What if inertial mass did NOT = grav. mass?

[smithpa9]

What if inertial mass did NOT equal gravitational mass?

How would our normal daily existence be different?

Most interesting and creative answer wins !

[Edit: This is intended to be a fun question to answer, but I'm hoping to learn something quite serious from it. The force of attraction between two charged particles is a function of their charge and distance apart. The resistance to that force is the inertial mass of the particles, which has little to do with the charge. The stronger the charges (assuming oppositely charged) and the shorter the distance, the greater the acceleration. The greater the inertial mass, the less the acceleration. Similarly with magnetic bodies. The acceleration due to magnetic attraction is resisted by the inertial mass, having little to do with magntism.

Not so for accleration due to gravity, where the attraction and resistance are both due to mass (given the equivalence of gravitational and inertial mass).

If they weren't equivalent, how would the simple act of walking, or throwing a baseball or driving a car, or flying a rocket to the moon, be different?]

 

[clj4]

What if inertial mass did NOT equal gravitational mass?

How would our normal daily existence be different?

Most interesting and creative answer wins !

Eotvos would be coming to haunt you (being Transylvanian, he was related to Dracula)

 

[Farsight]

Let's put aside charge for a moment, and relativistic speeds. Mass is a measure of resistance to change in motion.

When we talk about inertial mass, we're usually looking at a stationary mass and thinking about applying a force to get it moving. So we give it a push, and we measure the force applied and the time we applied it for, whereupon we see our object accelerating by a measured amount. We then calculate the inertial mass as a relation between the force, time and acceleration.

When we talk about gravitational mass, the force acting on our object is gravity instead of pushing. It's a different type of force, but it's the same object. How can that object exhibit a different resistance to change in motion to one flavour of force compared with another? It can't. Force is force.

OK, the magnitude of the force might vary with the distance from another mass, then catching a ball might come with a late rush and a click, like you were catching a magnet with a magnet. And then it would be real hard to throw the ball away. But that's a whole different kettle of fish from what you were asking. And I can't see how what you were asking makes sense. Even on the moon.

 

[pmb_phy]

Gravitational mass and inertial mass need not be equal, just proportional. If they are not proportional then the rate at which an object falls will depend on the inertial and gavitational mass. The equivalence principle woulf then no longer hold to be true.

Pete

 

[actionintegral]

Heavier objects would fall faster than lighter objects.

 

[Farsight]

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 ?

 

[neutrino]

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 ?

10 miles is barely any distance. But assuming an ideal situation, yes, they will have the same accelaration. Gm/r^2 will be same in both cases. It's symmetrical...you can't expect one to move faster than the other.

 

[Farsight]

I was wondering if they'd accelerate towards one another at a combined closing of rate of 19.6ms^-2.

 

[CarlB]

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 ?

The gravitational forces cancel, so of course there would be no acceleration towards each other or in any other direction, at least for the surfaces that are 10 miles apart.

By the way, holding the planets apart might be kind of tough on your wrists.

Carl

 

[smithpa9]

Farsight - I think Neutrino is right. Unless they were in mutual orbit, the gravitational acceleration between two Earth-sized planets positioned only 10 miles apart should be approx. 9.8 m/s^2 in total. Both would move towards the center of mass of the two (halfway in between) at about 4.9 m/s^2.

Reasoning: 9.8 m/s^2 is roughly the gravitation acceleration at the surface of the Earth for ANY OBJECT, regardless of its mass. 10 miles, compared to the 4,000+ miles radius of the Earth is not significantly different, so 9.8 is probably still correct to one decimal.

An apple, or an Elephant, or the Moon, or another planet Earth, all if positioned just 10 miles above the Earth, would fall at the same speed (ignoring air resistance). And they all would fall toward their common center of gravity. But with the apple and the Elephant, that center is essentially the center of the Earth since it's so much more massive. And the apple and elephant do most of the accelerating and the Earth barely budges.

But in your example of two Earths, they would both share the acceleration equally because of their identical masses.

 

[smithpa9]

pym_phy and actionintegral - very insightful responses !

So, if A) gravitational mass was greater than inertial mass, then Aristotle would have been right all along. . . heavier objects would fall faster than lighter ones.

But if B) inertial mass was greater than gravitational mass, then LIGHER objects would fall faster than HEAVIER objects ! Wouldn't that be weird !

Objects falling from great heights might break apart on their fall as the heavier/denser parts separated from the lighter/less dense parts because of the differential in gravitational forces and inertial forces. ??

Would carrying a 15 lb. bowling ball in one hand and an 8lb. one in the other would feel quite strange, as perhaps the 8lb. ball would be easier to keep from dropping, but would offer more resistance to the change of direction the forward and back motion of your arms as you walked?

Sounds ripe for an Isaac Asimov novel !

 

[clj4]

pym_phy and actionintegral - very insightful responses !

So, if A) gravitational mass was greater than inertial mass, then Aristotle would have been right all along. . . heavier objects would fall faster than lighter ones.

But if B) inertial mass was greater than gravitational mass, then LIGHER objects would fall faster than HEAVIER objects ! Wouldn't that be weird !

Objects falling from great heights might break apart on their fall as the heavier/denser parts separated from the lighter/less dense parts because of the differential in gravitational forces and inertial forces. ??

Would carrying a 15 lb. bowling ball in one hand and an 8lb. one in the other would feel quite strange, as perhaps the 8lb. ball would be easier to keep from dropping, but would offer more resistance to the change of direction the forward and back motion of your arms as you walked?

Sounds ripe for an Isaac Asimov novel !

Eotvos has demonstrated long ago the equivalence between the "two" masses. Reenactments of the Eotvos experiment have brought the error bars of the experiment towards 10^-20 (see the Eot-Wash experiment). So, if you continue to post, Eotvos' vampire will come hunting and haunting

And he'll get you!

 

[smithpa9]

Clj4. . . I think you may have misinterpreted the intent behind my question. :-) I have no quarrels with Eotvos, and certainly am not trying to deny that inertial and gravitational mass are equal. Of course they are equal.

I just thought it was in interesting question to contemplate what it would be like if we had been born into a universe where they were somehow not equal.

Plus, sometimes it helps me understand things better if I can contemplate what it would be like if the laws of nature were reversed somehow.

But I'll offer my offical appologies to Eotvos if in case I've offended him (or you).

:-)

 

[clj4]

Clj4. . . I think you may have misinterpreted the intent behind my question. :-) I have no quarrels with Eotvos, and certainly am not trying to deny that inertial and gravitational mass are equal. Of course they are equal.

I just thought it was in interesting question to contemplate what it would be like if we had been born into a universe where they were somehow not equal.

Plus, sometimes it helps me understand things better if I can contemplate what it would be like if the laws of nature were reversed somehow.

But I'll offer my offical appologies to Eotvos if in case I've offended him (or you).

:-)

Yes, I understood. I thought it was a contest for the funniest answers.

 

[smithpa9]

Oh yeah. . . . in that case, you clearly win !

:-)

 

[pmb_phy]

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 ?

No. The 9.8ms^-2 refers to test particles falling in the gravitational field of Earth assuming that the motion of the Earth can be neglected. Thus a coffee cup would fall at 9.8ms^-2 but Jupiter wouldn't. In fact Jupiter's mass is so much larger than Earth's that Earth would fall in Jupiter's gravitational field as if Jupiter was at rest. Then the Earth would fall at a different acceleration. The rate a body falls depends on where the body is. 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.

Pete

 

[Farsight]

pmb, or anybody: Supposing I dropped a test particle into a black hole and timed its initial fall, then calculated its acceleration at 9.8ms^-2. Now if I repeat this test replacing the test particle with a black hole identical to the other one, would the two black holes accelerate towards one another at 9.8ms^-2?

 

[pmb_phy]

pmb, or anybody: Supposing I dropped a test particle into a black hole and timed its initial fall, then calculated its acceleration at 9.8ms^-2. Now if I repeat this test replacing the test particle with a black hole identical to the other one, would the two black holes accelerate towards one another at 9.8ms^-2?

No.

Pete

 

[turbo]

There is a project under way at CERN to produce and study neutral anti-hydrogen. A key experiment in this project is the comparison of the gravitational infall rates of neutral antihydrogen and hydrogen. This is a critical test of the weak equivalence principal. Not a funny answer, but the most relevant I could come up with.

http://alpha.web.cern.ch/alpha/overview.html

 

[exponent137]

If inertial mass is not equal gravitational mass, general relativity changes in general objectivity. Therefore uniformly accelerated elevator is not equal to homogeneous gravitational field.

 

[vanesch]

Gravitational mass and inertial mass need not be equal, just proportional.

Indeed. In fact, we would make them equal, by redefining Newton's gravitational constant, G.

What needs to be, is that the ratio of inertial mass to gravitational mass is the same, for all bodies.

 

[vanesch]

I was wondering if they'd accelerate towards one another at a combined closing of rate of 19.6ms^-2.

Well, assuming that we can replace a spherical mass by a point mass (there's a theorem on that, but I'm not sure it applies in exactly this circumstance), we have two point masses of the mass of the Earth at a distance of 2 R (plus 10 miles).
Now, we know that the gravitational attraction of earth1 on earth2 is G M^2 / (2 R)^2, so filling this in into Newton's second law in the inertial frame which is the center-of-mass frame:

M.g1 = G M^2 / (2R)^2 or g1 = G M / 4 R^2

g1 is the acceleration with which earth2 is accelerating towards earth1.

Now, for a test particle at the surface of the earth, we have:

m g = G m M/R^2 or g = G M/R^2 = 9.81 m/s^2, so we find that g1 = g/4

By symmetry, g2, the acceleration by which earth1 is accelerating towards earth2, is also equal to g/4.

So we have that both Earth's accelerate towards the center of gravity with an acceleration of g/4 ; so from the PoV of the surface of one, the other is accelerating 2 g/4 = g/2.

So, as seen from the surface of earth1, earth2 is "falling" with an acceleration of g/2 = 4.9 m/s^2.

Under the assumption that we can use the theorem that a spherical mass can be replaced by its mass concentrated in its center, of which I'm not 100% sure that I can apply it here...

 

[clj4]

Well, assuming that we can replace a spherical mass by a point mass (there's a theorem on that, but I'm not sure it applies in exactly this circumstance), we have two point masses of the mass of the Earth at a distance of 2 R (plus 10 miles).
Now, we know that the gravitational attraction of earth1 on earth2 is G M^2 / (2 R)^2, so filling this in into Newton's second law in the inertial frame which is the center-of-mass frame:

M.g1 = G M^2 / (2R)^2 or g1 = G M / 4 R^2

g1 is the acceleration with which earth2 is accelerating towards earth1.

Now, for a test particle at the surface of the earth, we have:

m g = G m M/R^2 or g = G M/R^2 = 9.81 m/s^2, so we find that g1 = g/4

By symmetry, g2, the acceleration by which earth1 is accelerating towards earth2, is also equal to g/4.

So we have that both Earth's accelerate towards the center of gravity with an acceleration of g/4 ; so from the PoV of the surface of one, the other is accelerating 2 g/4 = g/2.

So, as seen from the surface of earth1, earth2 is "falling" with an acceleration of g/2 = 4.9 m/s^2.

Under the assumption that we can use the theorem that a spherical mass can be replaced by its mass concentrated in its center, of which I'm not 100% sure that I can apply it here...

If you start with R>> than the radius of earth1 , earth2 then your calculation would work perfectly because the volume integrals can be shown to be independent (almost) of the Earth radius.

There is a problem, though: we all know that the test probe will experience a given acceleration (say 9.81m/s^2) only in the vicinity of the Earth surface. So , now you need R to be very close to the Earth radius. Thus, your proof contains a contradiction.

So , we cannot solve this problem simplistically. What to do?
Here is the calculations leading to the correct solution (within Newtonian physics)
http://scienceworld.wolfram.com/physics/SphericalShellGravitationalPotential.html

 

[vanesch]

Here is the calculations leading to the correct solution (within Newtonian physics)
http://scienceworld.wolfram.com/physics/SphericalShellGravitationalPotential.html

Yes, well, what does it say ? For a full sphere, outside of the sphere ?
(so R > b, and a = 0)

We have that the potential is - 4 pi/3 b^3 rho G / R

Now, 4 pi / 3 b^3 rho is nothing else but the total mass of the sphere, so we have that outside of the sphere, at a distance R from its center, the potential is - M G / R which is exactly the same as if we lumped the entire mass at its center. That's the theorem I talked about.

What I'm slightly less sure of, is: does a sphere with density rho, bathing in such a potential, also act as if it were a point particle with its mass in its center ?
That is, if we sum all the forces acting on all the parts of a sphere, exposed to the potential of a point particle (but of course off-center), does this sum equal the force that would act on a point particle in the center of that sphere ? It is somehow the reciproke of the above theorem and my gut feeling would tell me that it is ok, but I should work it out.
This was what I'm not sure about.

 

[clj4]

Yes, well, what does it say ? For a full sphere, outside of the sphere ?
(so R > b, and a = 0)

We have that the potential is - 4 pi/3 b^3 rho G / R

Now, 4 pi / 3 b^3 rho is nothing else but the total mass of the sphere, so we have that outside of the sphere, at a distance R from its center, the potential is - M G / R which is exactly the same as if we lumped the entire mass at its center. That's the theorem I talked about.

What I'm slightly less sure of, is: does a sphere with density rho, bathing in such a potential, also act as if it were a point particle with its mass in its center ?
That is, if we sum all the forces acting on all the parts of a sphere, exposed to the potential of a point particle (but of course off-center), does this sum equal the force that would act on a point particle in the center of that sphere ? It is somehow the reciproke of the above theorem and my gut feeling would tell me that it is ok, but I should work it out.
This was what I'm not sure about.

There was a more severe problem, I am not sure you caught what i was saying.:

You are using R in two contradictory ways:

1. To calculate the attraction between the two "Earth's" (R is big, much bigger than the "earth" radius)

2. to calculate the acceleration of the test probe (R needs to be very close to the "earth" radius in order to get your predicted value of 9.81m/s62).

You cannot have both 1 and 2 to be true.

 

[vanesch]

There was a more sever problem, Iam not sure you caught what i was saying.:

You are using R in two contradictory ways:

1. To calculate the attraction between the two "Earth's" (R is big, much bigger than the "earth" radius)

2. to calculate the acceleration of the test probe (R ineeds to be very close to the "earth" radius in order to get your predicted value of 9.81m/s62).

You cannot have both 1 and 2 to be true.

Eh, what do you mean ? R is Earth's radius. So the two centers of the two Earth's are at a distance of 2 R (+ 10 miles), because the surfaces are touching each other almost.

The point is that the theorem allows us to replace the gravitational effect of a full sphere by the gravitational effect of all its mass in a point in its center, as long as we stay outside of the sphere of course.

So I replaced the two Earth's by their respective equivalent point masses in their centers which are now at a distance of 2R (+ 10 miles, which I neglect).

For the "active" part, I know the theorem is true (and you provided an explicit reference for it). For the passive part, I'm simply too lazy to work it out, I'm pretty convinced that it is in fact the same calculation, but there's some uncertainty left.

The point is that the "earth as a point mass" is not an approximation, it is exact, even for the gravitational effect 5 cm above its surface. So I do not need to be far away from Earth for it to work.

 

[clj4]

Eh, what do you mean ? R is Earth's radius. So the two centers of the two Earth's are at a distance of 2 R (+ 10 miles), because the surfaces are touching each other almost.

Then you have not calculated the attraction force correctly (look at your formula, you are using only 2R, there should be 2R+10). The bigger problem is that the formula holds only for distances between bodies that are MUCH LARGER than the bodies' radius.
You simply cannot use the "lazy" way. I think Pete said the same thing a few posts earlier. See post #16.

 

[Rach3]

Then you have not calculated the attraction force correctly (look at your formula, you are using only 2R, there should be 2R+10).

That's a tiny correction, vanesch threw it out for simplicity.

The bigger problem is that the formula holds only for distances between bodies that are MUCH LARGER than the bodies' radius.

It assumes point masses, but due to a geometric theorem originally discovered by Newton it also holds for spheres and spherically-symmetric balls.

 

[clj4]

That's a tiny correction, vanesch threw it out for simplicity.It assumes point masses, but due to a geometric theorem originally discovered by Newton it also holds for spheres and spherically-symmetric balls.

I really think that you need to recheck your calculations. The approximation applies when the distance between the centers of the spheres is MUCH BIGGER than the spheres' radiuses. This is not the case in the example and creates a "paradox". Someties "lazy" solutions are not the best thing, they waste more time in argument than applying the correct (albeit more complicated) math.

 

[Rach3]

For the "active" part, I know the theorem is true (and you provided an explicit reference for it). For the passive part, I'm simply too lazy to work it out, I'm pretty convinced that it is in fact the same calculation, but there's some uncertainty left.

It would be true for the attraction of a sphere to a point mass, by exactly the symmetry you are thinking of (just invoke Newton's 3rd); the sphere, attracted to a point mass, behaves as a point mass. Then by superposition, it would be true for the attraction of a sphere to two point masses as well - since you can just add the forces. So there you can just integrate over a whole bunch of points - and it is clear that a uniform sphere behaves as a point mass, in an arbitrary gravitational field (ignoring relativity, inhomogenities, etc.) Thus two ideal spheres attract each other as point masses.

In a real situation of course, tidal effects would rip the planets apart!

 

[vanesch]

It would be true for the attraction of a sphere to a point mass, by exactly the symmetry you are thinking of (just invoke Newton's 3rd); the sphere, attracted to a point mass, behaves as a point mass. Then by superposition, it would be true for the attraction of a sphere to two point masses as well - since you can just add the forces. So there you can just integrate over a whole bunch of points - and it is clear that a uniform sphere behaves as a point mass, in an arbitrary gravitational field (ignoring relativity, inhomogenities, etc.) Thus two ideal spheres attract each other as point masses.

This was also the kind of reasoning I did in my head...

In a real situation of course, tidal effects would rip the planets apart!

Sure, and as pointed out somewhere else, holding them, before the experiment, 10 miles apart, might put some load on your wrists too

I really think that you need to recheck your calculations. The approximation applies when the distance between the centers of the spheres is MUCH BIGGER than the spheres' radiuses. This is not the case in the example and creates a "paradox". Someties "lazy" solutions are not the best thing, they waste more time in argument than applying the correct (albeit more complicated) math.

As I said before, and as rach3 confirms, it is not an approximation. It is a theorem that gives you the exact field, as long as you have to do with spherical objects. It is indeed a theorem by Newton himself, and actually, I read that he held back the publication of his theory of gravity until he found that theorem (and invented the necessary calculus that goes with it). You quoted the theorem yourself in the reference you gave !

 

[vanesch]

I think Pete said the same thing a few posts earlier. See post #16.

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 ).

 

[Rach3]

I really think that you need to recheck your calculations. The approximation applies when the distance between the centers of the spheres is MUCH BIGGER than the spheres' radiuses. This is not the case in the example and creates a "paradox". Someties "lazy" solutions are not the best thing, they waste more time in argument than applying the correct (albeit more complicated) math.

There's nothing lazy about it - it's an application of Gauss' law and some symmetry arguments. I'm curious as to what "paradox" you have in mind?

 

[clj4]

There was a more severe problem, I am not sure you caught what i was saying.:

You are using R in two contradictory ways:

1. To calculate the attraction between the two "Earth's" (R is big, much bigger than the "earth" radius)

2. to calculate the acceleration of the test probe (R needs to be very close to the "earth" radius in order to get your predicted value of 9.81m/s62).

You cannot have both 1 and 2 to be true.

OK,

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?

 

[clj4]

In a more constructive way, if we start with the rigurous 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 viceversa) starting from an ARBITRARY distance z and from an initial speed say, zero. The acceleration will be a variable.
I will not have access to a computer for the next week, I am going away from any math and physics and emails and forums. What I wanted to say, is that we should aspire towards generic solutions.

 

[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&#039;(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&#039;(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).