# Gravitational field due to a cylinder

1. Apr 10, 2005

### asrodan

Calculate the gravitational field due to a homogeneous cylinder at an exterior point on the axis of the cylinder. Perform the calculation by computing the force directly.

I'm not sure if I did this right, here's what I did

z = distance from point to center of cylinder
r = distance from point to vertical center of cylindrical shell element
a = radius of shell element

(these three lines form a triangle with r as the hypotenuse)

$$\theta$$ = angle between r and z
R = radius of cylinder M = mass of cylinder $$\rho$$ = density of cylinder V = volume of cylinder L = length of cylinder

$$dg = \frac {-GdM} {r^2} = \frac {-G \rho dV} {r^2}$$

$$dV = 2 \pi aL da$$

$$g_x = g_y = 0$$ $$g_z = gcos( \theta)$$

$$cos( \theta) = \frac {z} { \sqrt {z^2 + a^2}}$$

$$g = \int_{0}^{R} dgcos( \theta) = -2 \pi G \rho l \int _{0}^{R} \frac {a z da} {(z^2 + a^2)^{3/2}}$$

$$g = 2 \pi G \rho lz ( \frac {1} { \sqrt {z^2 + a^2}})_{0}^{R}$$

$$g = 2 \pi G \rho l (\frac {z} { \sqrt {z^2 + R^2}} - 1)$$

Is this right?