Gravitational field due to a cylinder

In summary, to calculate the gravitational field due to a homogeneous cylinder at an exterior point on the axis of the cylinder, one can divide the cylinder into co-axial cylindrical shells and integrate the fields from these shells. However, it is important to consider the varying distances from each shell to the point where the field is required. An alternative approach is to find the field from a thin ring, then use it to find the field from a thin disc, and finally use this to find the field from the whole cylinder.
  • #1
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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)

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


[tex] dg = \frac {-GdM} {r^2} = \frac {-G \rho dV} {r^2} [/tex]

[tex] dV = 2 \pi aL da [/tex]

[tex] g_x = g_y = 0 [/tex] [tex] g_z = gcos( \theta) [/tex]

[tex] cos( \theta) = \frac {z} { \sqrt {z^2 + a^2}} [/tex]

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

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

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


Is this right?
 
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  • #2
You also need an integral along the length of the cylinder, as the force doesn't only come from one disc of the cylinder.
 
  • #3
asrodan said:
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
 
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What is the formula for the gravitational field due to a cylinder?

The formula for the gravitational field due to a cylinder is G * M * h / (a^2 + h^2)^(3/2), where G is the gravitational constant, M is the mass of the cylinder, h is the distance from the center of the cylinder, and a is the radius of the cylinder.

How does the mass of the cylinder affect the gravitational field?

The mass of the cylinder directly affects the strength of the gravitational field. The greater the mass, the stronger the gravitational field will be.

What happens to the gravitational field as the distance from the cylinder increases?

The gravitational field decreases as the distance from the cylinder increases. This is because the force of gravity is inversely proportional to the square of the distance between two objects.

Can the gravitational field due to a cylinder be negative?

No, the gravitational field due to a cylinder cannot be negative. The direction of the gravitational field is always towards the center of the cylinder, so it will always be a positive value.

How does the radius of the cylinder affect the gravitational field?

The radius of the cylinder has a direct effect on the gravitational field. As the radius increases, the gravitational field decreases. This is because the gravitational force is spread out over a larger area as the radius increases, resulting in a weaker force.

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