# Homework Help: Gravitational force on infinitely long rod

1. Oct 30, 2007

### kaze

A infinitely long rod, $$A$$, with linear mass density, say $$p$$, is placed along the z-axis. Another thin rod, $$B$$, of length $$L$$ and mass density, say $$p_{2}$$, is placed orthogonally to $$A$$ on the x-axis from $$x=a$$ to $$x=a+L$$. What is the gravitational force experienced by rod $$B$$ from rod $$A$$?
I have solved a similar problem which asks for the gravitational potential given a rod of finite length and a point a perpendicular distance away from the rod's midpoint (thornton and marion problem 5-7). I am trying to extend this but not having much luck. So if we were trying to find the gravitational potential of the end point on rod $$B$$ closest to $$A$$ we would get:

$$\phi =-G \int_{A} \frac{p}{r}ds =- p G \int_{A} \frac{1}{\sqrt{x^2+a^2}} dx = - p G \lim_{z\rightarrow \infty} \int_{-z}^{z} \frac{1}{\sqrt{x^2+a^2}} dx = - p G \lim_{z \rightarrow \infty} \ln(x+\sqrt{x^2+a^2})|_{-z}^{z}$$

This already goes to infinity. I was thinking that I could make $$\phi$$ a function of $$a$$ and then once I find a function for the gravitational potential, I'll integrate over rod $$B$$ to get the entire potential and hence the force by $$F=-\nabla U = -\nabla (m U ) = -m \nabla U$$

However, I guess I am stuck at step one since I cannot get the integral to work out. If you have any suggestions, I'd really appreciate the help. Thanks.

2. Oct 30, 2007

### rbj

the gravitational force experienced by rod B from rod A will be a finite force per unit length of rod B. since rod B is infinitely long, it would be an infinite force (acting on an infinite mass, and the attractive acceleration would be finite).

edit: WOW, this is my PF post #1000! a sorta milestone.

3. Oct 30, 2007

### Dick

But B isn't infinitely long. You shouldn't be discouraged by infinite potentials if their derivative is finite. Congrats rbj!