# Gravity outside a cylinder.

1. Apr 12, 2008

### yuiop

With reference to this web page http://principles.ou.edu/earth_figure_gravity/index.html [Broken]

I calculate the force of gravity tangential to the long axis of a cylinder to be proportional to 4GM/R where M is the mass per unit length of the cylinder. Does that seem correct?

Last edited by a moderator: May 3, 2017
2. Apr 12, 2008

### pam

F is perpendicular to the long axis of a cylinder.

3. Apr 12, 2008

### ZapperZ

Staff Emeritus
No it isn't.

Did you use the Gauss's Law equivalent to calculate g?

Zz.

Last edited by a moderator: May 3, 2017
4. Apr 12, 2008

### yuiop

I was trying to use the method demonstrated on that web page using an an imaginary Gaussian surface and the integral form of Poisson's equation. M is the mass contained within the imaginary volume of the Gaussian surface. For a cylinder I assumed the mass is proportional to $$2 \pi R^2 p L$$ where L, R are length and radius of the cylinder and p is the density per unit volume of the cylinder. The integral of the Gaussian volume is $$\pi r L$$.

Using the formula given,

$$g(\pi r L) = 4\pi GM$$

$$g = \frac{4GM}{r L} = \frac{4G (2 \pi R^2 p L)}{ r L} = \frac{8 \pi G R^2 p}{r}$$

Where R and p are constants.

I am not sure if that is correct. I was just trying to interpret the method shown. I would like to know what the correct solution is.

What I am trying to establish is if the force of gravity perpendicular to the long axis of a cylinder is proportional to 1/r rather than the usual 1/(r^2) when considering a spherical massive body.

Last edited: Apr 12, 2008
5. Apr 12, 2008

### yuiop

This paper seems to agree that the force of gravity perpendicular to a cylinder is proportional to 1/R.

http://www.pgccphy.net/103/gravity.pdf

Interestingly the force equation is independent of the length of the cylinder and so applies equally to the force of gravity perpendicular to the axis of symmetry of a thin disk. According to this, an object leaving a solar system that had a lot of mass at the perimeter of the system (in the form of a disk of meteorites for example) would experience greater gravitational acceleration than GM/(R^2) where R is the total mass of the solar system.

For a disk of uniform density the force of gravity on a test particle at a radius that is less than the radius of the disk appears to be independent of the distance from the centre of the disk. For a galaxy that happened to have an ideal disk shape and an even distribution of mass would have much greater orbital velocities consistent with F= GMm. For a more realistic galaxy with less than a perfect disk shape and greater density towards the centre the orbital velocities would be consistent with GMm/R > F > GMm/(R^2).

Last edited: Apr 12, 2008
6. Apr 12, 2008

### rcgldr

For an infinitely long cylinder, gravity diminishes relative to 1/r. For a infinitely large plane, gravity remains constant regardless of position. A long cylinder or a large plane closely approximate this if distances from the cylinder or plane are relatively small compared to the size of the cylinder or plane.

7. Apr 14, 2008

### yuiop

As an extension to the question in the OP what is the gravitational potential for a hypothetical body that has a gravitational field that attenuates relative to 1/r ?

8. Apr 16, 2008

### yuiop

I read somewhere that there may be difficulties with using the gaussian method to determine the force of gravity. Is it possible to calculate the gravity/distance relationship for a flat slab by another method such as integrating the total forces of infinitesimal "point" masses that make up up the slab?

I believe Newton used a similar method to calculate the gravitational forces of spherical shell when he concluded that the gravity of a large sphere is equivalent to that of a point particle of the same mass.