The question I want to ask is what point is there in using $$ F = G {m_1 m_2 \over r^2} $$ when we don't talk about point masses, but one of the masses is a sphere of radius r.

I'm currently not good enough at maths to calculate that myself.

In particular, let's take a vector form for point of mass: $$ \overrightarrow{F} = G {m_1 m_2 \over |r|^2 } {\overrightarrow{r} \over |r|} $$ and a differential form, (okay, bear with me, this is probably incorrect already: ) $$ d\overrightarrow{F} = G {\rho_1 m_2 \over |r|^2 } {\overrightarrow{r} \over |r|} dV $$

Now we should integrate the vectors over volume V, which is a sphere with center at distance r from the point of our calculations.

How would I go about such calculations?

If the scope of maths involved exceeds what's welcome on physics.se, could you help me rephrase the question for mathematics.se, or point me to resources that would let me solve this?

(assume uniform distribution of mass - asteroids etc.)


1 Answer 1


The way you would go about calculating the gravitational forces from a sphere of radius R is to start by looking at a homogeneous ring and calculating its gravitational pull on a fixed point at a certain distance from the circle's center on the rings center axis. Then you look at the gravitational forces from a spherical layer of an infinitesimal thickness, consisting of the rings from before with some infinitesimal width $\rho d\phi$ if $\rho$ is the radius of the spherical layer and $d\phi$ is the angular extent of the rings width.

You'll find that the attraction from such a spherical layer with mass $M$ is the same as for a point mass $M$ located at the center of the spherical layer. And since a filled homogeneous sphere is built up from such elementary spherical layers (each with mass $dM$) the total sphere will have the same property: for gravitational calculations with Newton's law, the sphere could just be replaced by a point mass located at its center.

The whole calculation is a bit long to write out completely here, but if this is too short, I could expand this answer. Also, if the sphere is not homogeneous, corrections need to be made. I'm not entirely sure how that's done though.

EDIT: the link provided by John Rennie explains in (much) more detail what I'm talking about.

  • $\begingroup$ I guess this, combined with the link is really sufficient for me. (huh, surprise-surprise, I thought lateral forces of a "slice of the surface" near the given point, acting much more towards the sides than towards the center (and pretty strong due to low "r^2") would completely ruin the "point" equation. I'd still love a "common sense" answer why my assumption was wrong.) $\endgroup$
    – SF.
    Commented Dec 9, 2012 at 14:32
  • $\begingroup$ @SF: Note that the field of a extended body only reduces to that of a point mass in the case of spherically symmetric distributions (or to zero for enclosing spherically symmetric distributions), so it is not that your intuition was silly, but that spherical distributions are special. $\endgroup$ Commented Dec 9, 2012 at 17:31

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