What if inertial mass did NOT = grav. mass?

[clj4]

I honestly don't think I was. You didn't stop spouting critique on my approach, first claiming it was an approximation (which it wasn't), then saying that I used a fancy theorem (replacing a sphere by a point particle) while an explicit calculation would do fine... just to find you doing the same thing
Yes, you now derived the force of gravity from its potential:
F = m a = m M G/ r^2 or a = M G / r^2.

If you now define z = r/R then we can rewrite this as:

a = M G / (z^2 R^2), and as g = M G / R^2, we can write this as:

a = g / z^2.

However, your z is clearly r / R (and not some absolute distance - this is not directly clear but it is the way to make sense).

From the symmetry, you deduce then that the relative acceleration is twice this: a = 2 g / z^2, where z is the distance between the two centers of gravity of the spheres divided by the radius of them.
It should be noted that by using the potential M G / R, you are implicitly using the same theorem as you attacked me for earlier, and you reduce just as well as me an entirely spherical body in near-field to a point potential. In other words, you do exactly the same as I do, except that you add a few lines to deduce Newton's force of gravity from the potential pf a point particle by looking at a differential change in kinetic and potential energy (in other words, you do delta (KE) = - delta(PE)
and from that, you find again that m.a = F).
Yes, that's exactly the result I also had, when you say that
z = (2 R + eps) / R = 2 + eps/R

(look at the post with my calculation).

While INITIALLY I criticised you for using the point theorem, I followed up with a criticism (posts 34,35) due to your using of a hack approach to the solution obtaining results by INIVIDUAL EXAMPLES rather than deriving a general formalism. Your proof is by example, it uses apprximations and it is not physically realizable (see again criticism at post 35 and try to understand it this time).

I'll let other people judge the two solutions side by side, this is an elementary problem that doesn't merit that many posts. If you want to continue solving problems by particular examples, feel free to do so but try to remember that physics does not produce such shoddy proofs.

 

[vanesch]

While INITIALLY I criticised you for using the point theorem, I followed up with a criticism (posts 34,35) due to your using of a hack approach to the solution obtaining results by INIVIDUAL EXAMPLES rather than deriving a general formalism. Your proof is by example, it uses apprximations and it is not physically realizable (see again criticism at post 35 and try to understand it this time).

If your criticism is that I solved the problem posed initially (acceleration of a second Earth at 10 miles above the Earth's surface), without going to a more general solution with arbitrary distance in which the "approximation" was to consider 10 miles small as compared to ~6000 km, then I think that's not justified. It was clear from the OP's question that he wanted the surfaces to be "close" and 10 miles, or 5 miles, or 20 miles, wouldn't make a difference.

I thought, however, given your post #27 and #29, that you were objecting to me using the "approximation" of a point particle, while it was a whole "sphere", and we should actually integrate over the sphere for each matter element in each sphere. I concluded that from your "The approximation applies when the distance between the centers of the spheres is MUCH BIGGER than the spheres' radiuses.", thinking that I could only assimilate an entire sphere to a point when its dimensions (radius) are much smaller than the distance between the points. As I pointed out, however, there's a theorem stating that this is not an approximation, but an exact result.

In your post #34, you start to change opinion. You're still a bit confused, and think that one should or, have two Earth's very far away (to use the point particle "approximation"), or have a small test particle and an earth. But the OP was: two equal, big spheres which nearly touch each other, and was not covered by either of your propositions. You're then nitpicking that I neglected the 10 km, and that I didn't write out a result in all generality, for small and large distances between equal, big, spheres.

In your post #35, you ask for the general solution for two equal spheres, at arbitrary distance. Now, given the gist of the calculation I presented, this requires of course only a very small change, and I give this to you in post #37. I give you the entirely general solution, followed by a few numerical applications:

The relative acceleration is, in all generality:
2/(2 + eps/R)^2 g

R is the radius of the Earth (or one of the spheres), eps is the distance between the two surfaces (big or small), g is the surface acceleration in the case of one sphere (9.81 m/s^2 in the case of the earth).

The solution is entirely general, and accurate, in the sense that no approximations are made if the two bodies are spherical.

In your post #40, you now realize that my solution is exact, that I didn't need to integrate over the spheres after all (thanks to a theorem you call a "clever hack"). But you still insist on the pedagogical unsoundness of such an approach, and then complain about the reality of the problem given (can't help that, it was the original question !).

And then you come up with your "more pedagogical" approach in post #49. All the criticism of me using the point-particle approach are now not applied to yourself anymore, as you write the potential between two spheres simply as V = M G / R, as if it were point particles (haha, in my case it was a "clever hack which was pedagogically unsound").
Next you confuse a bit the absolute distance between the centers with their ratio wrt R (z has the two functions).
And then you introduce a variational principle, where you start (committing 2 sign errors in a row so that it doesn't matter) to demonstrate that, from the variation of PE + KE = constant, we can derive a = dV/dz, which is nothing else but Newton's second law which I used directly.
Once you have obtained Newton's second law, and used the potential for point particles between spheres (hence using the very theorem you first didn't believe, and then called a hack), you simply write down the result
inertial frame acceleration of two point particles with mass M, at a distance z.R (because z is relative now), is given by g / z^2, so their relative acceleration is the double : 2 g / z^2.

In other words, except for some confusing re-derivation of Newton's second law, you use exactly the same construction as I did, and which you criticised first as being an approximation, next as being a clever hack, and finally as not general enough and pedagogically unsound.

I'll let other people judge the two solutions side by side, this is an elementary problem that doesn't merit that many posts. If you want to continue solving problems by particular examples, feel free to do so but try to remember that physics does not produce such shoddy proofs.

There was a specific problem to solve: two Earth's, with their surfaces at 10 miles distance. Of course this problem is unrealistic as tidal forces would rip them apart. But one can consider two spherical bodies in close proximity and ask for the relationship between their individual surface accelation, and their relative acceleration in the given context.

The entire trick was to see that one could use the theorem replacing the big spheres by point particles. From that point on, the problem became very simple. It was the essence of its solution.
The numerically interesting result is simply that two equal spheres in almost contact accelerate relatively with g/2, where g is their individual surface acceleration.
All the rest is superficial, and trivially understood, once one has worked out this problem. Apart from all your criticising, I have to say I don't see any pedagogical improvement in your approach, nor any larger generality.

 

[clj4]

If your criticism is that I solved the problem posed initially (acceleration of a second Earth at 10 miles above the Earth's surface), without going to a more general solution with arbitrary distance in which the "approximation" was to consider 10 miles small as compared to ~6000 km, then I think that's not justified. It was clear from the OP's question that he wanted the surfaces to be "close" and 10 miles, or 5 miles, or 20 miles, wouldn't make a difference.

I thought, however, given your post #27 and #29, that you were objecting to me using the "approximation" of a point particle, while it was a whole "sphere", and we should actually integrate over the sphere for each matter element in each sphere. I concluded that from your "The approximation applies when the distance between the centers of the spheres is MUCH BIGGER than the spheres' radiuses.", thinking that I could only assimilate an entire sphere to a point when its dimensions (radius) are much smaller than the distance between the points. As I pointed out, however, there's a theorem stating that this is not an approximation, but an exact result.

In your post #34, you start to change opinion. You're still a bit confused, and think that one should or, have two Earth's very far away (to use the point particle "approximation"), or have a small test particle and an earth. But the OP was: two equal, big spheres which nearly touch each other, and was not covered by either of your propositions. You're then nitpicking that I neglected the 10 km, and that I didn't write out a result in all generality, for small and large distances between equal, big, spheres.

In your post #35, you ask for the general solution for two equal spheres, at arbitrary distance. Now, given the gist of the calculation I presented, this requires of course only a very small change, and I give this to you in post #37. I give you the entirely general solution, followed by a few numerical applications:

The relative acceleration is, in all generality:
2/(2 + eps/R)^2 g

R is the radius of the Earth (or one of the spheres), eps is the distance between the two surfaces (big or small), g is the surface acceleration in the case of one sphere (9.81 m/s^2 in the case of the earth).

The solution is entirely general, and accurate, in the sense that no approximations are made if the two bodies are spherical.

In your post #40, you now realize that my solution is exact, that I didn't need to integrate over the spheres after all (thanks to a theorem you call a "clever hack"). But you still insist on the pedagogical unsoundness of such an approach, and then complain about the reality of the problem given (can't help that, it was the original question !).

And then you come up with your "more pedagogical" approach in post #49. All the criticism of me using the point-particle approach are now not applied to yourself anymore, as you write the potential between two spheres simply as V = M G / R, as if it were point particles (haha, in my case it was a "clever hack which was pedagogically unsound").
Next you confuse a bit the absolute distance between the centers with their ratio wrt R (z has the two functions).
And then you introduce a variational principle, where you start (committing 2 sign errors in a row so that it doesn't matter) to demonstrate that, from the variation of PE + KE = constant, we can derive a = dV/dz, which is nothing else but Newton's second law which I used directly.
Once you have obtained Newton's second law, and used the potential for point particles between spheres (hence using the very theorem you first didn't believe, and then called a hack), you simply write down the result
inertial frame acceleration of two point particles with mass M, at a distance z.R (because z is relative now), is given by g / z^2, so their relative acceleration is the double : 2 g / z^2.

In other words, except for some confusing re-derivation of Newton's second law, you use exactly the same construction as I did, and which you criticised first as being an approximation, next as being a clever hack, and finally as not general enough and pedagogically unsound.



There was a specific problem to solve: two Earth's, with their surfaces at 10 miles distance. Of course this problem is unrealistic as tidal forces would rip them apart. But one can consider two spherical bodies in close proximity and ask for the relationship between their individual surface accelation, and their relative acceleration in the given context.

The entire trick was to see that one could use the theorem replacing the big spheres by point particles. From that point on, the problem became very simple. It was the essence of its solution.
The numerically interesting result is simply that two equal spheres in almost contact accelerate relatively with g/2, where g is their individual surface acceleration.
All the rest is superficial, and trivially understood, once one has worked out this problem. Apart from all your criticising, I have to say I don't see any pedagogical improvement in your approach, nor any larger generality.

You talk a lot and argue even more. I will let the others look at the two solutions side by side. The potential is calculated thru volume integration, capisci?

 

[vanesch]

You talk a lot and argue even more. I will let the others look at the two solutions side by side. The potential is calculated thru volume integration, capisci?

Yes, that's the theorem I used. It's proof is the volume integration and the observation that outside of the matter distribution (r>b), the unit mass potential goes as -G M/r, as if the entire mass were concentrated in the center of the sphere. It is essentially expression (18) of the reference you gave. Once you know that, you just have to remember that spherical mass distributions can be replaced by their masses in the center.
But what we now have, is the potential of a point in the gravitational field of a sphere, and that this is the same potential as a point, in the gravitational field of another point.
Nevertheless, you now have to show, through an identical calculation btw, that a sphere, exposed to such a potential, undergoes a total force as if all its mass were also concentrated at its center in this potential (this was the thing I was initially slightly unsure about). So you also stopped half-way.

It is however rather easy to see this: once we established that sphere 1 (the "active" one) can be replaced by a point mass M1, the potential of each mass element of sphere 2 (the passive one) equals - G M1 rho/l dr^3 which needs to be integrated over sphere 2. It can now be recognized that the calculation is identical, and so the result will be that the potential energy between the two spheres equals - G M1 M2 / R ; in other words, they behave the same as if as well the active as the passive sphere are replaced by point masses at their center with mass equal the entire mass of the sphere.

And now the theorem is completely proven.
Well, almost. There's one more thing: that is: what if other sources of gravity are present ? But given that all sources of gravity can be reduced to a superposition of point masses, and that the above property has been established for each point mass contribution, we now have the general result: in a gravitational situation, a spherical mass distribution can always be replaced by a point mass at its center as long as other masses stay outside of its radius.

To come back to our dispute, however: when you ask people to put the two approaches next to each other, I ask:

What difference is one going to find if one "puts the two solutions side by side", given that they are identical ?
Given that we now both use the force between two point particles, at a distance which you call d = z R, and I call d = 2 R + eps, and that we both plug this in the force law:

M a = M^2 G / d^2 in an inertial frame,
so that a = M G / d^2, and hence the relative acceleration is 2 a, or:

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

where's the difference ?

 

[clj4]

Yes, that's the theorem I used.

No, you used a point equivalence theorem (http://scienceworld.wolfram.com/physics/PointMass.html) , I used the potential calculation (http://scienceworld.wolfram.com/physics/SphericalShellGravitationalPotential.html) followed by an application of variational mechanics.
You used some elementary calculation based on gravitational force, I used variational mechanics.
You used an example (with no mention of initial conditions and a very unrealistic assumption of holding planets 10km apart, how?), I used a pure formalism that works under any circumstances.
You derived the answer for a particular case, my solution explains the general case.

What difference is one going to find if one "puts the two solutions side by side", given that they are identical ?
Given that we now both use the force between two point particles, at a distance which you call d = z R, and I call d = 2 R + eps, and that we both plug this in the force law:

M a = M^2 G / d^2 in an inertial frame,
so that a = M G / d^2, and hence the relative acceleration is 2 a, or:

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

where's the difference ?

See above, the first paragraph of my answer. If you still don't get it, there is nothing more that I can do for you.

 

[vanesch]

No, you used a point equivalence theorem (http://scienceworld.wolfram.com/physics/PointMass.html) , I used the potential calculation (http://scienceworld.wolfram.com/physics/SphericalShellGravitationalPotential.html)

As I pointed out, the "potential calculation" is half of the proof of the theorem (and you forgot to point out the other half - the passive part). So you referred to half of a proof, I used the theorem.

followed by an application of variational mechanics.
You used some elementary calculation based on gravitational force, I used variational mechanics.

Your "application of variational mechanics" was the re-derivation of Newton's second law from energy conservation:

You did:

d KE + d PE = 0

Given that KE = 1/2 M v^2, d KE = M v d v
and d PE = - F dr

dividing by dt, we have: M v dv/dt - F dr/dt = 0 and given that dr/dt = v, we have:

M dv/dt = F, Newton's second law.

Usually, this is read in the other way, to arrive at conservation of energy

You usually do:

M dv / dt = F

so

M dv = F dt

Multiplying both sides by v = dr/dt we have:

M v dv = F dr

Now, in the specific case that F can be derived from a potential, U, so that
F = - dU/dr, we have:

M v dv = - dU/dr dr = - dU

which is a complete differential which we can integrate:

1/2 M v^2 = - U + constant, or:

1/2 M v^2 + U = constant

A constant which we call total energy, and the first term, which we call "kinetic energy".

You did the derivation back again to restore Newton's second law, that's all. But at no point you used any "variational calculus" to solve the problem at hand.

You used an example (with no mention of initial conditions and a very unrealistic assumption of holding planets 10km apart, how?), I used a pure formalism that works under any circumstances.

This is funny. The initial conditions do not change the relative acceleration (at least when they reach identical distances) !

Your "formalism" works in exactly the same circumstances as mine, given that its derivation and result are identical.

You derived the answer for a particular case, my solution explains the general case.

Can you say in what particular case your solution applies, and not mine ? Mine applies with two identical spheres of diameter R, at a distance (between the surfaces) eps.
Your formula applies with two spherical, identical masses M, when their centers are at a relative distance to their radius, z.

See above, the first paragraph of my answer. If you still don't get it, there is nothing more that I can do for you.

 

[clj4]

As I pointed out, the "potential calculation" is half of the proof of the theorem (and you forgot to point out the other half - the passive part). So you referred to half of a proof, I used the theorem.
Your "application of variational mechanics" was the re-derivation of Newton's second law from energy conservation:

You did:

d KE + d PE = 0

Given that KE = 1/2 M v^2, d KE = M v d v
and d PE = - F dr

dividing by dt, we have: M v dv/dt - F dr/dt = 0 and given that dr/dt = v, we have:

M dv/dt = F, Newton's second law.

Usually, this is read in the other way, to arrive at conservation of energy

You usually do:

M dv / dt = F

so

M dv = F dt

Multiplying both sides by v = dr/dt we have:

M v dv = F dr

Now, in the specific case that F can be derived from a potential, U, so that
F = - dU/dr, we have:

M v dv = - dU/dr dr = - dU

which is a complete differential which we can integrate:

1/2 M v^2 = - U + constant, or:

1/2 M v^2 + U = constant

A constant which we call total energy, and the first term, which we call "kinetic energy".

You did the derivation back again to restore Newton's second law, that's all. But at no point you used any "variational calculus" to solve the problem at hand.This is funny. The initial conditions do not change the relative acceleration (at least when they reach identical distances) !

Your "formalism" works in exactly the same circumstances as mine, given that its derivation and result are identical.
Can you say in what particular case your solution applies, and not mine ? Mine applies with two identical spheres of diameter R, at a distance (between the surfaces) eps.
Your formula applies with two spherical, identical masses M, when their centers are at a relative distance to their radius, z.

I know exactly what I did, I don't need it re-explained to me , especially in a long drawn style. Now, try to understand what I wrote for you, it is pretty elementary: there is a difference between formal derivation and derivation by example (what I call "hack")

 

[vanesch]

I know exactly what I did, I don't need it re-explained to me , especially in a long drawn style.

So that means that you agree with my explanation (and hence understanding) of what you did. In that case:

Now, try to understand what I wrote for you, it is pretty elementary.

... you will need to re-explain then, because I didn't get it

1) you re-derived (using variational calculus) Newton's second law from energy conservation which was initially derived from Newton's second law.

2) you used the formula for potential energy between 2 point masses and from that derived the force and hence acceleration of 1 point mass as the effect of another

3) From that you derived the relative acceleration by multiplying by 2, as a function of the ratio of the distance between the centers and their radius

4) you provided a link to a calculation which shows that the potential of a spherical body gives us the same potential as a point mass outside of its radius.

What I did was:

1) I referred to a theorem (which proof is twice the application of the calculation you provided a link to) which tells us that spherical bodies can be considered point masses concerning their gravitational interaction.

2) I used Newton's second law to get the force, and hence the acceleration of 1 sphere wrt the other

3) I multiplied by 2 to get the relative acceleration as a function of the distance between the surfaces and the radius


I fail to see the essential difference, which makes my approach "approximate", a "hack", a "shoddy proof", "by example", "unrealistic", and "pedagogically unsound" while yours is "general", "uses variational calculus", "is rigorous"...

Especially when the "variational calculus" is not applied to the problem at hand, but just re-derives Newton's second law, when your "rigorous potential calculation" is only worked out half (only the active potential is done, I had to fill in myself the passive part), and when the final expressions are identical.

Where is the difference, tell me, which allows for all this critique from your side, when the only visible difference is a different way of writing down the distance between the two spheres!

I think that the only difference is that you didn't know about the theorem, that you initially thought that I was making a stupid approximation (using point masses in near field where such an approximation is of course not sensible - or in far field when I confuse of course R as radius and 2 R as distance) instead of an exact calculation, that you thought that working this out by integrating over the two spheres (which you still forgot in your potential approach - that's the second half I talked about) would yield a different result, and once you realized that it was, in fact, correct, that you didn't know how to say so, and hence went nitpicking on a lot of irrelevant details, like neglecting 10 km, not writing it out for a general distance, using elementary forces and not a variational technique, using a theorem and not a derivation, ...

But at the end of the day, you do exactly the same as I did.

BTW, you didn't answer my question:

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

 

[clj4]

So that means that you agree with my explanation

Obviously not, you seem to have some serious comprehension problem.
How do you "hold" the two planets 10km from each other at the initial state of your solution?

 

[vanesch]

Obviously not, you seem to have some serious comprehension problem.
How do you "hold" the two planets 10km from each other at the initial state of your solution?

1) that was given in the problem, right ? I was answering the question in post #6:

Now that's got me thinking. Does anybody know the answer to this:

If we held two earth-sized planets ten miles apart and then let go, would their surfaces accelerate towards one another at 9.8ms-2 ?

2) you don't even have to "hold" them, even if they are already falling, or even orbiting, the result is the same (when the distance between their surfaces is 10 miles)

3) you were asking for generality, so consider now balls of 1 meter diameter, 1 cm apart. Now the problem is realistic.

But we're now back to my post #42 where I already said all that.

Face it, in the beginning of this thread you simply thought that replacing a sphere by a point was an approximation which only held when we were at distances much greater that the sphere's radius, which triggered your initial criticism. I too wasn't 100% sure that I could use it for the passive side - as I put as a caveat in my original answer in post #22. Turns out that it works, as rach3 argumented cleverly (using Newton's third), and as I argumented explicitly in post #54 how we could use the calculation of the external potential again to use now the total potential of a sphere bathing in a 1/r potential - something you omitted, but which is essential if you do not want to use the said theorem.
This calculation is then the basis for the proof of the theorem if one realizes that all gravitational potentials are superpositions of 1/r potentials.

From there on you started picking on this approach, without, in fact, proposing anything else.

You again didn't answer my question:

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

 

[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).

 

[Farsight]

Noted Jorrie. It's an interesting one this. That 2a means the falling is faster, but dropping two objects also says it isn't. Hmmn.

I wonder if the Earth would have a different orbit if the moon wasn't there?

 

[Jorrie]

Noted Jorrie. It's an interesting one this. That 2a means the falling is faster, but dropping two objects also says it isn't. Hmmn.

I wonder if the Earth would have a different orbit if the moon wasn't there?

I think there would have been a minor difference - instead of the Earth cycling around the Earth-Moon baricentre every 28 days or so, the Earth would have had a more standard elliptical orbit around the Sun.

IMO, the crucial point is that the mass of the orbiting object has negligible effect on the orbit, as long as its mass is small compared to the primary mass (the Sun in this case). So the Earth-Moon baricentre follows an orbit that is roughly independent of the mass of the Earth-Moon system.

 

[Farsight]

Thanks Jorrie.