# Gravitation force caused by uniform rod

1. May 9, 2007

### joex444

1. The problem statement, all variables and given/known data
What is the gravitational force caused by a thin uniform rod of length L on a point mass located perpendicular to the rod at it's center? Assume the point mass is a distance R perpendicular to the rod.

2. Relevant equations
$$g = - \frac{GM}{r^2}\hat{r} = - G\rho \int \frac{dV}{r^2}\hat{r}$$

3. The attempt at a solution

$$g = -G\rho \int \frac{Rdx}{\sqrt{x^2+R^2}^3}$$
$$x = Rtan(\phi)$$
$$g = -\frac{G\rho}{R} * ( \frac{x}{\sqrt{x^2+R^2}} )$$
Using limits of integration from -L/2 to +L/2 to yield:
$$g=-\frac{2G\rho}{R}*(L/\sqrt{L^2+4d^2})$$ after removing a factor of 4 from the (L/2)^2 from the square root.

Since the y components cancel, I took theta to be the angle nearest the point mass, and called rhat cos(theta) which I evaluated to R/sqrt(x^2+R^2) which could combine with the r^2 in the denominator from the gravity equation.

If this is wrong, I'll be glad to provide more steps to see where I went wrong. I ended up with a 1/(x^2+R^2)^(3/2) in the integral, so I did a trig substition to integrate it, and came up with the formula wikipedia has in its irrational integrals table, so I'm pretty sure that's alright, but wasn't sure if I set it up right.

Last edited: May 9, 2007
2. May 10, 2007

### maverick280857

Looks okay to me. This is how I did it (I get the same integrand form)

$$d\vec{f} = \frac{Gdm}{R^2+x^2}\left(-\cos\theta \hat{i} + \sin\theta \hat{j}\right)$$

where

$$\cos\theta = \frac{R}{\sqrt{R^2+x^2}}$$
$$\sin\theta = \frac{x}{\sqrt{R^2+x^2}}$$

The integrand over $\hat{i}$ is an odd, so it integrates to zero ($x$ ranges from $-L/2$ to $L/2$). You get

$$\vec{f} = -\hat{j}G\lambda R\int_{-L/2}^{L/2}\frac{1}{(x^2+R^2)^{3/2}}dx$$

Last edited: May 10, 2007