I'm just thinking about the qualitative and quantitative implications of a hypothetical orbital problem, but what I ended up does not make sense as it should.

In particular, assume that two objects are placed at two close circular orbits, and they are also connected to each other by a rigid tether-like stuff, whose length is $l$, as the following sketch depicts.

enter image description here

I assume that the distance of the upper (resp. lower) object from the center of the Earth is $r'$ (resp. $r$). What I am interested is the study of the overall stability of this system in view of the overtake rate of the object on the lower orbit (we know that the lower-orbit object is faster, thereby overtaking its peer over the passage of time).

Here is my analysis:

The overtake can be written as

$\Delta x = (v_{d}-v_{u})t.$

Given the gravitational parameter $\mu$, we have

$v_d = \sqrt{\dfrac{\mu}{r}},$

$v_u = \sqrt{\dfrac{\mu}{r + l \sin \theta}},$

which end up with

$\Delta x = \sqrt{\mu} (\dfrac{1}{\sqrt{r}} - \dfrac{1}{\sqrt{r + l \sin \theta}})t$.

I'm particularly interested in the $\dot{\Delta x} = 0$. Thus given $\theta = \theta(t)$, the total differentiation role reads

$\dot{\Delta x} = \dfrac{\partial \Delta x}{\partial t}\dfrac{\partial t}{\partial t}+ \dfrac{\partial \Delta x}{\partial \theta}\dfrac{\partial \theta}{\partial t}.$

In particular, one obtains

$\dfrac{\partial \Delta x}{\partial t} = \sqrt{\mu} (\dfrac{1}{\sqrt{r}} - \dfrac{1}{\sqrt{r + l \sin \theta}}),$


$\dfrac{\partial \Delta x}{\partial \theta} = \sqrt{\mu}(\dfrac{1}{2}l\dot{\theta}\cos \theta (r + l \sin \theta)^{-\dfrac{3}{2}}) t$.

Thus, applying $\dot{\Delta x} = 0$ condition yields the following dynamics corresponding to $\theta$.

$\dot{\theta} = \dfrac{2(r + l \sin \theta)^{\dfrac{3}{2}})[\sqrt{\dfrac{1}{r + l \sin \theta}}-\sqrt{\dfrac{1}{r}}]}{lt\cos \theta}$

Now, time for numerically integrating the equation above. In particular, the following Python script

import numpy as np
import math
from scipy.integrate import odeint
import matplotlib.pyplot as plt

def f(s,t):
    l = 20
    h = s[0]
    r = 1000
    dhdt = 2*((l*math.sin(h)+r)**(1.5))*((1/math.sqrt(l*math.sin(h)+r))-(1/r))/(t*l*math.cos(h))
    return dhdt

t = np.linspace(0.1,2000)

s = odeint(f,s0,t)

plt.plot(t,s[:,0],'r--', linewidth=2.0)



enter image description here

The dynamics look reasonable since one expects that $\theta$ eventually converges to the equilibrium $\theta = 0$. However, that $t$, jumping up and down in the denominator of $\dot{\theta}$, looks pretty weird because it literally implies that the rate of $\theta$ is infinitely large when the system is initiated! Did I mess during my computations?!

Any comments to clarify the situation are highly appreciated.


1 Answer 1


I'm afraid that your whole approach is wrong due to starting with contradictory assumptions. If the two bodies are connected by a rigid tether, then in general, they will not move in circular orbits, since in addition to Earth's gravity, forces from the tether will act on them. (As uhoh pointed out, there are some configuration where the bodies do move in circles.)

$\Delta x = (v_{d}-v_{u})t$ would be correct only if $v_d - v_u$ stayed constant, but even in your post, this difference changes with time, since it depends on $\theta$, which changes with time. $v_d - v_u$ is the momentary rate of change of $\Delta x$, so what you should have instead is $\dot{\Delta x} = v_d − v_u$.

The formulas for $v_d$ and $v_u$ are not correct because the bodies would not in general move in circular orbits.

And if $\dot{\Delta x} = 0$ over an interval of time, then $\dot{\theta}=0$ as well, since $\Delta x = l\cos\theta$.

  • $\begingroup$ Can you share your thoughts about the following follow ups? (i) If the rigidity assumption is relaxed, then the first issue would be (at least partially) resolved, and the effect of the connector's elasticity can be studied as perturbation, right? (ii) I didn't get why $\Delta x = (v_{d}-v_{u})t$ is correct only if the velocities are constant. $\endgroup$
    – user11175
    Commented Jun 4, 2019 at 17:53
  • $\begingroup$ (iii) And if Δx˙=0, then θ˙=0 as well, since Δx=lcosθ. First, we note that the $cos$ term may also provide $\dot{\Delta x} = 0$ without any obligation for $\dot{\theta}=0$. Additionally, if the rigidity assumption is relaxed, then derivations of $l$ also appear which do not necessarily end up with a sole solution like $\dot{\theta}=0$, right? $\endgroup$
    – user11175
    Commented Jun 4, 2019 at 17:53
  • $\begingroup$ "then they will not move in circular orbits" do you mean more specifically that they wouldn't necessarily move in independent circular orbits? There can be conditions under-which they move in circles; the center of mass can be in a circular orbit and the orientation could be zenith-nadir (i.e. rotating with the same period as the orbit like the ISS does) $\endgroup$
    – uhoh
    Commented Jun 4, 2019 at 22:05
  • $\begingroup$ @Roboticist (i) Maybe, I'm not sure. (ii) $v_d-v_u$ is the momentary rate of change of $\Delta x$ (i.e., what you would have instead is $\dot{\Delta x} = v_d-v_u$); if you want to find the total change over a time interval $t$, you need to multiply $t$ by the average rate of change over that interval. If the rate of change is not constant, they are not the same thing. $\endgroup$
    – Litho
    Commented Jun 5, 2019 at 7:21
  • 2
    $\begingroup$ @uhoh Yes, I meant that in general, for almost all initial conditions, they wouldn't. Note, by the way, that even in the configuration you describe, the bodies's speeds are different from the speeds of free bodies moving in circular orbits at the same altitudes. $\endgroup$
    – Litho
    Commented Jun 5, 2019 at 7:25

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