

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer

The curve for a chain hanging between two poles in a uniform gravitational field is known as the catenary.

Is there known an expression for the curve of a hanging chain on a planet of mass $M$ which has a gravitational field of $M/r^2$ ?

Let the rope have length $L$, uniform mass-density $\rho$ and negligible thickness. The poles are a distance $D$ apart, each a distance $R$ from the planets centre.

I am interested in solutions where the gravitational field may be taken to point straight down as well as when it point towards the center of the planet.

share|cite|improve this question
Note that near the surface of the Earth $r \approx 6400\text{ km}$. How long and high do you intend to have the arc be? How big a deviation from constant gravity will be observed? Do you think the effect will be larger or small then the size of the links? Then imperfections in the links? – dmckee Mar 29 '11 at 2:10
@dmckee There is no links or anything, its an ideal rope, this is a problem in theoretical physics – user1708 Mar 29 '11 at 3:47
Interesting solution when $L=2R$. – Carl Brannen Mar 29 '11 at 3:47
up vote 7 down vote accepted

This should be possible to solve in the same way we do the ordinary catenary problem, by variational calculus.

Suppose the angular separation between the endpoints is $\Delta$, where we could define $\Delta = \frac{D}{R}$ if I understand the problem correctly. Let the shape of the rope be given by a function $r(\phi)$, and write the potential energy of the rope as

$$U_G = \int_0^\Delta \frac{GM}{r}\lambda\sqrt{r'^2 + r^2}\mathrm{d}\phi$$

Here $\lambda$ is the linear mass density of the rope, and I've left the $\phi$ dependence of $r(\phi)$ implicit.

We also need a constraint that enforces the total length of the rope remain fixed at $L$. According to the method of Lagrange multipliers, we'll have to add a term

$$U_L = \int_0^\Delta l\sqrt{r'^2 + r^2}\mathrm{d}\phi$$

for a total of

$$U = \int_0^\Delta \biggl(\frac{GM}{r}\lambda + l\biggr)\sqrt{r'^2 + r^2}\mathrm{d}\phi$$

If we define $T$ to be the integrand,

$$T(r, r', \phi) = \biggl(\frac{GM}{r}\lambda + l\biggr)\sqrt{r'^2 + r^2}$$

this can be solved by the Euler-Lagrange equation,

$$\frac{\mathrm{d}}{\mathrm{d}\phi}\frac{\partial T}{\partial r'} = \frac{\partial T}{\partial r}$$

which gives

$$(r''r - r'^2)(GM\lambda + lr) = lr(r'^2 + r^2)$$

There does appear to be an analytic solution to this equation, but Mathematica makes it look rather icky to say the least... I'll have to come back to this and try to find some sensible solution and/or physical interpretation since I'm out of time right now.

share|cite|improve this answer

We have had a related problem with the brachistochrone problem in Schwarzschild coordinates here. This problem here has struck me as rather odd, for this seems to refer to geodesics, where to me the . However, it has occurred to me there might be a problem here of some relevancy. Suppose we have a string in some orbit around a black hole. We then impose some condition that this string has end points attached to a fixed surface of some kind --- a Dp-brane? This string is then in an orbit around the black hole and will assume some configuration in the curved spacetime.

So to set this up we start with the line element, which gives the unit 4-velocity, $$ 1~=~\Big(1~-~\frac{R}{r}\Big) \Big(\frac{dt}{ds}\Big)^2~-~ \Big(1~-~\frac{R}{r}\Big)^{-1} \Big(\frac{dr}{ds}\Big)^2~-~r^2\Big(\frac{d\theta}{ds}\Big)^2~-~r^2sin^2\theta\Big(\frac{d\phi}{ds}\Big)^2, $$ where $R~=~2GM/r$. We identify the $dt/ds$ with the energy $E$ and do some algebra to obtain an equation of motion $$ \Big(\frac{dr}{ds}\Big)^2~=~\frac{E}{m}~-~1~-~V(r) $$ for the potential function $$ V(r)~\simeq~\frac{1}{2}\Big(\frac{R}{r}~-~\frac{L^2}{2mr^2}~+~\frac{R}{mr^2}\Big) $$ The $L$ is the angular momentum of the object in orbit.

We now use this in the potential for the computation of the shape of the string $$ U_G~=~\int_0^\Delta V(r)\lambda\sqrt{r'^2~+~r^2}d\phi $$ $\Delta~=~D/r$ and $r’~=~dr/d\phi$ and employ the Lagrange multiplier $U_L$. The Euler-Lagrange equations are applied to the $$ {\cal L}~=~(V(r)~+~L)\sqrt{r’^2~+~r^2} $$

This is somewhat artificial for it assumes the string endpoints are held apart by some fixed distance. This could be relaxed by treating the endpoints are particle geodesic or orbits. In that case the angular separation of the fixed points varies and the problem is time dependent. This might be an interesting tool to use in some problems with strings and black holes.

share|cite|improve this answer

The explicit form of catenaries in a central force field proportional to a power of $r$ has been known since the 19th centruay. They are the so-called MacLaurin spirals which were discovered much earlier by this Scottish mathematician who showed that they were orbits of planets moving in a $r^\alpha$ field. They are the curves with polar equations $r^n \cos (n \theta) =1$ and their dilations and rotations (the index $n$ is related in a simple manner to the powers of $r$ involved). This is contained in the classic treatise on special curves by Teixeira Gomes. Unfortunately, I do not have access to a copy at the moment and so cannot give a more precise reference. A modern treatment can be found in the arXiv article 1102.1579 where it is shown that these and many other properties have their origin in the fact that the curves have equations of the form $rf(\theta)=1$ where $f$ is such that $f^2+f'^2$ is proportional to a power of $f$. In this paper, a further class of curves is introduced---the MacLaurin catenaries. They are characterised by the fact that they have parametrisations of the form $(F(t),f(t))$ where $f$ is as above and $F$ is a primitive of $f$. These have similar properties, now with respect to parallel forces proportional to a power of $y$. They thus generalise the catenaries---hence the name.

share|cite|improve this answer

Your Answer


