Newton’s Laws and Point Particles

[FS98]

Why are Newton’s laws often discussed as applying to point masses when Newton makes no references to point masses in the Principa. He seems to mention nothing other than bodies while describing each of his three laws, which are not necessarily point masses.

 

[phinds]

Why are Newton’s laws often discussed as applying to point masses when Newton makes no references to point masses in the Principa. He seems to mention nothing other than bodies while describing each of his three laws, which are not necessarily point masses.

It's a mathematical simplification that works because of the Shell Theorem

 

[sophiecentaur]

The point mass approximation related particularly to gravity and the only accurate observations in Newton's time involved celestial objects. These can nearly all use the point mass idea. The Moon, tides and synchronising are topics where the point mass assumption can't be made.
Since Archimedes' time, the behaviour of objects with distributed mass on Earth has been studied (principle of moments etc.)

 

[anorlunda]

@phinds mentioned the shell theorum, but it may help to expand on that.

It was Newton who proved the shell theorem. That makes calculations much simpler, and it gives the same answer as treating the mass as distributed. Who can resist using it?

In classical mechanics, the shell theorem gives gravitational simplifications that can be applied to objects inside or outside a spherically symmetrical body. This theorem has particular application to astronomy.

Isaac Newton proved the shell theorem (Newton, Isaac (1687). Philosophiae Naturalis Principia Mathematica. London. pp. Theorem XXXI.) and stated that:

1. A spherically symmetric body affects external objects gravitationally as though all of its mass were concentrated at a point at its centre.
2. If the body is a spherically symmetric shell (i.e., a hollow ball), no net gravitational force is exerted by the shell on any object inside, regardless of the object's location within the shell.

 

[FS98]

@phinds mentioned the shell theorum, but it may help to expand on that.

It was Newton who proved the shell theorem. That makes calculations much simpler, and it gives the same answer as treating the mass as distributed. Who can resist using it?

Wouldn’t this only work for spherical objects?

 

[anorlunda]

Wouldn’t this only work for spherical objects?

Yes. Or for nonspherical objects very far away. Think of the mountains on the Moon influence on Earth tides.

As the article says, "This theorem has particular application to astronomy."

 

[phinds]

Wouldn’t this only work for spherical objects?

Did you read what anorlunda said? In particular " ... applied to objects inside or outside a spherically symmetrical body. ..."

EDIT: I see he beat me to it.

 

[FS98]

Did you read what anorlunda said? In particular " ... applied to objects inside or outside a spherically symmetrical body. ..."

EDIT: I see he beat me to it.

I have a couple more questions about this.

Isn’t shells theorem about how the gravity of a sphere can be treated as if all the mass was at a single point rather than how bodies can be treated as particles according to Newton’s laws? Maybe I’m mistaken, but I feel like objects such as blocks are sometimes treated as if they were point particles when Newton’s laws are being applied.

Also, could you not apply the gravitational equation to objects like cubes without some error? Would using the distance between the centers of masses of two cubes not yield a perfectly correct answer?

 

[phinds]

I have a couple more questions about this.

Isn’t shells theorem about how the gravity of a sphere can be treated as if all the mass was at a single point rather than how bodies can be treated as particles according to Newton’s laws? Maybe I’m mistaken, but I feel like objects such as blocks are sometimes treated as if they were point particles when Newton’s laws are being applied.

Also, could you not apply the gravitational equation to objects like cubes without some error? Would using the distance between the centers of masses of two cubes not yield a perfectly correct answer?

Your questions have all already been answered in this thread. Read the replies again.

 

[Nugatory]

Maybe I’m mistaken, but I feel like objects such as blocks are sometimes treated as if they were point particles when Newton’s laws are being applied.

Yes, we often do that because it's a really good approximation in many problems.

Here's an exercise. First, calculate the gravitational force on a one kilogram spherical object ten meters above the Earth's surface. Because it is a sphere you can use the shell theorem and treat it as a point particle. Next, calculate the gravitational force on a one kilogram cube with same density and volume, also ten meters above the Earth's surface; you can't use the shell theorem and you have to allow for the $1/r^2$ variation in the force at different distances from the center of the Earth.

Now consider the difference between the two results (not much at all) and how much work it was to get the answer ($mg$ in one case, a remarkably unpleasant volume integral in the other). You will quickly see why we approximate masses as point particles whenever we can.

 

[FS98]

Your questions have all already been answered in this thread. Read the replies again.

Would it be correct to say that Newton’s three laws of motion apply to all bodies, but his law of gravitation applies only to point particles which can extend to spheres because of the shell theorem?

And would it be correct to say that the center of mass of a non-spherical object cannot be used to find the force of gravity without some degree of error?

 

[phinds]

Would it be correct to say that Newton’s three laws of motion apply to all bodies

yes

, but his law of gravitation applies only to point particles which can extend to spheres because of the shell theorem?

No. The Shell Theorem is not helpful for irregularly shaped bodies close to each other. For an intermediate case, see post #10

And would it be correct to say that the center of mass of a non-spherical object cannot be used to find the force of gravity without some degree of error?

Technically, yes but for objects that are very far apart relative to their size, the error is small enough to be ignored.

 

[FS98]

No. The Shell Theorem is not helpful for irregularly shaped bodies close to each other.

By irregular do you mean non-spherical? So Newton’s law of gravitation applies to point particles, extends to spheres at all distances, and approximates non-spherical objects at long distances?

 

[jbriggs444]

By irregular do you mean non-spherical? So Newton’s law of gravitation [...] approximates non-spherical objects at long distances?

I would understand the relevant concept to extend to objects which may or may not be spherical but which have a mass distribution which is not spherically symmetric.

 

[phinds]

By irregular do you mean non-spherical? So Newton’s law of gravitation applies to point particles, extends to spheres at all distances, and approximates non-spherical objects at long distances?

No. Why do you think Newton's law of gravitation does not apply to something just because it has an odd shape? That's a very strange point of view. Perhaps you are confusing ease of calculation with validity of the law. FINDING the gravitational attraction between two bodies is WAY easier if you can use the Shell Theorem but that doesn't mean there is some different law for irregular bodies close together. AGAIN, I suggest you reread post #10

 

[FS98]

No. Why do you think Newton's law of gravitation does not apply to something just because it has an odd shape? That's a very strange point of view. Perhaps you are confusing ease of calculation with validity of the law. FINDING the gravitational attraction between two bodies is WAY easier if you can use the Shell Theorem but that doesn't mean there is some different law for irregular bodies close together. AGAIN, I suggest you reread post #10

When I say that Newton’s law of gravitation doesn’t apply to non-spherical objects, I’m just saying that there is nothing that you can plug into the above equation to get the force due to gravity since there is no way to get a single value r for objects that are not pointlike or cannot be treated as if they were pointlike. You would have to do something more complicated as post #10 suggests.

 

[phinds]

You would have to do something more complicated as post #10 suggests.

Right. You got it.

 

[sophiecentaur]

You would have to do something more complicated as post #10 suggests.

Like breaking the body up into a series of small spheres, treating them all separately under gravity and then linking them all together to produce a rigid body.

 

[PeroK]

When I say that Newton’s law of gravitation doesn’t apply to non-spherical objects, I’m just saying that there is nothing that you can plug into the above equation to get the force due to gravity since there is no way to get a single value r for objects that are not pointlike or cannot be treated as if they were pointlike. You would have to do something more complicated as post #10 suggests.

There are at least three things to think about here.

1) When it comes to the Solar system the objects are spherical so the shell theorem allows us to consider these bodies as point masses.

2) Close to the Earth there is an almost uniform gravitational field. This allows us to consider objects as point masses as far as gravity is concerned. It doesn't matter which way you drop a long, thin object, for example. The $r^2$ in your equation is almost constant for all points on all objects close to the Earth.

3) This applies also to things like the International Space Station: the distance from any point on the station to the centre of the Earth is almost constant. Interestingly, though, if you take two small objects in the space station and set them at rest, they will slowly drift apart due to the different gravitational field strength at slightly different altitudes.

In terms of Newton's laws generally, in many cases you can treat objects as point particles for simplicity. Often, however, you have to consider them as rigid bodies that can spin about their centre of mass; or, as a non-rigid body - a fluid, for example. In these cases, the bodies can be considered as a large number of particles interacting with each other or a continuous mass distribution

 

[jbriggs444]

The $r^2$ in your equation is almost constant for all points on all objects close to the Earth.

As is the $\hat{r}$

 

[Herman Trivilino]

Why are Newton’s laws often discussed as applying to point masses when Newton makes no references to point masses in the Principa.

As others have pointed out, he knew he had to model the spherical Earth as what we would now call a point particle, so he was using what we now call the particle model. Perhaps it's just a language thing. language changes and Newton's writings are over 300 years old.

He seems to mention nothing other than bodies while describing each of his three laws, which are not necessarily point masses.

The jargon "point masses" certainly didn't exist 300 years ago. Moreover, Newton wrote in Latin, as was the practice of the day.

 

[FS98]

As others have pointed out, he knew he had to model the spherical Earth as what we would now call a point particle, so he was using what we now call the particle model. Perhaps it's just a language thing. language changes and Newton's writings are over 300 years old.
The jargon "point masses" certainly didn't exist 300 years ago. Moreover, Newton wrote in Latin, as was the practice of the day.

It seems like even more than just a language thing though. Newton’s uses both the words body and particle in the Principia (or at least in the English translation). When he described his laws he uses the word body. He defined momentum of a body as p = mv, then he said that the momentum of a body is equal to the sum of its parts. These two things seem to suggest that the v in p = mv is the velocity of the center of mass a body. If this is the case, I see no reason why point particles are necessary to describe the motion of objects through Newton’s laws if you already know the forces acting on them.

I see now why point particles make it easier to find the force due to gravity or electromagnetism on an object, but that’s different.

 

[Nugatory]

When I say that Newton’s law of gravitation doesn’t apply to non-spherical objects, I’m just saying that there is nothing that you can plug into the above equation to get the force due to gravity since there is no way to get a single value r for objects that are not pointlike or cannot be treated as if they were pointlike. You would have to do something more complicated as post #10 suggests.

Yes, if you are dealing with non-spherical objects then you have to do something more complicated than using the simplified form of Newton's law we use for spherical objects. But that "more complicated" thing is still Newton's law: You're just applying Newton's law to each infinitesimal element of the objects to get the force between each bit of the two objects and then summing (integrating) to get the total force.

It is not an accident that Newton invented calculus on the way to classical mechanics and gravitation.

 

[Herman Trivilino]

I see no reason why point particles are necessary to describe the motion of objects through Newton’s laws if you already know the forces acting on them.

I think they just make the analysis easier. So, for example, you can model a sphere as a point particle and pretend that all the forces acting on the sphere act at its center. That's much easier than modeling the sphere as a distribution of matter and looking at the forces that act on each piece of that distribution. The two schemes give the same result. That's the point of it (pardon the pun).

 

[FS98]

I think they just make the analysis easier. So, for example, you can model a sphere as a point particle and pretend that all the forces acting on the sphere act at its center. That's much easier than modeling the sphere as a distribution of matter and looking at the forces that act on each piece of that distribution. The two schemes give the same result. That's the point of it (pardon the pun).

But if we’re strictly speaking about Newton’s laws you don’t have to worry about mass distributions. As long as you know mass of the entire body and the force acting on it, you can find the acceleration of its center of mass, regardless of the mass distribution of the object.

 

[PeroK]

But if we’re strictly speaking about Newton’s laws you don’t have to worry about mass distributions. As long as you know mass of the entire body and the force acting on it, you can find the acceleration of its center of mass, regardless of the mass distribution of the object.

If you analyse, for example, a cylinder rolling down an inclined plane, then the acceleration does depend on the mass distribution: through a quantity known as the moment of inertia. Ironically, in this case, you need to know the moment of inertia to calculate the frictional force between the cylinder and the surface.

In general, modelling the motion of a rigid body is more difficult than the motion of a point particle. Modelling a particle as a small sphere, say, you have to take into account its rotational kinetic energy. If you model it as a point particle, it has only linear kinetic energy.

Likewise: if you have a cube of fluid and instanteously remove the sides, describe the motion of the fluid? Modelling a fluid is even harder, as you have unbalanced internal forces.

 

[FS98]

If you analyse, for example, a cylinder rolling down an inclined plane, then the acceleration does depend on the mass distribution: through a quantity known as the moment of inertia. Ironically, in this case, you need to know the moment of inertia to calculate the frictional force between the cylinder and the surface.

But the mass distribution is only required to find the force acting on the object. If you know that there is a certain force acting on a certain body, then you can find the acceleration of the center of mass of that body.

To find the force of gravity acting between two objects, the mass distribution is relevant, but once you know the forces acting on those objects, the mass distribution doesn’t matter anymore does it? AFAIK Newton’s second law should tell you the acceleration of the center of mass regardless of the mass distribution of that object, no point particle idealization required.

 

[Herman Trivilino]

But if we’re strictly speaking about Newton’s laws you don’t have to worry about mass distributions.

Yes, you do. If you have an extended body, for example, and you want to model it as a point particle, you need to know the location of the center of mass, and to find that you need to know the distribution of mass.

But the mass distribution is only required to find the force acting on the object.

You also need to know it if you want to understand how the object will respond. There's more than one force exerted on an extended object. This is what @Nugatory explains in Post #23.

If you know that there is a certain force acting on a certain body, then you can find the acceleration of the center of mass of that body.

AFAIK Newton’s second law should tell you the acceleration of the center of mass regardless of the mass distribution of that object, no point particle idealization required.

First of all, when you speak of using the center of mass to make a determination you are quite likely using the particle model.

And when you do something like the rolling cylinder @PeroK mentions in Post #26, you have two forces acting on the object, one acts at the center of mass of the rolling cylinder and the other acts at the point of contact between the cylinder and the surface upon which it rolls. How the center of mass moves is determined by both of those forces, even though one of them is not applied at the center of mass.

 

[PeroK]

But the mass distribution is only required to find the force acting on the object. If you know that there is a certain force acting on a certain body, then you can find the acceleration of the center of mass of that body.

To find the force of gravity acting between two objects, the mass distribution is relevant, but once you know the forces acting on those objects, the mass distribution doesn’t matter anymore does it? AFAIK Newton’s second law should tell you the acceleration of the center of mass regardless of the mass distribution of that object, no point particle idealization required.

For rigid body motion (and, in fact, more generally) you have

a) $\vec{F} = m\vec{a}$ for linear motion (of the centre of mass).
b) $\vec{\tau} = I \vec{\alpha}$ for rotational motion (e.g. about the centre of mass), where $\vec{\tau}$ is the torque and $\vec{\alpha}$ the angular acceleration..

If you are only interested in the motion of the centre of mass, then you need only consider the forces acting on an object. But, if you are also interested in the rotation of an object, then you need to consider where the forces are acting (e.g. relative to the centre of mass).

Once you get to rigid body motion, there is more than just $\vec{F} = m\vec{a}$.

On a general point, it's not a good idea in physics and maths to go back to the original sources and treat those texts with an almost religious reverence. Many people since Newton have developed and expanded his work - both theoretically and in terms of teaching the subject. A modern treatment of Newtonian mechanics has so much more going for it than the Principia.