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Chapter 2 of Kleppner & Kolenkow's An Introduction to Mechanics reads as follows -

Newton's Laws describe the behavior of point masses. In the case where the size of the body is small compared with the interaction distance, this offers no problem. For instance, the earth and sun are so small compared with the distance between them that for many purposes their motion can be adequately described by considering the motion of point masses located at the center of each.

Consider a system consisting of a block kept at rest on a table. Assume that friction is neglected. When drawing a force diagram for such a block, we assume it to be a point mass. (Also see: Section 2.4 of Kleppner & Kolenkow's book... Just to mention that such a step is followed in the book itself.) However, it doesn't seem to me that this assumption is valid, as the interaction distance between the block and the table is so small as compared to their sizes.

Then why is the point mass approximation valid in such a case?

Note: This chapter has not generalized Newton's laws to describe rigid bodies yet. So it would be great if you could answer my question without any reference to that (of course, if possible!).

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    $\begingroup$ When you get to the mechanics of extended bodies you'll get the tools to tell you if the physics of point particles is good enough in this case. Though, of course, that isn't very satisfying in the mean time. $\endgroup$
    – dmckee --- ex-moderator kitten
    Jul 18, 2016 at 21:37
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    $\begingroup$ I think K&K could have said a little bit more at this point. Best for now to take the point of view that you are modeling the book as a point particle. A model keeps properties that are important for the problem at hand, and ignore those that will turn out to be small, or those that are not of interest. "As simple as possible, but no simpler." $\endgroup$
    – garyp
    Jul 19, 2016 at 0:12
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    $\begingroup$ It's not true that Newton's laws require point masses. All that is required is that all mass elements of an extended body are moving at the same velocity and that they are experiencing the same acceleration. Point masses don't exist and they should never be mentioned in the teaching of classical mechanics. That's a shoddy way of teaching if there ever was one. What does exist is the particle approximation in which we acknowledge that extended bodies are not rigid, that rotational etc. degrees of motion exist but that we can often enough approximate it by the COM motion. Another doorstop. $\endgroup$
    – CuriousOne
    Jul 19, 2016 at 7:20
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    $\begingroup$ @CuriousOne: So if I understood correctly, the "particle approximation" says that the motion of extended bodies can be studied by studying the motion of their center of mass. Tell me if I got wrong somewhere. And this approximation is valid only when "all mass elements of the extended body are moving at the same velocity and that they are experiencing the same acceleration", right? This seems to be valid in my case! So is it that the book is actually using this "particle approximation" here? $\endgroup$
    – seavoyage
    Jul 19, 2016 at 16:08
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    $\begingroup$ The justification comes, as always, with comparison to experiment. For example, we usually model our atmosphere as a vacuum. That is, we ignore air resistance. Whether or not this is a good model depends on how accurate your measuring equipment is, and what your goals are. For university physics labs, it's usually justified. $\endgroup$
    – garyp
    Jul 19, 2016 at 16:15

2 Answers 2

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I think you are comparing two pretty different cases. Motion of the Earth with respect to the Sun (and vise versa) is different from motion of a block on a table with respect to it. In the former, both of the Earth and Sun experience rotational motion and cannot be assumed as a particle without approximation error. But in the later, the block doesn't experience rotational motion and thus can be exactly substituted with a particle in the COM and with the same mass.

In the first case (Earth and Sun), we neglect rotational motion affects. But in the second case (block), we neglect nothing because there is no affects due to rotational motion.

I think the book are talking about approximation not modeling.

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  • $\begingroup$ But it says that the planetary motion can well be accounted for by the point particle approximation, even when there is rotational motion, right? $\endgroup$
    – seavoyage
    Jul 19, 2016 at 16:25
  • $\begingroup$ @AkashBajaj Yes, but there will be an error that is negligible. $\endgroup$
    – lucas
    Jul 19, 2016 at 16:28
  • $\begingroup$ So what is the reason for there being "negligible error" in the case of planetary motion and there being "no error at all" in the block's case? $\endgroup$
    – seavoyage
    Jul 19, 2016 at 16:30
  • $\begingroup$ @AkashBajaj The reason is rotational motion. Planets experience rotational motion, but the block doesn't. $\endgroup$
    – lucas
    Jul 19, 2016 at 16:31
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Your intuition is correct. When considering a block resting on the table the reaction forces do not go through the center of mass (point mass approximation location). See this answer and the accompanying diagram for a more realistic picture.

block on table

You do find though that for a rigid body, the point approximation is entirely appropriate when modeling the response of the center of mass. For example this video shows the motion of a tennis racket when thrown across a room. When viewed as a whole body the motion seems very complex, but when the lights turn off and only the center of mass is tracked the result is much simpler (a parabola).

The laws of motion state that the net force acting on a rigid body equals mass times acceleration of the center of mass. When tracking the center of mass it is not only appropriate, but required to consider all the forces acting as if they were passing through the center of mass. Only when rotations are examined the location of the forces (away from the COM) become relevant.

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