When is the Earth, for example, not considered a point in relation to gravity?

I am thinking that if one was asking a question below the crust continously to the core, then this would be an example.

Is there a question that would motivate a different set of answers.

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    $\begingroup$ Is your question inspired by Newton's shell theorem? $\endgroup$ – Qmechanic Feb 7 at 12:21
  • $\begingroup$ Another example is when considering tidal forces. $\endgroup$ – Lewis Miller Feb 7 at 13:30
  • $\begingroup$ My question is related to (1), that for objects outside the shell, the mass could be considered to be concentrated in the center of the sphere. However, I am working through the kinematics chapter of Halliday/Resnick 2nd ed., and a question was asked about when objects should be modeled as points versus “composite bodies”[mine]. I could not grasp the question fully because the terms I am familiar with relate to how to model a radiation source as a point versus field. $\endgroup$ – bentaisan Feb 7 at 13:43

An object is considered as a particle whenever this gives predictions of sufficient accuracy for whatever one is trying to understand. In order to check the accuracy of the predictions given by the point particle model, one does a rough estimate of what a more complete model would say.

For example, in the case of the Earth, for positions at depth $d$ below the surface, a reasonable way to estimate the acceleration due to gravity is to treat the Earth as if it had constant density and was spherical. In this case the acceleration due to gravity is only owing to the material up to radius $R-d$, where $R$ is the radius of the Earth. So we have:

point particle model of Earth (so the whole mass contributes) says the acceleration due to gravity at depth $d$ is $$ g_p = \frac{G M}{(R-d)^2} $$ where $M$ is the mass of the Earth.

The better model says $$ g = \frac{G M (R-d)^3/R^3}{(R-d)^2} $$ because $M(R-d)^3 / R^3$ is the amount of mass in the Earth up to radius $(R-d)$. This better model is still not perfect, of course, but we can use it to judge how bad the point particle model is. To do this, we compare the two answers: $$ \frac{g}{g_p} = \frac{(R-d)^3}{R^3} $$ This can also be written $$ \frac{g}{g_p} = \frac{(R-d)^3}{R^3} = \left( 1 - \frac{d}{R} \right)^3 = 1 - 3 \frac{d}{R} + 3 \frac{d^2}{R^2} - \frac{d^3}{R^3} \simeq 1 - 3 \frac{d}{R} $$ where the last step is accurate when $d \ll R$. Thus we find, for example, that when you are at a depth of 1 percent of the radius of the Earth, then the point particle model over-estimates the acceleration due to gravity by about 3 percent. At a depth of 10 percent of the radius of the Earth, the point particle model becomes very inaccurate.

The above example treating the Earth's gravity can be seen as an illustration of what more generally one has to do when assessing models in physics. For any given model, one has to ask oneself what has been left out or ignored. If possible, you should develop a model which you have good reason to think is more accurate, and then you can assess the accuracy of the first, simpler, model, by a rough (or "quick-and-dirty") comparison with the more accurate model.

  • $\begingroup$ Just to make sure, but is the << symbol mean “floor”? $\endgroup$ – bentaisan Feb 7 at 13:49
  • $\begingroup$ That symbols means "is much less than" or "is much smaller than". $\endgroup$ – Andrew Steane Feb 7 at 14:11

If your object is planet Earth the main macroscopic effect you couldn't explain treating it as a point is precession. In a simplified model of Earth as a gyroscopic body (endowed with a polar axis of symmetry) precession rate is proportional to $(C-A)/C$, where $C$ is the polar moment of inertia and $A$ the equatorial one ($C>A$ for a flatened Earth). In the limit of a pointlike Earth moments of inertia lose meaning (or all vanish, if you like).

Note that in my answer the gravitational field of interest isn't Earth's. Precession is due to the action of Sun's and Moon's fields on Earth. If Earth had an exact spherical symmetry it would behave like a point mass. But it's flattened, and that makes the difference.

  • $\begingroup$ I seem to recall that precession has epicycles. Obviously a "wobbling" axis can't be considered a point because by definition it is an imaginary line. I also think that the Jakarta earthquake put an additional mode in the precession epicycles. $\endgroup$ – bentaisan Feb 7 at 20:48

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