A rope in tension between Earth and Moon

[dauto]

If we start with the extended rope as described, neglect the influence of the sun and assume a circular orbit for earth, all forces are radial and the rope stays as it is, orbiting Earth once per ~27 days.

Relativistic effects are negligible here, so I'm not sure if this is the right forum.

That question started elsewhere and I told the author to post it here because after I stated that relativistic effects were negligible the author said the relativistic effects were the whole point of the question. Those effects, negligible as they may be, are exactly what (s)he is trying to understand.

 

[PAllen]

So, we can rephrase the OP as: given a classical solution, what corrections would there be due to GR, and what would their order of magnitude be?

There are at least two corrections: to the inverse square law, and due to Earth rotation (frame dragging). My guess would be GR corrections to inverse square law (though exceedingly tiny) would be larger than those due to frame dragging. But I have computed nothing in making this statement.

 

[A.T.]

My guess would be GR corrections to inverse square law (though exceedingly tiny) would be larger than those due to frame dragging.

Would the corrections to inverse square law deflect the rope perpendicularly to the rope, or just affect the stresses in the rope.

 

[Nugatory]

No, the rope would not point toward the COM of the Earth. Because there would be a small force(very small) from frame dragging, which would cause a non-radial force on the end of the rope. This force would cause a small deflection which would cause the rope to point somewhere other than the COM of Earth.

Is that correct?

If I understand you properly, one end of the rope is tethered at the moon and the other end is free and reaching all the way down to the top of the Earth's atmosphere. If that's the problem, the rope will NOT hang straight up and down, and this has nothing to do with any relativistic effects - it's a pure classical problem.

 

[mfb]

Are they negligible(veeeery small) or are they non existant? I was wondering if there was any force that would deflect the rope.

I guess frame-dragging gives a small tangential deviation. The deviations from the inverse square law are still radial forces.

However, keep in mind this is a purely hypothetical setup. In our solar system, nearly everything else would lead to larger deviations - the non-circular orbit of moon, the thin upper atmosphere of earth, the non-spherical mass distribution of earth, gravitational forces from the sun, sunlight pressure, gravitational effects from some other planets, maybe even non-uniform thermal radiation from the cable, ...

@Nugatory: The ideal, non-relativistic cable has an analytic solution for the tension.

 

[8BitTRex]

If that's the problem, the rope will NOT hang straight up and down, and this has nothing to do with any relativistic effects - it's a pure classical problem.

Can you explain what would cause it to not hang straight? What non-radial force component is there in this problem? I can't think of anything that would cause the rope to deflect.

 

[PeterDonis]

I guess frame-dragging gives a small tangential deviation.

Only if there is radial motion. If each small piece of the rope remains at a constant radius, all frame-dragging will do is slightly change the radial proper acceleration required to hold the piece of the rope at its current radius. This was the topic of a thread on PF some time ago, I'll see if I can dig up the link.

 

[mfb]

Only if there is radial motion. If each small piece of the rope remains at a constant radius, all frame-dragging will do is slightly change the radial proper acceleration required to hold the piece of the rope at its current radius. This was the topic of a thread on PF some time ago, I'll see if I can dig up the link.

Ah okay.

 

[8BitTRex]

I guess frame-dragging gives a small tangential deviation. The deviations from the inverse square law are still radial forces.

However, keep in mind this is a purely hypothetical setup. In our solar system, nearly everything else would lead to larger deviations - the non-circular orbit of moon, the thin upper atmosphere of earth, the non-spherical mass distribution of earth, gravitational forces from the sun, sunlight pressure, gravitational effects from some other planets, maybe even non-uniform thermal radiation from the cable, ...

yeah I understand that. I thought it was an interesting problem, however unrealistic, to develop my intuition, thanks for the reply.

 

[PAllen]

Would the corrections to inverse square law deflect the rope perpendicularly to the rope, or just affect the stresses in the rope.

I'm thinking the perfect, non-stationary, two body system is not stable in GR (whatever initial non-stationary initial conditions you assume, there cannot be exact stability in GR). GW would cause orbiital decay, thus change of angular speed. This would produce tangential effects on the rope. The GW themselves also would produce tangential effects. All of this would be mind-bogglingly small, but that seems to be the intent of the question - to think about a perfect setup in an isolated universe.

 

[bahamagreen]

If I understand your odd description correctly, is it like this picture?
http://i.imgur.com/Y08NsU7.png

I'm having a hard time picturing it.

You have the curve lagging behind; I think it would lead forward (just rotate the drawing 180 degrees around the line between the Earth and Moon)... the leading curve is on the same side as the direction of rotation in your picture.

There is a small curve between the Moon and L1 in the lagging direction.

In general, I think the dominant influence on the rope at any point along its length is the difference between the orbital period of a free mass at that altitude versus the constraint suffered by the rope to be at that altitude and limited to the orbital velocity of the Moon.

Below the Lagrange point L1, the rope wants to orbit the Earth progressively faster with decreasing altitude, and will try to do so, forming a curve in front of the Earth-Moon line.

 

[8BitTRex]

In general, I think the dominant influence on the rope at any point along its length is the difference between the orbital period of a free mass at that altitude versus the constraint suffered by the rope to be at that altitude and limited to the orbital velocity of the Moon.

Why would the orbital period of a free mass create a force on the suspended rope? If angular momentum is conserved, wouldn't forces be radial? What would cause the tangential influence you are discussing?

 

[PeterDonis]

there would be a small force(very small) from frame dragging, which would cause a non-radial force on the end of the rope. This force would cause a small deflection which would cause the rope to point somewhere other than the COM of Earth.

Is that correct?

No. Your initial condition specified no radial motion, and in that case frame dragging does not add any tangential motion; it only changes the radial force required to hold a piece of the rope at constant altitude. See my previous post in response to mfb.

 

[8BitTRex]

No. Your initial condition specified no radial motion, and in that case frame dragging does not add any tangential motion; it only changes the radial force required to hold a piece of the rope at constant altitude. See my previous post in response to mfb.

To clarify, because the answers seem to vary...

The rope would ALWAYS point radially (in my ideal problem), classically and relativistically, because there is no tangential force that would act on it??


Is that right?

 

[PeterDonis]

The rope would ALWAYS point radially (in my ideal problem), classically and relativistically, because there is no tangential force that would act on it??

I think there's general agreement that, if the classical solution is indeed an equilibrium with the rope oriented purely radially and at rest in the rotating frame, there are no relativistic effects that would introduce tangential motion.

I'm not sure there's general agreement that the classical solution is in fact an equilibrium with the rope oriented purely radially and at rest in the rotating frame.

 

[PAllen]

I think there's general agreement that, if the classical solution is indeed an equilibrium with the rope oriented purely radially and at rest in the rotating frame, there are no relativistic effects that would introduce tangential motion.

I'm not sure there's general agreement that the classical solution is in fact an equilibrium with the rope oriented purely radially and at rest in the rotating frame.

Actually, I disagree with this. Suppose such an initial equilibrium existed. Then the moon's angular velocity starts to increase due to [GR mandated] orbital decay - this produces a tangential force on the rope.

 

[bahamagreen]

Why would the orbital period of a free mass create a force on the suspended rope? If angular momentum is conserved, wouldn't forces be radial? What would cause the tangential influence you are discussing?

Imagine two masses in orbit tied together with a rope, one mass in a higher orbit than the other.

The orbital period of the higher orbit mass is longer, so it will try to slow down, but the orbital period of the lower orbit mass is shorter, so it will try to speed up.

Neither mass can orbit the period it would if free, since they are tied together... the upper orbit mass is pulling back to go slower, the lower orbit mass is pulling forward to go faster.

The result is that the two masses each experience an acceleration toward each other, and the tension in the rope increases, and the angle of the rope shifts so the lower end is forward and the upper end is behind.

This same effect is in play with the OP's question, with the additional complications that the rope passes through the Lagrange point L1, and the Moon end is fixed.

 

[8BitTRex]

The orbital period of the higher orbit mass is longer, so it will try to slow down, but the orbital period of the lower orbit mass is shorter, so it will try to speed up.

What is the mechanism that causes it to speed up?

 

[Nugatory]

What is the mechanism that causes it to speed up?

An object traveling in a circular trajectory is experiencing an acceleration in the radial direction of $-v^2/r$ - if it weren't for that acceleration it would be moving at a constant velocity in a straight line tangent to the circle. This requires (by $F=ma$) a force equal to $-mv^2/r$. But the gravitational force is equal to $-mK/r^2$ (where $K=Gm_e$) so the speed of an object in orbit is given by $mv^2/r=mK/r^2$, or $v=\sqrt{K/r}$.

However, we also have $v=r\dot{\theta}$ for both objects, so a bit more algebra gets us to $\dot{\theta} = \sqrt{k/r^3}$, which is to say that the angular velocity of the two objects is different - if they started lined up radially, they won't stay that way.

The same considerations apply to the different sections of your rope. Their natural trajectory under the influence of gravity will not keep them all aligned radially.

 

[PeterDonis]

Then the moon's angular velocity starts to increase due to [GR mandated] orbital decay - this produces a tangential force on the rope.

Hm, yes, I guess if we're going to dot all the I's and cross all the t's, we have to take into account *all* of the effects involved. But that also includes the tidal effects on the Earth-Moon system, which slow the Earth's rotation and reduce the Moon's angular velocity (and increase its orbital radius), thereby putting a tangential force (in the opposite direction) on the rope. This is a classical effect, true, but it is certainly much, much larger than the effect of gravitational wave emission on the Moon's orbit (and also, I'm pretty sure, much larger than any effect on the rope due to frame dragging). So if we're going to that level of detail, then even classically, there is not a true equilibrium until the Earth and Moon are completely tidally locked (i.e., the Earth always faces the same side to the Moon just as the Moon always faces the same side to the Earth).

Are there any other effects we've missed?

 

[mfb]

Imagine two masses in orbit tied together with a rope, one mass in a higher orbit than the other.

The orbital period of the higher orbit mass is longer, so it will try to slow down, but the orbital period of the lower orbit mass is shorter, so it will try to speed up.

You start with wrong initial conditions. Neither mass is in its original orbit - the lower mass is slower and the upper mass is faster than a corresponding orbit in this distance. This is given by the initial conditions. The corresponding radial forces are transmitted via the rope.

Alternatively, you can see the whole system as a single object orbiting the central body with a bound rotation - once per orbit.

 

[PeterDonis]

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

Their natural trajectory under the influence of gravity will not keep them all aligned radially.

But gravity is not the only force on each section of the rope; there is also the tension in the rope. (And centrifugal force, if we work in the rotating frame in which the Moon is at rest, as I will below--but that's really included in the "natural trajectory".) On thinking this over, I think there *is* an equilibrium configuration, if the tension in the rope varies in the right way with altitude.

Consider a small piece of the rope, and ask what it takes to keep it on a circular trajectory around the Earth, but with the Moon's angular velocity (as opposed to the--larger--correct orbital angular velocity for its altitude). If we work in the rotating frame in which the Moon is at rest, then the condition for equilibrium is just that all the forces on the small piece of rope sum to zero. There are three such forces: gravity inward, centrifugal force outward, and tension in the rope, which can act both inward and outward. So we have (writing formulas for acceleration instead of force, since the mass of the small piece of rope is the same for all three forces and can be divided out)

$$
a_{total} = 0 = a_{grav} + a_{centrifugal} + a_{tension}
$$

Substituting $a_{grav} = - G M_E / r^2 + G M_M / \left( R_M - r \right)^2$ (where $M_E$ is the mass of the Earth and $M_M$ is the mass of the Moon) and $a_{centrifugal} = \omega_M^2 r$, where $\omega_M^2 = G M_E / R_M^3$ is the angular velocity of the Moon in its orbit ($R_M$ is the radius to the Moon's center), we have

$$
a_{tension} = \frac{G M_E}{r^2} - \frac{G M_M}{ \left(R_M - r \right)^2} - \frac{G M_E r}{R_M^3} = G M_E \left( \frac{1}{r^2} - \frac{1}{R_M^2} \frac{r}{R_M} \right) - \frac{G M_M}{ \left(R_M - r \right)^2}
$$

The rope will be in equilibrium if the proper acceleration due to tension in the rope satisfies this formula. Note that at the bottom of the rope, the RHS is clearly greater than zero, so $a_{tension}$ is directed outward; but there will be a "turning point" at which $a_{tension}$ goes to zero, and above that point it will be directed inward.

[Edit: Added the following to clarify the distinction between "proper acceleration due to tension" and tension.]

If we model the rope as having some constant mass per unit length $\mu$, then the proper acceleration $a_{tension}$ is actually proportional to the *gradient* of the tension in the rope (i.e., it is due to the *net* force due to tension on each piece of the rope, which is the difference between the force from the piece above and the force from the piece below). The formula is ($T$ is the tension in the rope as a function of radius $r$):

$$
\frac{dT}{dr} = \mu a_{tension}
$$

Plugging in the formula above for $a_{tension}$, we can easily integrate the result to obtain ($r_0$ is the radius of the bottom end of the rope, and we have the boundary condition that the tension at that point is zero, i.e., $T(r_0) = 0$)

$$
T = \mu G \left[ M_E \left( \frac{1}{r_0} - \frac{1}{r} - \frac{r^2 - r_0^2}{2 R_M^3} \right) - M_M \left( \frac{1}{R_M - r} - \frac{1}{R_M - r_0} \right) \right]
$$

This obviously satisfies the boundary condition.

 

[PeterDonis]

Neither mass is in its original orbit - the lower mass is slower and the upper mass is faster than a corresponding orbit in this distance. This is given by the initial conditions.

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.

 

[mfb]

Neither mass is in its original orbit - the lower mass is slower and the upper mass is faster than a corresponding orbit in this distance. This is given by the initial conditions.

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.

The quoted part applies to bahamagreen's modified scenario with two masses connected by a rope. It also applies to an idealized point-like moon with a non-zero rope mass (but then the center of gravity is very close to the moon) or a really massive rope. Anyway, there is always some fraction of the mass slower and some fraction faster than the corresponding orbital velocity.

 

[Nugatory]

I think there *is* an equilibrium configuration, if the tension in the rope varies in the right way with altitude.

I agree that there is an equilibrium configuration, and that it requires the tension in the rope to vary with altitude. 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.

Consider the end of the rope when we've started in the straight up and down configuration. The boundary condition is, as you say:

we have the boundary condition that the tension at that point is zero, i.e., $T(r_0) = 0$)

The tension at that point is zero, so the only force acting on the tip of the rope is gravitational. Thus, the tip of the rope cannot have the same $\dot{\theta}$ value as the other end of the rope which is tracking the moon's orbit, so the straight up and down condition cannot be maintained.

What I'm seeing in your solution is the correct expression for the radial component of the tension in the rope at equilibrium as a function of $r$. If we add the requirements that angular momentum in the equilibrium configuration is equal to the angular momentum in the initial up-and-down configuration and that energy (PE plus KE) is conserved, I think that is enough to completely determine the equilibrium configuration - which is not straight up-and-down.

88BitTRex, this is a really fun problem even without relativity.

 

[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).