What is the pressure in the middle of a bar floating in space due to its own gravity?

Question

Imagine a long bar, floating in space away from other sources of mass. Gravity will cause the bar to try to compress due to its own mass. So what force would I measure if I cut the bar in the middle and put a force meter there? Assume the bar has length L, cross-sectional area A, and mass M (and any other constants you think you need).

To a first order approxiamtion, this is easy. Split the bar in half and treat as 2 masses separated by L/2 (so the force is GMM/((L/2)^2).

But if this is true, you should get the same answer if you split the bar in 4 parts - but actually then the force is bigger. In fact, the more bits you split the bar into, the bigger force you get.

Same problem if I try integration - I always end up with one limit being 1/0 (ie infinite) at the middle of the bar.

If anyone can resolve this for me I'd be really grateful, I've asked a bunch of clever people and no-one's solved it yet!

MooseBoys is looking the best so far .... but still no answer to my question!

I'd taken the simplification you suggested (ie assume long, thin bar ~= a line with mass per unit length). But it's where to go from here - I always seem to end up with 1 over 0 in my answer.

Answer 1

The self-gravitational pressure at the centre of a symmetric bar (which is what you are trying to measure) is exactly 0, regardless of the number of pieces.

This is the same reason that things are weightless (but NOT massless) at the centre of the earth. The gravitational pull is equal in all directions, resulting in a net force of 0.

The same applies to your bar. The actual greatest force will be at either extremity, where you have the entire mass of the bar exerting gravitational pull. As soon as you move inside the bar, the pull lessens because the mass behind is pulling against the mass ahead, until you reach the centre, where the forces cancel completely.

Answer 2

Your first order approximation is, in fact an exact solution, but you forgot to divide each mass in half. The force comes out to be f = GM²/L².

If you split the bar at some location other than the middle, say at a fraction K from 1 end, you wind up with
f = GM²K(1-K)/((K/2 +(1-K)/2)² *L²)
which has a maximum at K = .5 and is zero at K = 1.0
.........exactly what you'd expect..........
Since the force is obviously zero at the ends, the force decreases from a maximum at the center to zero at the ends. This does not mean the gravitational field there is zero, just that the attracted mass (M*(1-K)) is going to zero.

I have no idea how you get a bigger force by splitting the bar into more pieces......

Since you specify pressure, not force, the above values of f need only to be divided by the x-sectional area of the bar.

Answer 3

ianmacpherson55, you are correct that there will be zero net force, however there will certainly be a nonzero pressure. Just look at Jupiter. The pressure at the center of the planet would crush your bones to powder.

One important thing you must realize about pressure is it is force per area. The pressure of a point with a normal vector parallel with the bar will be much higher than the pressure of that same point with a normal vector perpendicular to the bar. The only reason gas pressures do not need to specify a normal is because the pressure is uniform, regardless of the normal.
In order to calculate pressure at the center of the bar, you will need to take the limit of a 5-layer embedded integral. First, you will be taking the integral of all net forces (dot) normal vectors, then integrating all of that across a plane, whose surface area is approaching zero. For force calculation, this is not the force of gravity acting on the point, but rather the force of gravity of a non-center point due to all the other non-center points. In reality, this will be an 8-layer integral.
You can simplify the answer by assuming roughly uniform pressure across a cross-section of the rod (this will be true only if the rod is much longer than it is thick). And can further approximate by transforming a rod into equivalent mass line, with density units of kg/m rather than kg/m^3.

Answer 4

Don't forget that in your first approximation, when you split the bar in two, each mass is also half the total.

Answer 5

The integration needed seems to be covered in the reference web page.

Answer 6

When you cut the bar in half you ruined the experiment.