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So we have a uniform rod of mass M and length L and we have placed a point mass of mass m, which is distance a away from the midpoint of the rod.

Now setting the coordiates axes to work with, I put the rod along the Y axis with its mid point coinciding with the Origin while the point mass m is put along the X axis a distance units away from the rod.

The point mass is P, and a differential mass element dm of the rod situated at y=length(OQ) where Q is the position of the mass element, is taken, and its corresponding force was found to be

$ \displaystyle{ d \vec{F} = \frac{G m dm} {{y}^{2}+{a}^{2}} (- cos \theta \hat{i} + sin \theta \hat{j}) } $

where $\theta$ is the angle OPQ, Now using the linear mass density,

$ \displaystyle{ \frac{dm}{dx}=\frac{M}{L} } $

and writing $cos \theta$ and $ sin \theta $ in terms of y,

$\displaystyle{ cos \theta = \frac{a}{\sqrt{{y}^{2}+{a}^{2}}} } $

and

$\displaystyle{ sin \theta = \frac{y}{\sqrt{{y}^{2}+{a}^{2}}} }$

the Forces comes out to be,

$\displaystyle{ {F}_{x} = - \frac{GmMa}{L} \int _{-\frac{L}{2}}^{\frac{L}{2}} { \frac{dy} {{({y}^{2}+{a}^{2})}^{\frac{3}{2}}} } } $

and,

$\displaystyle{ {F}_{y} = \frac{GmM}{L} \int _{-\frac{L}{2}}^{\frac{L}{2}} { \frac{y dy} {{({y}^{2}+{a}^{2})}^{\frac{3}{2}}} } } $

Which gave the following results:

$\displaystyle{ {F}_{y} = 0 } $

as expected, due to the symmetry.

and,

$\displaystyle{ {F}_{x} = - \frac{GmM}{a \sqrt{{a}^{2} + {(\frac{L}{2})}^{2}}} } $

which I didn't expect to come out since I believe that if the it was assumed that the whole mass of the rod is concentrated at its mid point then the force should have been

$\displaystyle{ {F}_{x} = - \frac{GmM}{{a}^{2}} } $

So I am curious as to why or why not this result should follow.

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  • $\begingroup$ I've added the homework-and-exercises tag. In the future, please use this tag on this type of question. $\endgroup$ – Ben Crowell Feb 16 at 18:42
  • $\begingroup$ @BenCrowell ok i will 👍 $\endgroup$ – Aditya Feb 16 at 18:56
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Newton invented integral calculus to prove the point mass premise for spherical objects. Gravitational force as a point mass is in general not valid for other shapes, although it is approximately true at large distances. At $a>>L$, your result approaches what you would expect.

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  • $\begingroup$ Take a look at :mathpages.com/home/kmath139/kmath139.htm "There is so strong an objection against the accurateness of this proposition that without my demonstrations... it cannot be believed by a judicious philosopher to be any where near accurate." $\endgroup$ – Vincent Fraticelli Feb 16 at 18:58
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Consider two forces of magnitude $F$ pulling on an object. If they pull at an angle $\theta$ to each other, the net magnitude will be $\sqrt{(F + F\cos\theta)^2 + (F\sin\theta)^2} = F\sqrt{(1 + \cos\theta)^2 + \sin^2\theta} = F\sqrt{2+2\cos\theta}$. As you can see, as the angle of separation increases, the force decreases. If the whole mass of the rod is concentrated at the midpoint, then the gravitational pull is all in one direction, leading to the greatest force. However, if the mass is spread out as in your scenario, gravitational pulls from different parts of the rod will be at different angles, making the net force weaker. In addition, some of the mass of the rod will be farther away from the point mass, which will weaken the pull as well. Sure enough, $$-\frac{GmM}{a\sqrt{a^2 + \big(\frac{L}{2}\big)^2}}$$ has a smaller absolute value than $$-\frac{GmM}{a^2}$$ because the denominator is larger when $L > 0$. As $L$ increases, the fraction will further decrease in magnitude because the gravitational pulls are at greater angles and portions of the mass are at even greater distances. So, this result makes perfect sense.

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