant potential and which, hence, doesn't feel any force.

Amen.

 

[vanesch]

I see you edited your post to add some more wise words :!)

So far , you haven't produced any calculations for the problem I gave you, just words and some silly little faces.

Well, opinions can differ of course, but if one can solve a problem by thinking a bit about it, instead of jumping into long calculations, then I find that much more instructive and elegant than "brute force".

So, no, to find the answer to your easy problem, I didn't need to do any calculations.

BTW, the potential method uses differentiation (as in variational mechanics) , not integration. I hope that you can tell the difference.

Well, the potential method uses differentiation, once one has calculated the potential, which is obtained by integration. I know Lagrangian mechanics, thank you. However, if - as you did - the differentiation is simply with respect to the cartesian coordinates of a point, one simply falls back on Newton's second law, so not much is gained in this game: I'd say that it is a far cry to call taking the gradient of a potential to find the force "variational mechanics". I will grant it of course that in more difficult problems, variational techniques are very useful. But let us not forget that initially, the question to solve was simply: two Earth's separated by 10 miles, what's their relative acceleration. I failed to see the point to go through a very general and elaborate approach, when a little reflection could give the answer in 3 lines.

 

[clj4]

Ok, accepted

First of all, it is another theorem if you like, that inside a hollow spherical shell, the potential is constant. This can be shown easily by using Gauss' theorem: By symmetry, the field vector must be radial and only function of r. Its flux integral over a centered sphere is thus equal to its magnitude times the surface of the sphere. And (Gauss), this flux integral is equal to the total mass inside, which is zero. So its magnitude is 0. Hence, there is no field, nowhere, inside a hollow spherical body. The sphericity is needed to be able to require the symmetry of the field vector.

Now, if this is too sophisticated a reasoning, you can also bluntly work it out, using your own favorite page:
http://scienceworld.wolfram.com/physics/SphericalShellGravitationalPotential.html

Look at equation 29, and you will see that, for R < a (inside the hollow sphere), the potential is not a function of R (nor of theta or phi for that matter), so is constant.

It's another theorem to remember: inside a hollow spherical body or mass distribution, there is no field (and hence the potential is constant)

The small sphere bathing in a constant potential, it doesn't feel any force.
If you want to verify that, you can integrate over it , but I can also use my favorite theorem where I replace that smaller sphere by a point mass, which is in a constant potential and which, hence, doesn't feel any force.

Amen.

Amazing!
So, if one drops sphere2 in a well (or mineshaft) it will not fall. It will just hover at the top of the well. I decided to test this. My 4 year old is very interested in physics so I told him that we were going to do an experiment together. I asked permission to take his favorite ball and to throw it in the well in our garden. I told him that a respected physicist has demonstrated that the ball will not fall. We ran the experiment and he was in tears, his favorite ball was gone.
I managed to fish it out with the bucket, when it came out, it was wet. This was proof that it has fallen towards the center of sphere1.

Incredible how simple experiments manage to defy the best written mathematical "proofs".

 

[vanesch]

Amazing!
So, if one drops sphere2 in a well (or mineshaft) it will not fall. It will just hover at the top of the well. I decided to test this. My 4 year old is very interested in physics so I told him that we were going to do an experiment together. I asked permission to take his favorite ball and to throw it in the well in our garden. I told him that a respected physicist has demonstrated that the ball will not fall. We ran the experiment and he was in tears, his favorite ball was gone.
I managed to fish it out with the bucket, when it came out, it was wet. This was proof that it has fallen towards the center of sphere1.

Incredible how experiments manage to defy the best written mathematical "proofs".

I think I'm going to stop discussing with you because you're obviously trolling. We were talking about a HOLLOW sphere. The Earth is not hollow, as you might know.

You can't accept the facts, do you?
Ok, since you asked for it, try using your method to solve the following situation:

-you have a hollow sphere of interior radius r and exterior radius R
-somewhere inside the first sphere there is a second ball of radius a
-find out the sphere2 acceleration in its motion towards the sphere1 center

In fact, there is an application of the two theorems together for a particle falling in a shaft drilled in a massive sphere:

If the particle is at distance a from the center of the Earth (or any other massive sphere) of radius R > a, then the spherical shell with r between R and a does not contribute to the force on the test particle, and the force felt by the particle is the same as if we removed it. What remains, is the part of the Earth within radius a: the total mass of this can be placed at the center (because it is a sphere), as if it were a point particle, and this mass M(a) gives you the acceleration of the test particle by G M(a) / a^2

Extra: if we assume the Earth of constant density (which it is not exactly), then M(a) ~ a^3 (for a < R of course), so we see that the acceleration of the test particle in the shaft goes as a^3 / a^2 ~ a.
In other words, a particle in a shaft in a sphere of uniform density undergoes a harmonic force like Hooke's law.

I'm surprised that a person of your wisdom and caliber has never done this classical problem!

 

[clj4]

I think I'm going to stop discussing with you because you're obviously trolling. We were talking about a HOLLOW sphere. The Earth is not hollow, as you might know.

No need to get personal, I challenged you to use your approach, you produced a ridiculous answer.

In fact, there is an application of the two theorems together for a particle falling in a shaft drilled in a massive sphere:

Yes, it is a high school problem, this is why I asked you to use your methodology to solve it. It is much easier to solve with the variational method. I am still seeing very little math and a lot of prose from you.

If the particle is at distance a from the center of the Earth (or any other massive sphere) of radius R > a, then the spherical shell with r between R and a does not contribute to the force on the test particle, and the force felt by the particle is the same as if we removed it. What remains, is the part of the Earth within radius a: the total mass of this can be placed at the center (because it is a sphere), as if it were a point particle, and this mass M(a) gives you the acceleration of the test particle by G M(a) / a^2

Extra: if we assume the Earth of constant density (which it is not exactly), then M(a) ~ a^3 (for a < R of course), so we see that the acceleration of the test particle in the shaft goes as a^3 / a^2 ~ a.
In other words, a particle in a shaft in a sphere of uniform density undergoes a harmonic force like Hooke's law.

Contrast all of the above with using the correct formula (29) (the second one, not the third you chose). The whole point of the discussion is that for each particular case you need to go thru gyrations in solving the problem. Applyng the variation of potential, one uses the same formalism (provided one chooses the appropriate potential).

I'm surprised that a person of your wisdom and caliber has never done this classical problem!

I did, in high school. I used your ad-hoc approach because I didn't know the theory of variations at that time.

The fact that you resort a second time to personal attacks in one post demonstrates that you ran out of scientific arguments.

 

[vanesch]

No need to get personal, I challenged you to use your approach, you produced a ridiculous answer.

I produced the correct answer: inside the hollow big sphere, the small sphere will not feel any force.

Contrast all of the above with using the correct formula (29) (the second one, not the third you chose). The whole point of the discussion is that for each particular case you need to go thru gyrations in solving the problem. Applyng the variation of potential, one uses the same formalism (provided one chooses the appropriate potential).

You mean, for each different problem, you need to think a few seconds about it, and then find the answer immediately, or, you have to sit down and calculate for 20 minutes to find the same answer ?

I think it was Feynman who said: never start a calculation before you know the answer to it

Again, you were in error. You thought that your little sphere was going to be attracted to the center of the hollow sphere or something of the kind (explaining your post #72), while it is obvious, given the property of constant potential within a spherical shell, that it is not going to feel any force.

Next, you were again in error, because you gave as a "counterexample" to the correct answer that the little sphere inside a hollow sphere didn't feel any force, the dropping of a ball in a well of a massive sphere. Why would you expect these answers to different problems to yield similar answers ? Because you didn't realize that the ball inside the hollow sphere wasn't going to undergo any force when you formulated the problem, and when I gave the answer. It is only when I explicitly showed you why, that you felt kinda stupid and needed to produce another argument.

I did, in high school. I used your ad-hoc approach because I didn't know the theory of variations at that time.

Well, I do know Lagrangian mechanics, but when I can solve a simple problem without it, by just thinking about it, that's much more elegant and simple, and insightful. As I said somewhere else: don't you use Kirchhoff's laws anymore, now that you've learned Maxwell's equations, to solve a simple circuit ?

The fact that you resort a second time to personal attacks in one post demonstrates that you ran out of scientific arguments.

No, because you're trolling, and because you're not free of personal attacks either, insinuating I am ignorant of some very elementary facts.

When confronted with a watertight argument, you change the discussion. This happened already several times now:
- first you didn't know about the sphere = point theorem, and thought I was using a silly approximation, so you attacked me on that
- next, when you realized that this was not the case, you went on with me not giving a "general solution, but only numerical examples" (while the original question was for a specific numerical case)
- when I gave you in 3 lines the general formula between two identical spheres, you attacked my "pedagogy" and hailed "a more general approach" - although up to that point you didn't produce any.
- when pressed for it, what you called "variational techniques" was a re-derivation of Newton's second law, followed by *exactly the same calculation as mine*
- when I asked you for the difference, you said that your formula was way more general than mine (although it was identical)
- when I asked you for an example in which your formula worked and not mine, for at least 4 times, you never gave an answer
- I explained you that when you depart from spherical symmetry, that problems become suddenly way more complicated, and that you didn't handle this complication either
- then you challenged me with a trivial problem of a sphere within a hollow shell
- I gave you immediately the correct answer, but you thought that this was not the right answer, so you challenged my "calculations"
- when I explained to you how, with a simple reasoning, I arrived at it, you found a "counter example" which hadn't anything to do with your original question (the silly ball of your kid experiment).
- finally, the problems to which, apparently, you ignored the correct answers (no matter your bragging about variational techniques), were classified as high school problems (which they are, indeed) as if that were a reason not to use simple techniques and a bit of insight to solve them.

 

[clj4]

I produced the correct answer: inside the hollow big sphere, the small sphere will not feel any force.
You mean, for each different problem, you need to think a few seconds about it, and then find the answer immediately, or, you have to sit down and calculate for 20 minutes to find the same answer ?

I think it was Feynman who said: never start a calculation before you know the answer to it

Again, you were in error. You thought that your little sphere was going to be attracted to the center of the hollow sphere or something of the kind (explaining your post #72), while it is obvious, given the property of constant potential within a spherical shell, that it is not going to feel any force.

Next, you were again in error, because you gave as a "counterexample" to the correct answer that the little sphere inside a hollow sphere didn't feel any force, the dropping of a ball in a well of a massive sphere. Why would you expect these answers to different problems to yield similar answers ? Because you didn't realize that the ball inside the hollow sphere wasn't going to undergo any force when you formulated the problem, and when I gave the answer. It is only when I explicitly showed you why, that you felt kinda stupid and needed to produce another argument.
Well, I do know Lagrangian mechanics, but when I can solve a simple problem without it, by just thinking about it, that's much more elegant and simple, and insightful. As I said somewhere else: don't you use Kirchhoff's laws anymore, now that you've learned Maxwell's equations, to solve a simple circuit ?
No, because you're trolling, and because you're not free of personal attacks either, insinuating I am ignorant of some very elementary facts.

When confronted with a watertight argument, you change the discussion. This happened already several times now:
- first you didn't know about the sphere = point theorem, and thought I was using a silly approximation, so you attacked me on that
- next, when you realized that this was not the case, you went on with me not giving a "general solution, but only numerical examples" (while the original question was for a specific numerical case)
- when I gave you in 3 lines the general formula between two identical spheres, you attacked my "pedagogy" and hailed "a more general approach" - although up to that point you didn't produce any.
- when pressed for it, what you called "variational techniques" was a re-derivation of Newton's second law, followed by *exactly the same calculation as mine*
- when I asked you for the difference, you said that your formula was way more general than mine (although it was identical)
- when I asked you for an example in which your formula worked and not mine, for at least 4 times, you never gave an answer
- I explained you that when you depart from spherical symmetry, that problems become suddenly way more complicated, and that you didn't handle this complication either
- then you challenged me with a trivial problem of a sphere within a hollow shell
- I gave you immediately the correct answer, but you thought that this was not the right answer, so you challenged my "calculations"
- when I explained to you how, with a simple reasoning, I arrived at it, you found a "counter example" which hadn't anything to do with your original question (the silly ball of your kid experiment).
- finally, the problems to which, apparently, you ignored the correct answers (no matter your bragging about variational techniques), were classified as high school problems (which they are, indeed) as if that were a reason not to use simple techniques and a bit of insight to solve them.

I don't have the time to waste to read all the prose that you produce every time.
So here is a straight challenge;

1. Assume the Earth to be made of
-crust (density \rho , thikness h)
and
-core (density \rho&#039; , thikness R-h where R is the Earth's radius)

2. A well of depth h&lt;d&lt;R is drilled into the Earth, starting at the Earth's surface, oriented radially

3. A ball of mass m is dropped into the well

What is the instantaneous acceleration of the ball as it falls into the well? Spare me the editorials, I don't read them. Just the math, please. Use your "pedagogy" , I will give you the variational method for contrast, hopefully this will help you understand the difference between a clever hack and a general solution. If this doesn't do it, nothing will.

 

[vanesch]

I don't have the time to waste to read all the prose that you produce every time.
So here is a straight challenge;

1. Assume the Earth to be made of
-crust (density \rho , thikness h)
and
-core (density \rho&#039; , thikness R-h where R is the Earth's radius)

2. A well of depth h&lt;d&lt;R is drilled into the Earth, starting at the Earth's surface, oriented radially

3. A ball of mass m is dropped into the well

What is the instantaneous acceleration of the ball as it falls into the well? Spare me the editorials, I don't read them. Just the math, please.

Ok, rather simple.

The mass of the core M_core = rho' (R-h)^3 4/3 pi

As long as the particle is in the well in the crust, at a distance r from the center, so r > R - h, the core contributes entirely (theorem), and of the crust, only the part between (R-h) and r contributes, which has a mass:
M_crust(r) = 4/3 pi {r^3 - (R-h)^3} rho.
Using the theorem again, this means that the acceleration of our particle, as long as it is within the crust, is given by:

a_crust(r) = G (M_crust(r) + M_core)/r^2

(you can work this further out if you want, substituting M_crust(r) and M_core)

Next, when the particle enters into the core (so r < R - h), we can forget about the crust (theorem). The mass below the particle is now:
M_core(r) = rho' r^3 4/3 pi

so the acceleration of the particle is given by:

a_core(r) = G M_core(r)/r^2 = G rho' r 4/3 pi

 

[clj4]

Ok, rather simple.

The mass of the core M_core = rho' (R-h)^3 4/3 pi

As long as the particle is in the well in the crust, at a distance r from the center, so r > R - h, the core contributes entirely (theorem), and of the crust, only the part between (R-h) and r contributes, which has a mass:
M_crust(r) = 4/3 pi {r^3 - (R-h)^3} rho.
Using the theorem again, this means that the acceleration of our particle, as long as it is within the crust, is given by:

a_crust(r) = G (M_crust(r) + M_core)/r^2

(you can work this further out if you want, substituting M_crust(r) and M_core)

Next, when the particle enters into the core (so r < R - h), we can forget about the crust (theorem). The mass below the particle is now:
M_core(r) = rho' r^3 4/3 pi

so the acceleration of the particle is given by:

a_core(r) = G M_core(r)/r^2 = G rho' r 4/3 pi

Contrast the above with using the potential formula (adjusted for the two domains of definition), followed by calculating the differential.
This was the point all along, not that your solutions are incorrect but that you need to produce a separate argument for each instance.
It is also very easy to make mistakes with your approach : notice that the units in the right hand of your formulas do not match acceleration.

 

[Farsight]

vanesch: ...so the answer is 2a.

Thanks vanesch. Clj4, do you concur with that? It does mean that heavier objects do in fact "fall" faster. Obviously we wouldn't notice the difference between say dropping a bullet and dropping a cannonball. We only start noticing a difference when the dropped mass is significant in comparison to the mass of the earth. And if we dropped them at the same time we wouldn't notice any difference at all. Interesting.

I don't think this is particularly relevant to the original quesiton, which I think was satisfactorily answered some time ago. Nor however is your ongoing debate, which IMHO is based on a misunderstanding. You're a couple of good guys, and I'd like to see both of you move on from this.

 

[clj4]

Thanks vanesch. Clj4, do you concur with that? It does mean that heavier objects do in fact "fall" faster. Obviously we wouldn't notice the difference between say dropping a bullet and dropping a cannonball. We only start noticing a difference when the dropped mass is significant in comparison to the mass of the earth. And if we dropped them at the same time we wouldn't notice any difference at all. Interesting.

I don't think this is particularly relevant to the original quesiton, which I think was satisfactorily answered some time ago. Nor however is your ongoing debate, which IMHO is based on a misunderstanding. You're a couple of good guys, and I'd like to see both of you move on from this.

The discussion is not about the results (on which we agree), it is about the approach (on which we disagree, see above). The danger of using hacks (no matter how clever) is that they may produce wrong results if the problems become sufficiently complex.

 

[vanesch]

Contrast the above with using the potential formula (adjusted for the two domains of definition), followed by calculating the differential.

Of course, because for this case, you've got the solution already printed out. If you have to do it from scratch, it is simply more work. Imagine you're in a place with no books or internet access, just a piece of paper and a pencil: then you'd have to do all the integrals all over, for the different cases, just to reproduce the potential expression in the different domains. Not too terribly difficult of course, but just remembering a few theorems gives you the same result, without all the hassle. I agree that for some more complicated problems, one has to go that way - but usually, then these problems are so difficult that in any case there's no closed-form solution.

I still claim that, for these simple problems, if you start from scratch on a piece of paper, there's much less work (and more insight to win) just thinking a bit about the specific case, than to bring out the full machinery.

Of course, I'd concur that, for a sufficiently complex problem, I have no difficulties bringing out more sophisticated tools, but I fail to see the point in doing so for simple situations. But, as I said, opinions can differ. Nevertheless, even in more complicated cases, there's a point in trying to handle specific cases by other methods: it provides a cross-check of the more general result.

This was the point all along, not that your solutions are incorrect but that you need to produce a separate argument for each instance.

Well, you did insinuate quite a few times that my solutions were wrong
But besides that, if the argument is simple I find that way more elegant and instructive than a long calculation. Opinions can differ however.

It is also very easy to make mistakes with your approach : notice that the units in the right hand of your formulas do not match acceleration.

I would like to object. It is easy not to make mistakes if you can copy the formula from an internet source or a book. But if you have to redo the integrations by hand on a piece of paper, then that's more error-prone than using the theorems, which are applicable in this case.

And I would like to know where I made a dimensional mistake
From Newton, we have F = G m M / R^2, so G M / r^2 must be an acceleration, right ?

a_crust(r) = G (M_crust(r) + M_core)/r^2
and
a_core(r) = G M_core(r)/r^2

both are of this form.

Next, a mass is a density times a length^3, well,
M_crust(r) = 4/3 pi {r^3 - (R-h)^3} rho
and
M_core(r) = rho' r^3 4/3 pi

are both of this form.

So where was I wrong ?

 

[clj4]

Well, you did insinuate quite a few times that my solutions were wrong

No, I said your approach is "hacky". Do you get the difference?

And I would like to know where I made a dimensional mistake
From Newton, we have F = G m M / R^2, so G M / r^2 must be an acceleration, right ?

Correct, I misread one of your formulas.

 

[Daverz]

This lecture video might be of interest:

http://www-conf.slac.stanford.edu/ssi/2005/lec_notes/Long/default.htm

It's somewhat technical, but I found it interesting how much experimental research is going on in this area.

 

[Jorrie]

Thanks vanesch. Clj4, do you concur with that? It does mean that heavier objects do in fact "fall" faster. Obviously we wouldn't notice the difference between say dropping a bullet and dropping a cannonball. We only start noticing a difference when the dropped mass is significant in comparison to the mass of the earth. And if we dropped them at the same time we wouldn't notice any difference at all. Interesting...

I tend to agree with the 2a of Vanesh. However, I do not agree with your statement that it shows 'heavier objects fall faster.' I believe that the Earth and the Moon falls towards the Sun with exactly the same average acceleration!

I think what the 2a "shows" is that if the Sun was more massive, the Earth and Moon would both have 'fallen faster' towards the Sun. In fact, all three would have fallen faster towards their baricentre (if this is the right word).