A rope in tension between Earth and Moon

[PeterDonis]

The tension at that point is zero, so the only force acting on the tip of the rope is gravitational.

No. The language that it's natural to use here makes it hard to see what's actually going on, so let me first state it in terms of $a_{tension}$, the proper acceleration due to the gradient in the tension in the rope; $a_{tension}$ is certainly *not* zero at the bottom end of the rope, in fact it's at its maximum value (that's obvious from the formula I gave for it). That means there must be a nonzero force in addition to "gravitational" force (which, as I pointed out previously, really means "gravity" plus "centrifugal" force if we are working in the rotating frame, as I was) on the rope even at its bottom tip; there has to be, because there is nonzero proper acceleration there, and "gravitational" forces produce zero proper acceleration.

In other words, the tension in the rope is zero at the mathematical point at the very bottom tip of the rope; but there is a positive gradient in the tension there, which is larger than anywhere else in the rope, so the piece of rope just above the bottom tip has nonzero tension, and therefore exerts nonzero force on the piece of rope right at the bottom tip. That's why $a_{tension}$ is nonzero (and maximum) at the bottom end of the rope even though the tension itself goes to zero there.

 

[PeterDonis]

there is always some fraction of the mass slower and some fraction faster than the corresponding orbital velocity.

Ah, that's right, we have to include the effect of the Moon's mass. Hm, I need to re-write some of the formulas from my previous post. [Edit: Went back and corrected the formulas to include the effect of the Moon's mass.]

 

[A.T.]

But I don't see how that equilibrium position is straight up and down, so that all points on the rope have the same $\theta$ value.

Forget about $\theta$. Go to the rotating frame where $\dot{\theta}=0$ for all rope parts at $t=0$. The rope is at rest in this frame and all forces on the rope parts are still radial. Nothing deflects it sideways.

 

[Nugatory]

No. The language that it's natural to use here makes it hard to see what's actually going on, so let me first state it in terms of $a_{tension}$, the proper acceleration due to the gradient in the tension in the rope; $a_{tension}$ is certainly *not* zero at the bottom end of the rope, in fact it's at its maximum value (that's obvious from the formula I gave for it). That means there must be a nonzero force in addition to "gravitational" force (which, as I pointed out previously, really means "gravity" plus "centrifugal" force if we are working in the rotating frame, as I was) on the rope even at its bottom tip; there has to be, because there is nonzero proper acceleration there, and "gravitational" forces produce zero proper acceleration.

In other words, the tension in the rope is zero at the mathematical point at the very bottom tip of the rope; but there is a positive gradient in the tension there, which is larger than anywhere else in the rope, so the piece of rope just above the bottom tip has nonzero tension, and therefore exerts nonzero force on the piece of rope right at the bottom tip. That's why $a_{tension}$ is nonzero (and maximum) at the bottom end of the rope even though the tension itself goes to zero there.

Got it - thx

 

[bahamagreen]

Actually, the entire rope is slower than the corresponding orbital velocity at the distance of any individual piece of the rope. The point where the angular velocity just matches orbital velocity is at the center of the Moon, which is at a larger radius than any part of the rope.

What about Lagrange point L1?

Wiki states

"A satellite at L1 would have the same angular velocity of the Earth with respect to the sun and hence it would maintain the same position with respect to the sun as seen from the earth."

(this is wrt Earth and Sun, but same applies to Earth and Moon)

also

"The location of L1 is the solution to the following equation balancing gravitation and centrifugal force:

M1/((R-r)^2) = M2/(r^2) + ((((M1/(M1+M2))xR)-1)x((M1+M2)/(R^3))

where r is the distance of the L1 point from the smaller object, R is the distance between the two main objects, and M1 and M2 are the masses of the large and small object, respectively.

 

[D H]

[Edit: Added the effect of the Moon's mass.]

You didn't quite finish doing that. Two issues:

1. The Earth is accelerating gravitationally toward the Moon. You have an accelerating reference frame. One way to overcome this is to make the origin of your frame the Earth-Moon barycenter, and then make a change of variables so as to go back to measuring r from the center of the Earth. An equivalent approach is to keep the origin at the center of the Earth and add a fictitious acceleration to account for the acceleration of the Earth toward the Moon.

2. The angular velocity of the Moon is G(M_E+M_M)/R³, not GM_E/R³.It helps to work in a system where R_M=1 and G(M_E+M_M)=1 and to denote the ratio of the Moon's mass to that of the Earth as k (k≈0.0123). With this, the gradient of the tension is
$$\frac {dT}{dr} = \frac{\mu}{1+k}
\Bigl( \Bigl(\frac 1 {r^2} - r\Bigr) - k \Bigl(\frac 1 {(1-r)^2} - 1\Bigr) \Bigr)$$

The tension in the rope is thus
$$T(r) = \frac{\mu}{1+k}(r-r_0)\Bigl(\Bigl(\frac 1 {r_0r}-\frac{r+r_0}2\Bigr) -k*\Bigl(\frac 1 {(1-r)(1-r_0)} - 1\Bigr)\Bigr)$$

 

[PeterDonis]

You didn't quite finish doing that.

You're right, thanks for the corrections!

 

[D H]

If my calculations are correct, a lunar space elevator would need a counterweight unless the cable (rope) reached almost 3/4 of the way from the Moon to the Earth. A uniform rope longer than that would be under tension the entire length. With a shorter rope, the tension at the attachment point at the surface of the Moon is negative. Ropes don't stand up well against negative tension (i.e., compression).