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Consider the Earth (mass $M$, radius $R$, rotating about its own axis at $\Omega$) and the moon (mass $m$, radius $r$, with axial rotation equal to $\omega_m$), whose centre of masses are $d$ apart. They rotate around their barycentre at $\omega_e$ and $\omega_m$ radians/second respectively (I believe $\omega_m=\omega_e$ to conserve momentum).

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We now tether a point on the Earth's surface to a point on the Moon's surface using a light, inextensible (and very strong) piece of string. Assume the Earth and the Moon are rigid and indestructible.

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What happens?

It would be nice to find a solution of the general type, but for a first analysis it may be easier to assume that $M \gg m$, so that the Earth's rotation around the barycentre is negligible and the barycentre $\approx$ the Earth's centre of mass. Another (perhaps reasonable) assumption is that all rotation takes place in one plane and that the moon's orbit is circular, and that both bodies have uniform density.

Here are some of my thoughts based solely on intuition, feel free to ignore in answering the question:

  1. The moon is immediately pulled in (as $\Omega_e>\omega_m$) and sticks to the Earth like a barnacle. I think this may occur if $\Omega_e \gg \omega_m$ and $M\gg m$.
  2. If $\omega_m$ is large enough, the moon as before will initially be pulled in, then be thrown outwards again as $\omega_m(t)$ increases.
  3. If $m\approx M$, the Earth and Moon will form a dumbbell-type system (i.e. the string wouldn't have had much of an effect, as that is the case anyway in 'binary' systems where $M=m$).
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Would the close-voter care to explain how I could improve this question? I must commend how quickly you have read through the question. – Alyosha Jul 5 '13 at 21:20
Unless it's made of Adamantium, go with option 4; the string breaks. :-P – Jim Jul 5 '13 at 21:23
BTW: "the moon (mass m, radius r, no axial rotation)" The "no axial rotation" part is incorrect. It rotates as fast as it orbits so that we always see the same face. – dmckee Jul 6 '13 at 0:31
Even if the string is made of Adamantium and doesn't break, it surely gets ripped out of the ground, even if it's anchored miles deep into the mantle. – Nathaniel Jul 6 '13 at 4:47
The earth orbits around its axis ( the moon is not continually in the same spot of the sky) . The string would wind around the earth and draw the moon in, since its rotation is slower and its mass smaller. No such string can exist in reality. – anna v Jul 6 '13 at 14:18

Note this is a hard problem, which depends on so many factors. Ergo giving complete answers which go through all the possibilities is a bit hard. For instance, the case changes dramatically depending on where you attach the pinpoint on Earth(i.e. north pole or equator) or whether Moon is in its apogee or perigee or somewhere else of its orbit. I will write some raw thoughts and give some rough numbers. I think actual computer simulations are needed for more detailed answers.

Qualitative Facts:

  1. When we attach the string, if Moon is in its apogee the destruction will obviously be minimum.
  2. Since Earth and Moon are (almost)sphere's, due to Moon's orbit inclination and its axis tilt; the string will wrap around Earth a finite number of times. I will try to estimate it in the next section.
  3. Even if the string initially wraps around the Earth, due to the constant torque applied on Earth, Earth's rotation will slow down and the string will eventually unwrap.
  4. Although Earth and Moon won't collide the changes in tidal forces and daytime length will be dramatic.
  5. The changes in Earth's period around Sun(year) will be negligible.
  6. The string's maneuver on Earth will change(at most half of) Earth's face considerably in short terms.
  7. The string's tension can presumably affect tectonic plates movements(continental drift) and cause severe earthquakes.
  8. Depending on where we attach the string, it will also change Earth's axis of rotation; i.e. the equator and the Pole Star will change.

Quantitative Estimations:

These are estimations which can presumably be made on the back of a cocktail napkin.

  • How many times will the rope rap around Earth? Or equivalently, what will be the length of the tangled rope around Earth?

The mean inclination of the lunar orbit to the ecliptic plane is $\theta = 5.145°$. Therefore, the rope will only rap a finite number of times around Earth. Assuming the pinpoint to be on Equator, I will find an upper limit and a lower limit, then take their geometrical mean as a reasonable guess (A valid technique in Fermi problems)for the amount of string rapped around earth. The upper limit comes from rapping a rope around a cylinder of radius and height $R_E$.

$$d_{up}= \frac{R_E}{\sin{\theta}} \approx 11 R_E \approx 7.1 \times 10^{7} \text{m}$$ The lower limit is basically half of equator's length: $$d_{down}=\pi R_E \approx 2.0 \times 10^7 \text{m}$$ $$d \approx \sqrt{d_{up}d_{down}} = \sqrt{11 \pi}R_E \approx 5.9 R_E \approx 3.8 \times 10^7 \text{m}$$

i.e. Moon will get about $10 \%$ closer.

  • The new day: Or what will be the new definition of one hour!

After attaching Moon and Earth together, and after all the wobbly motions settle down and the system reaches its steady routine motion again (the rope is no longer wrapped around Earth); Moon will be again in its initial distance from Earth; however, Earth and Moon will be rotating with the same angular velocity $\omega$. We can estimate this by conservation of Momentum. The answer may depend on whether we attach the string at Moon's apogee or perigee, so I will estimate both cases.

The initial angular momentum around system's center of mass, will be(ignoring axial tilt): $$L=\frac{2}{5}M_E {R_E}^2 \omega_E + M_E {r_E}^2 \Omega + \frac{2}{5}M_M {R_M}^2 \omega_M + M_M {r_M}^2 \Omega$$ where $M_E$ and $M_M$, $R_E$ and $R_M$, $r_E$ and $r_M$, $\omega_E$ and $\omega_M$ are Earth and Moon's mass, radius, distance to COM and angular frequency around themselves respectively. But we know: $$\omega_M = \Omega \\ M_E r_E = M_M r_M$$ $$\Rightarrow L=\frac{2}{5}M_E {R_E}^2 \omega_E + M_M \Omega \left( \frac{2}{5}R_M^2 + r_M(r_M + r_E)\right) \\ \approx \frac{2}{5}M_E {R_E}^2 \omega_E + M_M \Omega \left( r_M^2 \right) $$

Writing the angular momentum afterwards: $$L=\frac{2}{5}M_E {R_E}^2 \Omega' + M_E {r_E}^2 \Omega' + \frac{2}{5}M_M {R_M}^2 \Omega' + M_M {r_M}^2 \Omega' \\ \approx \left( \frac{2}{5}M_E {R_E}^2 + M_M r_M^2\right) \Omega'$$

$$T'=\frac{2 \pi}{\Omega'} \approx \frac{2\pi \left( \frac{2}{5}M_E {R_E}^2 + M_M r_M^2 \right)}{\frac{2}{5}M_E {R_E}^2 \omega_E + M_M \Omega r_M^2 } = \frac{2 \pi}{\omega_E} \frac{\frac{2}{5} \frac{M_E}{M_M}+\left(\frac{r_M}{R_E} \right)^2}{\frac{2}{5} \frac{M_E}{M_M}+\left(\frac{r_M}{R_E} \right)^2 \frac{\Omega}{\omega_E}}$$

Taking the values from here and here and here, we'll get:

$$T' \approx 22.1 \text{day}$$ Which is really long. For perigee and apogee, the day-times will respectively be: $$T'_{p}=21.6 \text{day} \\ T'_{a}=22.5 \text{day}$$ The difference is not that significant though.

To be completed

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I will add some estimations about the rope tension and tidal forces; then I will make some conclusions. Also, I bet there are a zillion typos and grammatical errors, please feel free to correct them. – Ali Jul 20 '13 at 18:46

The string breaks. Though on an elementary level, it appears as though the orbit of the moon about its barycentre is circular, its actually elliptical. So the distance between the centres of the earth and the moon is not constant, so the string breaks under the tremendous tension developed in it.

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I think the OP understands this: what is sought is the following: the constraint changes the motion and he/she is trying to get their head around a sound quantitative model describing this change. I'll come back to an answer to this when I have more time or if someone else doesn't answer first. – WetSavannaAnimal aka Rod Vance Jul 13 '13 at 5:03
That is the reason Ali in his answer said that tie the string when the moon is at apogee..: P – Cheeku Jul 14 '13 at 22:45

Nothing would change at all ...Earth and moon are already tied by the force of gravity..The nature of gravity is spheric around the spheric masses with almost uniform distribution (earth and moon) so the only difference is the fixed point on earth in gravity case is like thread attached to slider with slider recesses on a certain latitude all around. However the missing part is you require thread to be inextensible. For that case consider motion of binary star systems where gravity is strong enough to make the virtual thread inextensible.

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protected by Qmechanic Jul 28 '13 at 13:17

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