I previously mentioned the problem of the catenary shape in a uniform gravitational field in the "Natural Extremes" series of articles, and in that post, I posed a question to the readers: what would the shape of a catenary be in the gravitational field of a point mass (Earth)? To be honest, when I posed this question, I did not yet know how to solve it, but now I do. Looking back, several months have passed; time flies...
Following the same logic as before, we still adopt the "Axiom of Equilibrium," which states that the total potential energy is minimized. From celestial mechanics, we know that the potential energy between any two point masses is -\frac{Gm_1 m_2}{r}. For the catenary problem in this case, we can place the Earth at the origin of the coordinates, and let the two fixed points of the catenary be (x_1,y_1) and (x_2,y_2), with the total length of the chain being l. That is: \int_{x_1}^{x_2} \sqrt{dx^2+dy^2}=l
Assume the linear density of the catenary is 1 (dimensionless treatment, where length corresponds to mass). Then the gravitational potential energy of a small segment of the catenary from point (x,y) to (x+dx,y+dy) can be written as: dE=-GM\frac{\sqrt{dx^2+dy^2}}{\sqrt{x^2+y^2}}
The total potential energy is clearly: E=-GM\int_{x_1}^{x_2} \frac{\sqrt{dx^2+dy^2}}{\sqrt{x^2+y^2}}
It is necessary to select appropriate coordinates to simplify the integrand; we adopt a polar coordinate system: \frac{\sqrt{dx^2+dy^2}}{\sqrt{x^2+y^2}}=\frac{\sqrt{r^2 d\theta^2+dr^2}}{r}=\frac{\sqrt{r^2 \dot{\theta}^2+1}}{r}dr
where \dot{\theta}=\frac{d\theta}{dr}.
Thus, E=-GM\int_{r_1}^{r_2} \frac{\sqrt{r^2 \dot{\theta}^2+1}}{r}dr=-GM\int_{r_1}^{r_2} F(\dot{\theta},r)dr
Substituting into the Lagrange equation (treating r as the independent variable and \theta as the function), we obtain: 0=\frac{\partial F }{\partial \theta}=\frac{d}{dr}\left(\frac{\partial F }{\partial \dot{\theta}}\right)=\frac{d}{dr}\left(\frac{r\dot{\theta}}{\sqrt{r^2 \dot{\theta}^2+1}}\right)
Integrating, we obtain \frac{r\dot{\theta}}{\sqrt{r^2 \dot{\theta}^2+1}}=Const.
And simplifying, we get: r^2 \dot{\theta}^2=C_1^2 \quad ( C_1 > 0 )
Requiring C_1 > 0 is merely for convenience of description, i.e., \pm \dot{\theta}=\frac{C_1}{r} \Rightarrow C_2 \pm \theta=C_1 \ln r
We adopt the convention that when \theta=C_1 \ln r-C_2 < 0, the negative sign is taken above, and vice versa; then it can be written as: C_2 + |\theta|=C_1 \ln r
It can be further simplified into the form r=Ae^{B|\theta|}. Its shape is shown in the figure below: