# The Naked Scientists Forum

### Author Topic: Newton made an error and Einstein copied it  (Read 24668 times)

#### gem

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##### Newton made an error and Einstein copied it
« Reply #50 on: 01/05/2010 15:09:04 »
Gem,

I'm still thoroughly confused as to what you're saying Newton claimed.  So let's assume the earth is a uniform sphere.

1) Are you saying that Newton claimed that the earth's gravity measured above the earth's surface is proportional to the inverse square of the distance between the center of the earth and the object?

#### gem

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##### Newton made an error and Einstein copied it
« Reply #51 on: 01/05/2010 15:10:08 »
I know that newtons theory takes inverse square law from the centre of the earth and it is the value attraction at earths surface that is used, sorry for not making that clearer.

#### JP

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##### Newton made an error and Einstein copied it
« Reply #52 on: 01/05/2010 15:28:27 »
It's still a bit unclear.  The value of attraction at the earth's surface is used for what?  It sounds like you're saying that Newton's claim is that the force of gravity, F, on an object is given by

F=Fsurface/r2,

where Fsurface is the force of gravity on that object at the earth's surface and r is the distance from the center of the earth to the object.  Is this the equation that you're saying Newton used?

#### gem

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##### Newton made an error and Einstein copied it
« Reply #53 on: 01/05/2010 16:35:49 »
Geezer your last post is right on the crux of the matter.

so to show what is wrong with the maths.
I think that you will find the vector sum arrow in red in JP's diagram , will increase in magnitude relative to the attractive force arrows in black for the 'point mass' object lying along the ring's axis, as the distance from the surface of the sphere increases for all calculations done for any pair of point Masses contained within the sphere acting on the said point mass object. IE a greater percentage of the potential gross force

And this will cause inverse square law violation.

I think what you are saying here is that as the object gets further from the earth, the angles will reduce, so the component of the force acting between the centers (the attractive force times the cosine of the angle) will increase with distance.

Assuming I got that right, then yes, I agree it will. However, the gravitational attraction will also diminish while the distance increases as the inverse of the square of the distance, so it's necessary to account for the reduced force associated with distance at the same time as the increased effectiveness of the force associated with the reduction in angle.

absolutely right.

and, it turns out that, regardless of the distance between the Earth and the object, when the forces are all integrated, they produce the same force that would be produced if all the mass of the Earth was concentrated in one point at its center.

absolutely right again i posted this statement much earlier.

'I do not disagree with Shell theorem proving A spherically symmetric body affects external objects gravitationally as though all of its mass were concentrated at a point at its center at various distances in space.'

so it's necessary to account for the reduced force associated with distance at the same time as the increased effectiveness of the force associated with the reduction in angle.

The theorem takes that into account,

this is the crux of the matter i don't believe that at various distance the increased effectiveness of the force associated with the reduction in angle has been taken into account.

I believe what has been done is show , regardless of the distance between the Earth and the object, when the forces are all integrated, they produce the same force that would be produced if all the mass of the Earth was concentrated in one point at its center.
And concluded that therefore it is reasonable to apply inverse square law to a measured value of gravitational acceleration.[so this is the error]

Because as i have pointed out earlier the gravitational acceleration we measure is a vector sum only IE a net figure, at the background of that net force there is a gross force that will manifest a greater percentage of its potential force the more acute gets the angle of interaction.

This is will be demonstrated [as in JPs diagram] by the vector sum arrow in red  increasing in magnitude relative to the attractive force arrows in black for the 'point mass' object lying along the ring's axis, as the distance from the surface of the sphere increases for all  calculations for all point mass pairs calculated from where they are in reality.

#### Geezer

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##### Newton made an error and Einstein copied it
« Reply #54 on: 01/05/2010 20:24:45 »
Gem,

Are you saying that the inverse square law only works for point masses but it is in error when the object has real dimensions (like the Earth for example)?

#### LeeE

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##### Newton made an error and Einstein copied it
« Reply #55 on: 01/05/2010 21:27:27 »
Gem:

If Newton et al. got it wrong then it would also seem that the currently accepted mass of the Earth is also wrong.

If we treat the Earth as a point mass and work out the acceleration at the radius of the Earth's surface we get the same value of acceleration that is actually measured at the Earth's surface...

Using a= -(G*M)/r2

where...
G = the gravitational constant = 6.67428e-11
M = the mass of the Earth (kg) = 5.9736e24
r = Earth's equatorial radius (m) =  6.3781e6
a = the resulting acceleration (towards the center of the Earth)

we get an acceleration of -9.800718 m/s2

which, when the centripetal reduction due the the equatorial rotation is taken into account, is just about what is actually measured (the average acceleration at the equator is surface 9.780327 m/s2).

However, if the accepted mass of the Earth is incorrect then all the stuff we've launched up into orbit wouldn't be where it's supposed to be: geostationary satellites would drift and GPS would be wildly inaccurate.

#### Geezer

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##### Newton made an error and Einstein copied it
« Reply #56 on: 02/05/2010 01:47:38 »

If Newton et al.

Don't you mean Al as in Al Einstein?

#### LeeE

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##### Newton made an error and Einstein copied it
« Reply #57 on: 02/05/2010 12:25:17 »
Oops! nope - I meant Ethal.

#### gem

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##### Newton made an error and Einstein copied it
« Reply #58 on: 03/05/2010 17:37:16 »
Gem,

Are you saying that the inverse square law only works for point masses but it is in error when the object has real dimensions (like the Earth for example)?

Yes i believe  a unified field theory requires it.

Which then calls into question the validity of big G.

As its name of gravitational constant becomes a bit of a misnomer because with different density's of of mass but the same value of mass within a homogeneous sphere the angles of attraction will vary, [due to variation in the lengths of radius ] causing tests of gravitational constant to vary.

If Newton et al. got it wrong then it would also seem that the currently accepted mass of the Earth is also wrong.

Yes it would seem so, if you look at the average angles of attraction between the large mass spheres in Cavendish's original experiment and the small mass spheres there respective sizes being approximately 12 inch spheres and 2 inch spheres.

Then their angle of attraction between there respective point masses would be more acute than the comparison that was made between the small lead sphere and the earth.

Meaning that there will have been a greater pro rata value of attraction in the experiment causing a slight under value of the density of the earth.

[/ftp]
However, if the accepted mass of the Earth is incorrect then all the stuff we've launched up into orbit wouldn't be where it's supposed to be: geostationary satellites would drift and GPS would be wildly inaccurate.

Geostationary satellites are not used for GPS they are used for communication.

However if your point is one of the altitude that they are supposedly stationary at  and what the value of gravitational acceleration is at that distance in space from the earth then your point is a valid one .

But it is not as easy as just applying the equations as there are large variations in the values of attraction that already exceed the effect i am suggesting.

Because while a geostationary orbit should hold a satellite in fixed position above the equator, orbital perturbations, such as by the Moon, IE the earths orbit around its barycentre, and from the fact that the Earth is not an exact sphere cause slow but steady drift away from the geostationary location. Satellites correct for these effects with station-keeping maneuvers

where...
G = the gravitational constant = 6.67428e-11
M = the mass of the Earth (kg) = 5.9736e24
r = Earth's equatorial radius (m) =  6.3781e6
a = the resulting acceleration (towards the center of the Earth)

we get an acceleration of -9.800718 m/s2

which, when the centripetal reduction due the the equatorial rotation is taken into account, is just about what is actually measured (the average acceleration at the equator is surface 9.780327 m/s2).

Yes the variation of  measured g  with latitude at sea level ranges from about 9.78 m/s at the equator to over 9.83 at the poles, so allowing for the fact that the radius is 21k/m less at the poles you should get a gravitational acceleration force that is approximately 3 millimetres per/second squared less.

However the recorded measured values don't confirm this and this is thought to be down to the fact that the earth is not a homogeneous sphere and gets denser the closer to the center.

Although big G seems to be close to the values at earths surface it is not exact,So not the last word on the matter.

And as i have said earlier in this post it is as the distance from the mass increases that this gross force will manifest itself but it will obviously be diluted by the inverse square of the distance but as a percentage of the force it will not be inconsiderable.
« Last Edit: 03/05/2010 17:39:38 by gem »

#### JP

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##### Newton made an error and Einstein copied it
« Reply #59 on: 04/05/2010 03:55:10 »
Gem, could you clearly write out the equation you think Newton erroneously used, indicating what all the constants and variables mean?  Then could you point out what needs to change in his equation in order to make it right?  It's easy to understand the general idea of what you're saying but extremely hard to nail down mathematical details that would lead to disagreements with Newton's theory based on your explanations with no math to back them up.

As it is, when I try to apply the vector sum techniques you mention, I get something that agrees with Newton.

#### Geezer

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##### Newton made an error and Einstein copied it
« Reply #60 on: 04/05/2010 05:04:14 »
Gem,

Are you saying that the inverse square law only works for point masses but it is in error when the object has real dimensions (like the Earth for example)?

Yes i believe  a unified field theory requires it.

Gem,

The Shell Thorem proves that a spherical object of uniform density produces exactly the same gravitational force as a point object of the same mass. If the gravitational forces they produce are different as you say, there must be something wrong with the math in Shell Theorem, but you also agree that Shell Theorem is correct.

Apparently the foundations laid down by Newton are still quite secure.

#### gem

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##### Newton made an error and Einstein copied it
« Reply #61 on: 04/05/2010 22:19:52 »
There is a difference in the following two statements.

The Shell Theorem proves that a spherical object of uniform density produces exactly the same gravitational force as a point object of the same mass.

I think you will find that shell theorem does not actually prove that, what it proves is.

a spherically symmetric body affects external objects gravitationally as though all of its mass were concentrated at a point at its centre ,after allowing for the partial cancellation due to the vector nature of the force.

IE the value of the cancellation is not included.

In your statement above there is no cancellation so no variation in cancellation at different distances.
So what you posted here does not get considered.

so to show what is wrong with the maths.
I think that you will find the vector sum arrow in red in JP's diagram , will increase in magnitude relative to the attractive force arrows in black for the 'point mass' object lying along the ring's axis, as the distance from the surface of the sphere increases for all calculations done for any pair of point Masses contained within the sphere acting on the said point mass object. IE a greater percentage of the potential gross force

And this will cause inverse square law violation.
I think what you are saying here is that as the object gets further from the earth, the angles will reduce, so the component of the force acting between the centers (the attractive force times the cosine of the angle) will increase with distance.

Assuming I got that right, then yes, I agree it will. However, the gravitational attraction will also diminish while the distance increases as the inverse of the square of the distance, so it's necessary to account for the reduced force associated with distance at the same time as the increased effectiveness of the force associated with the reduction in angle.

So do you still agree it will?.

JP the error i am highlighting is i don't believe it is possible for the gravitational force to diminish according to the inverse square law. so i am not saying there is a error in his equation.
The error is to think it proved that you could apply inverse square law because of it.

Are you aware of any aspect of accounting for a increased value of the force?

so it's necessary to account for the reduced force associated with distance at the same time as the increased effectiveness of the force associated with the reduction in angle.

The theorem takes that into account,

Does it ?

And if we all agree that the value of gravitational attraction we measure at the earths surface is a vector sum what percentage is the amount of cancellation prior to that net figure at that point?.

What figure does shell theorem give to it, IE what does shell theorem state the gross force is.
« Last Edit: 04/05/2010 22:27:07 by gem »

#### Geezer

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##### Newton made an error and Einstein copied it
« Reply #62 on: 04/05/2010 23:08:47 »

a spherically symmetric body affects external objects gravitationally as though all of its mass were concentrated at a point at its centre ,after allowing for the partial cancellation due to the vector nature of the force.

Gem,

All you are saying is that a body in the Earth's gravitational field is subject to multiple forces. Each force can be resolved into two components. One component can be considered to act between the center of the Earth and the body. The other component can be considered to act at right angles to the first force.

The "cancellation" that you refer to only applies to the components that act at right angles to the direction of the gravitational attraction. Because they act at right angles to the gravitational force, they can have no influence on the gravitational force at all. There is no cancellation of any of the components that produce the gravitational effect.

I suspect the problem here is not so much with any error that Newton (or Einstein for that matter) made, but more to do with your lack of understanding of how to properly add vectors.
« Last Edit: 04/05/2010 23:58:40 by Geezer »

#### gem

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##### Newton made an error and Einstein copied it
« Reply #63 on: 05/05/2010 19:06:07 »

There is no cancellation of any of the components that produce the gravitational effect.

I thought we already covered partial cancellation due to the vector nature of the force,you are contradicting 'JP' 'wikipedia'and it would seem yourself.

I think what you are saying here is that as the object gets further from the earth, the angles will reduce, so the component of the force acting between the centers (the attractive force times the cosine of the angle) will increase with distance.
Assuming I got that right, then yes, I agree it will.

And this
so it's necessary to account for the reduced force associated with distance at the same time as the increased effectiveness of the force associated with the reduction in angle.

The theorem takes that into account,

Because forces are vectors, the affect of an individual force upon an object is often canceled or partially canceled by the affect of another force.

The theorem takes that into account,

Could you show where increased effectiveness of the force associated with the reduction in angle is actually accounted for ?.

#### Geezer

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##### Newton made an error and Einstein copied it
« Reply #64 on: 05/05/2010 19:38:45 »

Could you show where increased effectiveness of the force associated with the reduction in angle is actually accounted for ?.

Sure. The force is proportional to the cosine of the angle. As the angle decreases, the cosine of the angle increases. You'll see the theorem takes that into account.

#### JP

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##### Newton made an error and Einstein copied it
« Reply #65 on: 06/05/2010 02:44:09 »
Gem,

Everyone here agrees with you that vector addition is how to properly account for gravitational forces from objects that aren't points.  What we disagree with you on is that Newton didn't account for this vector addition.  He did.  There's a mistake somewhere in your calculations in between your first (correct) statement that the forces are vectors and must be added as such and your (incorrect) conclusion that this shows that Newton made a mistake.  Since you're trying to make a mathematical point, we can't do much more than tell you that you've made a mistake somewhere unless you're willing to actually show us your calculations.

By the way, Geezer is pointing out the correct way to do vector addition that agrees with what Newton said.

#### Geezer

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##### Newton made an error and Einstein copied it
« Reply #66 on: 06/05/2010 06:02:27 »
Gem,

Perhaps you are thinking that it will require more mass in the case of sphere to achieve the same gravitational force that a point mass would produce because, in the case of a sphere, some of the force produced by its mass is not contributing anything to the gravitational force?

It is quite true that some of the mass of the sphere is producing forces that tend to stretch an object within its gravitational field "sideways" (tending to make the object "fatter" ). That being the case you might conclude that the gravitational force has to be somewhat reduced.

However, realize also that, in the case of the sphere, a great deal of its mass is closer to the object than would be the case for a point mass, and because the force exerted is proportional to the inverse of the distance squared, this conveniently compensates for all the "lost" components of the vectors. Consequently a point mass ends up exerting exactly the same gravitational force that a sphere (of the same mass) would exert.

It's not intuitively obvious (to me anyway) that this would be the case, but the math seems to prove that it is quite true.

As JP says, if there is a flaw in this, the only way to get at it is to find a flaw in the math. I suspect it's not feasible to solve this problem by inspection, although there may well be alternative mathematical methods that could be applied. I think Physbang mentioned that Newton did use an alternative method.

#### gem

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##### Newton made an error and Einstein copied it
« Reply #67 on: 06/05/2010 22:33:11 »
OK all i know i must be testing your patience, because it seems like i am agreeing with newtons shell theorem on one hand but saying there is an error on the other.

But hang in there and i will try to isolate what i perceive to be the error.

So, shell theorem says that the magnitude of the gravitational force is the same as that of a point mass in the centre of the shell with the same mass taking into account all the vector sums and all the cancellations.

And yes i believe it does, so whats the problem?.

The problem is the gravity force that we measure at the surface of the earth has two aspects  magnitude, and a direction.

And because newton showed with shell theorem that the magnitude of the force was equal to a point mass coming from the centre that has been the assumed to be the direction also.

But shell theorem does not prove direction it proves magnitude of the force.

In your diagram the points A and B if they had opposite points in the other half IE a mirror image you would have two cones under your body of mass on earths surface one cone of average interaction for the northern hemisphere and one cone of average interaction for the southern hemisphere.

So the average direction cone of the force will be some where between the two.

And it is this aspect of the direction of the force that will cause variation.
In that at various distance the increased effectiveness of the force associated with the reduction in angle has to be taken into account.

Because at earths surface there is a force field that is equal to the magnitude of all of earths mass equivalent to that of a point mass in the centre.

Because of its direction we are unable to measure all of its force as weight or gravitational acceleration
« Last Edit: 06/05/2010 22:36:42 by gem »

#### Geezer

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##### Newton made an error and Einstein copied it
« Reply #68 on: 07/05/2010 01:24:29 »
Gem,

I give up.

#### JP

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##### Newton made an error and Einstein copied it
« Reply #69 on: 07/05/2010 04:48:29 »
Gem,

The shell theorem works by specifically taking the vector nature of the forces into account when adding them.  The reason you keep claiming it's wrong is precisely the reason why it's right.  We've made many attempts to explain this to you, but you continue to post the same argument over and over again without actually looking at how your argument is precisely what makes the derivation work.  I highly recommend that you take some time to try to read and understand a derivation of the shell theorem from the basic principles of adding vectors of the gravitational field.  The wikipedia link (http://en.wikipedia.org/wiki/Shell_theorem) is one of the best online sources I've seen for it, although basic physics textbooks generally have a similar derivation.  The full derivations aren't necessarily easy since they generally involve calculus, but you can at least see how the vector addition comes in without needing calculus.

I'm going to lock this thread at this point, since it is demonstrably false that the shell theorem (and Newton) are wrong based on the reasoning you're providing, and because the thread is going in circles now.  If you have questions about the derivation of the shell theorem, feel free to start another thread to ask them.

Cheers,
JP (moderator)

#### The Naked Scientists Forum

##### Newton made an error and Einstein copied it
« Reply #69 on: 07/05/2010 04:48:29 »