# Gravitational force exerted by a rod on a point mass

I have doubts with the solution of a certain problem. I will give the entire solution below and will lay out my doubts as well.

A point mass $m_1$ is separated by a distance $r$ from a long rod of mass $m_2$ and length $L$.The objective is to find the total gravitational force exerted by the rod on the point mass.

This is how a particular author solved this question in a book.

The total mass of the rod was differentiated with respect to the total length of the rod, and each mass piece was called $dm$ and each length piece was called $dx$. Hence this equation was formulated:

$\large \frac{m_2}{L}\ =\ \frac{dm}{dx}$

Also, the distance between the point mass and each individual $dm$ piece was taken as $x$. Then the gravitational force between the point mass and each mass piece is given by:

$F\ =\ \large \frac{Gm_1dm}{x^2}$

$dm$ was substituted using the first equation, now the new equation becomes:

$F\ =\ \large \frac{Gm_1m_2}{Lx^2}\small dx$

This is then integrated from $x\ =\ r$ to $x\ =\ r + L$.

My question is this: how can we integrate $x$ with respect to $dx$?
$dx$ represents the tiny length pieces of the rod, and $x$ represents the distance from the centre of the point mass to any point along the rod. How can we integrate $x$ with respect to $dx$, it does not make any physical sense to me? I am new to differentiation and integration, but I understand the basics well enough. If there is something wrong with this solution please tell me the right way to solve this problem, otherwise tell me how this solution makes sense.

• You are right. One cannot integrate $x$ wit respect to $dx$. Maybe you are misinterpreting the author. A link to that solution would help. Commented Sep 7, 2013 at 14:54
• @udiboy. It is not on a site, it is from one of the coaching center study materials. The final answer comes out to be $F\ =\ \frac{Gm_1m_2}{r(L + r)}$. Is this what you would get if you solved the problem the right way? Commented Sep 7, 2013 at 17:21

I don't think that you really understand integration. Let me clear this up for you. In that question there is a rod of length l. You know how to calculate gravitational force between two point masses but not in continuous mass bodies.

If you apply the formula to find the gravitational force you don't know what to take the distance as because it is continuous body. Let us apply the formula $F\ =\ \large \frac{Gm_1m_2}{r^2}$ by taking r as the distance between the two bodies. Obviously we get a wrong answer. Now let us consider the rod to be made up of two bodies each of mass $m_2/2$ an length $l/2$ . Now we can define the force as $F\ =\ \large \frac{Gm_1m_2/2}{r^2} + \frac{Gm_1m_2/2}{(r+l/2)^2}$ . Again we get a wrong answer. Now let us divide it into $N$ parts. Now the force can be represented as $F\ =\ \large \frac{Gm_1m_2/N}{r^2} + \frac{Gm_1m_2/N}{(r+l/N)^2} +\frac{Gm_1m_2/N}{(r+2l/N)^2}+...... \frac{Gm_1m_2/N}{(r+(N-1)l/N)^2}+\frac{Gm_1m_2/N}{(r+l)^2}$ Or $F\ =\ \large \frac{Gm_1m_2}{N}[\frac{1}{(r)^2} +\frac{1}{(r+l/N)^2} + \frac{1}{(r+2l/N)^2} + .....\frac{1}{(r+(N-1)l/N)^2}+\frac{1}{(r+l)^2}]$ Now if we increase the value of $N$ we get a more accurate answer as the divisions become more smaller and more like point masses. Now if we make the value of $N$ very very high then we would get an accurate answer. This is where differentiation ad integration comes in. $\large \frac{m_2}{N}$ represents $dm$ and $\large \frac{l}{N}$ represents $dx$. Now you know why $dm\ =\large \frac{m_2dx}{l}$

Let us now apply this in the above expression. So now we have .. $F\ =\ \large \frac{Gm_1m_2dx}{l}[\frac{1}{(r)^2} +\frac{1}{(r+dx)^2} + \frac{1}{(r+2dx)^2} + \frac{1}{(r+3dx)^2} + .....+\frac{1}{(r+l)^2}]$ This is same as integrating $\ \large \frac{Gm_1m_2dx}{lx}$ from $x=r$ to $x=r+l$
$Note:$ Every dm mass is not at the same distance from the point mass. We can integrate $x$ with respect to $dx$ as $dx$ is a small change in $x$ . You add all the values and you add $dx$ to $x$ in every next value. Hope this helps