So question goes like

"What is the magnitude of gravitation force of the particle due to rod?"

In the figure the particle is of mass $m$ and from distance $d$ from end of the rod and the rod has length $l$ and has mass $M$.

In solution of the answer they use simple integration that we use normally. My question is that why can't we use concept of centre of mass?? Gravitational force is centre to centre force.. So if we assume that all the mass of rod is concentrated in centre of rod (it is uniform rod) Then the answer should be

$$ F= G \frac{ Mm}{(d+L/2)^2} $$

So where i am wrong here in applying the concept?

  • $\begingroup$ Can you show that the average force is from the center? $\endgroup$
    – JEB
    Oct 27, 2023 at 3:59
  • $\begingroup$ What kind of answer are you looking for? For many people, the fact that the two answers are different indicates that you can't use the concept of center of mass. $\endgroup$
    – Allure
    Oct 27, 2023 at 4:01
  • $\begingroup$ I am asking why are they different? Cause as far as i know concept of centre of mass should work.. $\endgroup$
    – jalok2008
    Oct 27, 2023 at 4:27
  • $\begingroup$ @Allure i am asking why are we getting wrong ans because as far as I know concept of centre of mass should work.. why it is not working.. $\endgroup$
    – jalok2008
    Oct 27, 2023 at 4:28
  • $\begingroup$ @jalok2008 that's the question though isn't it - "as far as I know concept of centre of mass should work" - since it doesn't work, you must've remembered wrong. The question becomes who/what made you think it should work? Can you describe that? $\endgroup$
    – Allure
    Oct 27, 2023 at 5:30

3 Answers 3


(This answer is based on the Newtonian model, ignoring relativity. Relativity would yield the same results, but from a very different point of view.)

Using the center of mass instead of treating every atom of an extended body individually (the integral approach) is a simplifying model that really hits the exact model only in very rare situations.

But in many, many situations it gives a very good approximation of reality. So, it's very often a useful model. As with all models, it's crucial to know its limitations. You can surely use it if the gravitational field that we are dealing with, is (nearly) constant in magnitude and direction over the whole geometry of our object.

For example, for bodies smaller than a kilometer on Earth's surface, the field variation is less than 0.01% from its average (the main factor being that gravity of a sphere decreases with 1/r²). And we know the value of Earth's gravitational field at its surface (g = 9.81m/s²). And mathematicians have shown that the gravitational field of a spherical object is exactly the same as a point-mass gravitational field (as long as you stay outside the sphere).

So, if we replace both extended objects (Earth and our 1km body) by point-masses at their center of mass, we can expect the results to deviate from the exact integral results by 0.01% or less. If that's okay, we use the center-of-mass model. If not, we use the integral.

Now for the rod: The gravitational field of a rod is significantly different from that of a point mass, unless you are very far away. So, in your given problem, there's no justification for replacing the rod with a point mass and you have to resort to the integral. Luckily, the "particle" seems to be a mass concentrated in one point, so there's no need to do the double integral over both geometries.

Gravitational force is centre to centre force.

This is your misconception. From its nature, it's a mutual force between individual particles. E.g. every atom of Earth individually attracts every atom of your body (and it attracts every other atom of Earth, as well as every atom of your body also gravitationally attracts every other atom of your body - gravity does not "know" to which body an atom belongs). Treating gravity as "centre to centre" already implies that the center-of-mass simplification is applicable, which is not the case here.

  • $\begingroup$ Thank you very much this helps alot!! $\endgroup$
    – jalok2008
    Oct 27, 2023 at 8:52
  • Centre of mass is used to isolate rotational motion from translation part in terms of CM frame. For uniform gravity, CM is equivalent to centre of gravity.

  • On the surface of our planet earth, we can regard the gravity as uniform field acting on "small" size scale objects. CG is good enough for calculating motion under the influence of gravity.

  • We can apply the shell method for spherically symmetrical object. Hence a spherical object can be regarded as a point mass locating at its own CM.

  • For large non-spherical objects up to stellar scale, you cannot simply replace everything by CG. For instance, Earth's precession and nutation are caused by the gravitational forces of the Moon and Sun acting on the non-spherical Earth.

  • Your problem is just a calculus exercise of finding the gravity of a geometrical object.

  • You may refer to some mathematical results for triaxial ellipsoid in my older post of Math SE.


To make things easier imagine that you had a massless rod of length $L$ and point masses $M$ at each end of the rod.
The centre of mass of the rod is in the middle.

Now imagine the following situation.

Point mass $m$ distance $d$ from one end of the rod and distance $d+L$ from the other ends of the rod, ie all the masses are on a straight line.

The total gravitational force on the rod due to mass $m$ is $\dfrac{GMm}{d^2} + \dfrac{GMm}{(d+L)^2}$ and this need to be equal to $\dfrac{2GMm}{(d+x)^2}$ where $x$ is the position of the centre of gravity of the rod from one end of the rod.

$\dfrac{GMm}{d^2} + \dfrac{GMm}{(d+L)^2} = \dfrac{2GMm}{(d+x)^2} \Rightarrow x= \dfrac{\sqrt 2 \,d(d+L)}{\sqrt{(d+L)^2+R^2}}-d\ne \dfrac L2$.

However, if $d=1000$ and $L=1$ then $x\approx 0.499625$, ie if $d\gg L\Rightarrow x\to \dfrac L2$, the centre of the rod, as then the rod and the masses on the ends of it finds themselves in an approximately uniform gravitational field.

If a rod with uniformly distributed mass is broadside to the mass $m$ then the parts of the rod closest to mass $m$ will have a bigger influence than the parts which are further away.


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