Gravitational field due to a cylinder

[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?

 

[mfb]

You also need an integral along the length of the cylinder, as the force doesn't only come from one disc of the cylinder.

 

[Steve4Physics]

Is this right?

From your working it looks like you are dividing the cylinder into a set of co-axial cylindrical shells, finding the field from a single shell and then integrating the fields from the shells.

But your calculation is wrong. You have ignored the fact that different parts of a cylindrical shell are different distances from the point (P) where the field is required. You can’t simply use the distance from the shell’s centre to P (because the field does not vary linearly with distance).

Consider an alternative approach:
- find the field from a thin ring;
- use this to find the field from a thin disc;
- use this to find the field from the whole cylinder