How would a rope around Earth fall?

Question

I have searched and saw a similar question but i dont know if it's exactly the same.

If I had a rope around earth, let's say a few meters above (and supposing a flat surface, and suddenly dropped it at the same time from all of the points, how would it fall? Or what would happen? I know gravity’s pull is not exactly the same around earth, but if it was the case?

If the rope is a few meters above ground, it's circumference is bigger than earth’s. Imagine it's a solid circumference and doesn't bend or whatever, it's pretty solid. How would it fall? It's supposed to fall all around earth at the same time but since its circumference is bigger, it can't touch surface all at the same time. If its just a rope, where does the circumference start deforming if gravity pulls equally? What parts of the rope touch the ground earlier?

supposing a flat surface The surface of the Earth isn’t flat, so what are you talking about? Do you mean a smooth, spherical surface? — G. Smith, Jul 31 '20 at 00:49
If it were suspended 1 m, it would fall in $\sqrt{2/g}$ seconds (where $g$ is the acceleration of gravity) and get $(2\pi)^{-1},\mathrm{m}$ shorter, corresponding to a compressive strain of... just about zero. What did you think would happen? — Chemomechanics, Jul 31 '20 at 05:36
Chemomechanics, i didnt know what really would happen, i hd clear it would fall but disnt know about the extra large. So it would get shorter as it falls? — Emilio Lasheras, Jul 31 '20 at 07:05
General tip: Let's not have posts look like revision histories. — Qmechanic, Jul 31 '20 at 08:34
A rope would simply fall as it has no(t much) internal strength. A solid ring of metal with enough strength might ... remain - gravity difference is not too strong. — planetmaker, Jul 31 '20 at 08:55
The rope will buckle, but don't forget that, if the rope is 1 m above the earth, it's only 6.18 m longer than the earth's circumference (40,000,000 m) so the buckling will be invisible to the naked eye. — hdhondt, Jul 31 '20 at 09:50
Does this answer your question? Build a ring around Earth, then remove the supports Also see https://physics.stackexchange.com/q/41254/123208 — PM 2Ring, Aug 01 '20 at 05:22

Tfovid · Answer 1 · 2020-07-31T09:50:16.767

0

Under idealized conditions, if the rope were instead a hula hoop that's perfectly centered around the Earth, it wouldn't fall at all but simply experience tidal forces all around its circumference. As soon as it's center displaced from that of the Earth, then it would oscillate along the line joining the two centers of gravity. If you're instead talking of a (non-rigid) rope, it would buckle at the any points where the tidal forces are greatest but its center of gravity would follow the same fate as the hula hoop as per Gauss' law for gravity.

That's the best I could come up with without getting into the math.

edited Jul 31 '20 at 09:50

answered Jul 31 '20 at 08:55

Tfovid

1,315

3

Unless I've done my math wrong, it wouldn't oscillate. Specifically, the potential energy of the rigid ring is $$U = - \frac{2GMm}{\pi R} \frac{K(4q/(1+q)^2)}{1+q}$$where $M$ and $m$ are the mass of the Earth & the hoop; $R$ is the radius of the hoop; $q = d/R$, where $d$ is the distance between the center of the Earth and the center of the hoop; and $K$ is the complete elliptic integral of the first kind. Plotting this function reveals that $q = 0$ is a maximum of the potential energy, and so the equilibrium is unstable. – Michael Seifert Jul 31 '20 at 19:48
@MichaelSeifert I just presumed that oscillatory behavior by reducing the hoop to its center of gravity, but I guess that's not a correct ansatz? – Tfovid Jul 31 '20 at 21:53
I think what's going on is that in a non-uniform gravitational field, the center of mass is not equivalent to the center of gravity. – Michael Seifert Jul 31 '20 at 23:00
As Michael said, the ring won't oscillate. A ring centred on a point mass is in an unstable equilibrium. See Why is Larry Niven's Ringworld Unstable?. – PM 2Ring Aug 01 '20 at 05:21

Agnius Vasiliauskas · Answer 2 · 2020-08-01T08:45:47.730

Given this situation

In case rope geometric center is at the Earth com, then elementary Hooke force change due to rope length contraction is : $$ dF_H = k~2\pi~ dr $$ And elementary gravity force change due to change in elementary radius is : $$ dF_G = -2G\frac{Mm}{r^3} dr $$ Rope will start to fall down uniformly until Hooke restoring force will compensate gravity force, namely when : $$ F_H = F_G $$

Expressing in integrals would be : $$ \int^{R_2}_{R_1} k~2\pi~ dr = \int^{R_2}_{R_1} -2G\frac{Mm}{r^3} dr $$

Integrating gives :

$$ k~2\pi~ (R_2-R_1) = G~Mm \left(\frac {1}{{R_2}^2} - \frac{1}{{R_1}^2}\right) $$

Refactoring right side in alternate form gives :

$$ k~2\pi~ (R_2-R_1) = G~Mm \frac {-(R_2 - R_1) (R_1 + R_2)}{{(R_1 R_2)}^2} $$

Simplifying, and expressing left side in terms of radiuses, gives :

$$ \frac {(R_1 + R_2)}{{(R_1 R_2)}^2} = - \frac{k~2\pi}{G~Mm} $$

Inverting both sides of equation and multiplying by $1/{R_2}^2$ gives :

$$ \frac{{R_1}^2}{R_1 + R_2} = -\frac{G~Mm}{{R_2}^2} \frac{1}{k~2\pi} $$

Noticing in the right side of equation there's a gravity force $F_G$ and multiplying both sides by $1/R_1$ gives :

$$ \frac{R_1}{R_1 + R_2} = -\frac{F_G}{k~2 \pi R_1} $$

It's not hard to spot that in the right side denominator there is a circumference of final contracted rope state, so :

$$ \frac{R_1}{R_1 + R_2} = \frac{F_G}{-k~ l^{~\prime}} $$

Right side denominator is Hooke restoring force, which we get compressing final rope state at equilibrium into singularity, namely $F_{l^{~\prime} \to 0} = -k~ l^{~\prime}$, so we can re-write equation as :

$$ \frac{R_1}{R_1 + R_2} = \frac{F_G}{F_{l^{~\prime} \to 0}} $$

Assuming $R_1 \ll R_2$, and inverting ratios, we get :

$$ \boxed {\frac{R_2}{R_1} = \frac{F_{l^{~\prime} \to 0}}{F_G} }$$

This is a very beautiful relation, I didn't expected that. In addition to that $R_1$ could be called as
Hooke radius, for noticing some analogies to Schwarzschild radius and black-holes (singularity participates here !). So this problem is a nice classic example on the road to relativity.

Up to a sign factor, Hooke's law in this case is a curl whereas gravity is a divergence, and yet you start off equating the two $F_H = F_G$. Am I misunderstanding something? — Tfovid, Aug 01 '20 at 08:58
@Tfovid I could reformulate relationship as $F_H / F_G = \alpha$, will it change anything in principle? I.e. one can measure ratios between ANY two quantities, despite the roots of their origin (curl, gradient, etc., etc.). If my goal is just checking some correlations, then I think it's pretty legal. — Agnius Vasiliauskas, Aug 02 '20 at 06:48

How would a rope around Earth fall?

2 Answers2