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?

Answer 1

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.

Answer 2

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.