# A rope in tension between Earth and Moon

by 8BitTRex
Tags: earth, moon, rope, tension
Physics
PF Gold
P: 5,518
 Quote by PAllen 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?
Mentor
P: 10,853
 Quote by bahamagreen 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.
Physics
PF Gold
P: 5,518
[Edit: Added the effect of the Moon's mass.]

 Quote by Nugatory 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.
Physics
PF Gold
P: 5,518
 Quote by 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.
Mentor
P: 10,853
Quote by PeterDonis
 Quote by 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.
Thanks
P: 2,972
 Quote by PeterDonis 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.
Physics
PF Gold
P: 5,518
 Quote by Nugatory 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.
Physics
PF Gold
P: 5,518
 Quote by mfb 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.]
P: 3,553
 Quote by Nugatory 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.
Thanks
P: 2,972
 Quote by PeterDonis 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
P: 503
 Quote by PeterDonis 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.

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.
Mentor
P: 14,476
 Quote by PeterDonis [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(ME+MM)/R3, not GME/R3.

It helps to work in a system where RM=1 and G(ME+MM)=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
$$T(r) = \frac{\mu}{1+k}(r-r_0)\Bigl(\Bigl(\frac 1 {r_0r}-\frac{r+r_0}2\Bigr) -k*\Bigl(\frac 1 {(1-r)(1-r_0)} - 1\Bigr)\Bigr)$$
Physics