What if inertial mass did NOT = grav. mass?

[Jorrie]

In a more constructive way, if we start with the rigorous formula of the potential from the wolfram page, we can determine the acceleration as a function of the distance between the planets for ANY value. That is for "earth1" falling towards "earth2" (and vice versa) starting from an ARBITRARY distance z and from an initial speed say, zero. The acceleration will be a variable.

I agree that we should strive for generality, but my feeling is that the approximation, if it is indeed an approximation, as used by vanesch, is a good one. If we ignore tidal gravity and the two Earths stay spherical, the surfaces of the two "Earths" will accelerate relative to their mutual centre of gravity by very close to +g/4 and -g/4 respectively. This means that the one surface accelerates towards the other surface at roughly g/2.

 

[vanesch]

I will try for one last time. In your derivation R has two different meanings.
In the first acception R is in effect half the distance between the twi "Earth's" (forget the 10km, let's just assume that the distance between the planets is very big wrt the radius of the planets such that we produce the solution for a realistic case).

In the second acceptance (the case of the probe), R is in effect the radius of one of the "Earth's". So, your solution only works if the two "Earth's" are at a distance $$2R+\epsilon$$ with $$\epsilon$$ very small. Sure, you are going to answer, you chose $$\epsilon=10km$$, which is indeed very small but the solution is far from being general. This is why I do not like it. The OP did not mention anything about the two planets being very close, we should always give general solutions, not particular ones. Does this make sense, now?

Duh

Ok, epsilon can now be big or small, doesn't matter, it is the distance between the two surfaces (previously taken to be 10 miles).

So the acceleration of earth1 (in the inertial frame) is:

g1 = G M / (2R + eps)^2 = G M / {(2 + eps/R)^2 R^2 } = g / (2 + eps/R)^2

g2 = g1.

So, *as seen from the surface of one earth* (and hence not in the inertial frame of their center of gravity), the other one is "falling down" with an acceleration of g1 + g2 = 2 g1

which equals: 2/(2 + eps/R)^2 g

For eps << R and considered neglegible (as in "eps = 10 km"), we have that this is 2/ (2)^2 g = g/2.

For eps = R for instance (so now the surface of the second Earth hovers about 6000 km above the surface of the first), this becomes:
2/(3)^2 g = 2/9 g

etc...

 

[vanesch]

I agree that we should strive for generality, but my feeling is that the approximation, if it is indeed an approximation, as used by vanesch, is a good one. If we ignore tidal gravity and the two Earths stay spherical

The point is indeed, that if the rigidity of the material making up those Earth's is good enough not to deviate from spherical form under the tidal forces, the calculation is exact.
In reality of course, there would not only be a serious deformation, probably the entire Earth would be torn apart.

 

[Jorrie]

The point is indeed, that if the rigidity of the material making up those Earth's is good enough not to deviate from spherical form under the tidal forces, the calculation is exact.

I agree with the exactness for a 'test particle' outside a homogeneous, spherical body. But are we sure about the case where the 'test particle' is also an extended body?

 

[clj4]

Duh

Ok, epsilon can now be big or small, doesn't matter, it is the distance between the two surfaces (previously taken to be 10 miles).

So the acceleration of earth1 (in the inertial frame) is:

g1 = G M / (2R + eps)^2 = G M / {(2 + eps/R)^2 R^2 } = g / (2 + eps/R)^2

g2 = g1.

So, *as seen from the surface of one earth* (and hence not in the inertial frame of their center of gravity), the other one is "falling down" with an acceleration of g1 + g2 = 2 g1

which equals: 2/(2 + eps/R)^2 g

For eps << R and considered neglegible (as in "eps = 10 km"), we have that this is 2/ (2)^2 g = g/2.

For eps = R for instance (so now the surface of the second Earth hovers about 6000 km above the surface of the first), this becomes:
2/(3)^2 g = 2/9 g

etc...

Why use proper physics when we can persist in using hacks (albeit clever ones)?

Why use the correct solution based on the rigurous potential expression, which works for any situation when we can produce a hack and insist on using it?

In your examples, how do you propose to bring two planets within 10 km (or even 6000km) from each other while KEEPING them at rest?

What I am reacting to is your pedagogical approach, I have give the rigurous approach but you insist on your clever hack. Try solving the problem as if were real , not a contrived situation.

 

[pmb_phy]

Yes, except that Pete forgot that the center of the second Earth is not at distance R, but at distance 2 R (plus 10 miles ).

I didn't forget anything. I told Farsight that a second Earth would not fall at a rate of 9.8 m/s^2 as it was described by his post. Farsight seemed to believe that the location of the surface would accelerate as if all the mass were centered at the Earth's surface and has the dimensions which can be considered a particle. That is not true. The Earth is an extended body in a tidal field and as such one has to take into effect the tidal forces. Equivalently the radii of the falling body must be neglegible with respect to the tidal field so that the tidal forces would not significantly affect the rate of fall. For these reasons one has to be careful. I think my statement

A body such as the Earth would not fall as if the entire mass was located at its center unless the field was uniform enough so that spacetime curvature could be ignored.

was confusing. I stated that a body like Earth would not fall as if the body was a point object. As this is written (i.e. I screwed up ) it appears is if I was addresing the stated question in this sentance. I was not. I was simply stating a fact which is relavent to what Farsight had in mind.

Sorry for the confusion.

Pete

 

[vanesch]

Why use the correct solution based on the rigurous potential expression, which works for any situation when we can produce a hack and insist on using it?

Indeed, why would we do something like that, when there's a theorem that helps us ?
Remember that the theorem simply states:
"for spherically symmetrical mass distributions, their gravitational effects outside of their radius are entirely equivalent to the case where the entire mass is concentrated in a mass point at their centre".

Your complaint sounds a bit like:
"why use a clever hack like Kirchhof's laws to solve a circuit, while we could, in all generality, write out a rigorous solution as a function of the EM fields ?" or:
"why using conservation of energy and momentum to solve a mechanical problem, while we could, in all generality, solve Newton's equation using all forces present ?"
Answer: because it makes life much simpler ! Because it trades computational effort for insight. That's why.

It's not a hack, it's a theorem.

In your examples, how do you propose to bring two planets within 10 km (or even 6000km) from each other while KEEPING them at rest?

Because that was the question to answer:

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 ?

What I am reacting to is your pedagogical approach, I have give the rigurous approach but you insist on your clever hack. Try solving the problem as if were real , not a contrived situation.

I must have missed your rigorous approach and solution in this thread

I did solve it as if it were real (within, of course, the limits of the problem set out) ; now if that annoys you, just consider two spherical bodies of radius 5 km and radial density profile rho(r) of your liking, at a distance epsilon from each other, and write down the relative acceleration of their surfaces as a function of the acceleration a test particle would suffer on the surface of one of them, when alone.

Now, for my approach, nothing changes. The formula I deduced, in 3 lines, are still valid. I would like to see your "more general" approach in this. And then come back and criticize again my "pedagogical approach" and my lack of generality

 

[vanesch]

The Earth is an extended body in a tidal field and as such one has to take into effect the tidal forces. Equivalently the radii of the falling body must be neglegible with respect to the tidal field so that the tidal forces would not significantly affect the rate of fall. For these reasons one has to be careful.

Does nobody then know that theorem, that you can replace a spherical body by a point mass in its center ?

http://scienceworld.wolfram.com/physics/PointMass.html

 

[vanesch]

I stated that a body like Earth would not fall as if the body was a point object.

Well, things are even more confusing.

Consider two point masses A and B with masses m and M at a distance R.
The gravitational force acting on both of them is of course G m M / R^2.

This means that A will accelerate with acceleration g1 = G M/R^2 (in an inertial frame), while B will accelerate with acceleration g2 = G m / R^2 (in an inertial frame).

This means that the relative acceleration is g_rel = g1 + g2 = G (m + M)/R^2

If A is a test particle, then m ~ 0, so the relative acceleration is G M/R^2 = g.

However, if A has the same mass as B, m = M, then g_rel is 2 G M/R^2 = 2 g.

So even if we were dealing with point masses, the relative acceleration would not be the same, as a function of the mass of the particle we let go.

However, the acceleration of A is the same wrt an inertial frame: g1 = G M / R^2, whether its mass is m ~ 0, or whether it is M or 10 M.

 

[Farsight]

I asked my question because I thought the 9.8ms² was basically telling us how much spacetime "curvature" is created by the mass of the earth. I was thinking bowling balls and rubber sheets, and wondering whether if you had two Earth masses very close together, there would be twice as much curvature in that region. And that would mean that more massive objects do in fact fall faster.

I think the two Earth's was the wrong question. How about this:

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?

 

[pmb_phy]

Does nobody then know that theorem, that you can replace a spherical body by a point mass in its center ?

http://scienceworld.wolfram.com/physics/PointMass.html

I know of no such theorem for the body which is free-fall in a g-field. That may be true for the active gravitational mass of a spherically symmetric gravitating body when a test particle is moving outside the body but it does not work when the body is the body that is in free-fall. Tidal accelerations will alter its course away from the worldline of a test particle. If you disagree with this then perhaps I can create a new web page on this since the topic arises quite often. If so then can you assist me if it proves to be a tough calculation? It'd be fun to make such a page.

Pete

 

[Rach3]

I know of no such theorem for the body which is free-fall in a g-field. That may be true for the active gravitational mass of a spherically symmetric gravitating body when a test particle is moving outside the body but it does not work when the body is the body that is in free-fall.

But that's precisely Newton's third law! The force experienced by the point mass attracted to a sphere (call it F_point) is precisely (-1) the force experienced by the sphere (F_sphere = -F_point); so if F_point is the same if we replace the sphere with a point mass, then F_sphere is also the same!

Tidal accelerations will alter its course away from the worldline of a test particle.

Tidal accelerations do not affect COM motion, they are relative to the COM.

 

[Jorrie]

I asked my question because I thought the 9.8ms² was basically telling us how much spacetime "curvature" is created by the mass of the earth. I was thinking bowling balls and rubber sheets, and wondering whether if you had two Earth masses very close together, there would be twice as much curvature in that region. And that would mean that more massive objects do in fact fall faster.

At certain points there will be close to twice as much curvature, so that a test particle will fall with twice the acceleration. But this has nothing to do with how fast a massive object falls! According to GR, all objects fall the same in the same gravitational field.

I I think the two Earth's was the wrong question. How about this:

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?.

Here you open another can of worms. Although you now have "point masses", the problem is that if two black holes are in close proximity to each other, Newton does not hold and the GR calcs are probably very tough. I know the Schwartzschild coordinate acceleration of a test particle starting from rest at radial distance r outside a black hole with mass M will be:

a = -[1-2GM/(rc^2)] GM/r^2

Does anyone know how to do this for two black holes in close proximity? I am pretty sure we can't just add the calculated accelerations like in the Newton case. Isn't there a "commandment" in relativity that says: "thou shall not add your acceleration directly to that of thy neighbor?"

 

[clj4]

Now, for my approach, nothing changes. The formula I deduced, in 3 lines, are still valid. I would like to see your "more general" approach in this. And then come back and criticize again my "pedagogical approach" and my lack of generality

Sorry, I have been away for some time. Anyway, there is no need for you to be arrogant. Here is the general proof:

From: http://scienceworld.wolfram.com/physics/SphericalShellGravitationalPotential.html

the potential is

V=MG/R , for more clarity, I will write it as V=Mg/z where z> radius of the "earth"

Applying a variational principle:

$$\delta(Mv^2/2)=\delta(V)$$

$$Mv*\delta(v)=Mg*\delta(z)/z^2$$

$$v*\delta(v)=g*\delta(z)/z^2$$

$$v*\delta(v)/\delta(t)=g*\delta(z)/\delta(t)*1/z^2$$

Since $$v=\delta(z)/\delta(t)$$ and $$a=\delta(v)/\delta(t)$$

it follows that $$a(z)=g/z^2$$

The two bodies "fall towrads each other" with the acceleration dependent on the instantaneous distance between them:

$$a'(z)=2g/z^2$$ valid for any z.

 

[vanesch]

Sorry, I have been away for some time. Anyway, there is no need for you to be arrogant.

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

Here is the general proof:

From: http://scienceworld.wolfram.com/physics/SphericalShellGravitationalPotential.html

the potential is

V=MG/R , for more clarity, I will write it as V=Mg/z where z> radius of the "earth"

Applying a variational principle:

$$\delta(Mv^2/2)=\delta(V)$$

$$Mv*\delta(v)=Mg*\delta(z)/z^2$$

$$v*\delta(v)=g*\delta(z)/z^2$$

$$v*\delta(v)/\delta(t)=g*\delta(z)/\delta(t)*1/z^2$$

Since $$v=\delta(z)/\delta(t)$$ and $$a=\delta(v)/\delta(t)$$

it follows that $$a(z)=g/z^2$$

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).

The two bodies "fall towrads each other" with the acceleration dependent on the instantaneous distance between them:

$$a'(z)=2g/z^2$$ valid for any z.

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).

 

[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