# Moon tied to Earth

1. Nov 13, 2009

Imagine that we go to the moon, and tie one end of a rope to the surface of the moon, and then take the other end of the rope back to earth. We cant tie it down of course, becuase it would snap as the moon moves away. So instead we make the rope long enough to be a few feet off the ground, and let it go.

What happens next?

Since the gravity of Earth is stronger, I assume the rope would just dangle above the ground, although moving across the sky as the moon drags it along.

The foundation on the moon might have to be very strong to hold the rope, becuase the moon is pulling one way, and Earth the other.

2. Nov 13, 2009

### JANm

My intuitive answer is that the rope will make some sort of spiral. I never calculated the moon-o-statical orbit. I know that for the earth that is 26000 km. So a satelite further away moves slower than the earth surface and a satelite closer moves faster. The moon moves around one time each month (it literally faces the earth). The center of the earth moon system lies within the earth. 3800 km of the centre. If you take the distance of the moon d_ME= 1 1/3 lightsecond, then I think there is enough data to calculate the moon-o-statical orbit.
A rope shorter than the ...-o-statical orbit moves to slow to leave it upright, yet a rope reaching beyond the o-statical orbit moves fast enough to centrifugate more than then the gravitational pull. I am still not certain what would become the shape of a rope let us say twice as large as the moon-o-statical orbit...
greetings JanM

3. Nov 14, 2009

### eachus

The moon rotates too slowly for a traditional! space elevator to work. However, a rope through either the L1 or L3 Lagrange point is quite practical. Since the rope/beanstalk/skyhook would be much shorter than on Earth, and subject to 1/6 the gravity, even more conventional materials than carbon nanotubes can be used. For example, such a space elevator made of Kevlar would only take a few shuttle loads to set up.

The two tricky parts are the moon's libration, and that its orbit around the earth is elliptical. For a skyhook through the L1 point, the latter can be dealt with by making it longer than necessary and anchoring it well on the moon's surface. But this is where the libration comes in. Since the moon has libration in both latitude and longitude, connecting the beanstalk to a track wouldn't work. Besides, the track would take a lot more construction materials, and work, than the beanstalk itself. A simple solution is to use the beanstalk to lift the anchor on the moon, use the pendulum effect to sway the anchor past the sub-Earth point, then put it down again. Seems complicated, but it is probably the easiest way to make a lunar beanstalk work.

Note also that any plans for visiting Mars without setting up a beanstalk are silly. The weight of cable needed is small compared to any Mars lander/orbiter, and Deimos provides a useful anchor. Again, a beanstalk dropped from Deimos, would not remain above any particular point on the Martian surface. For exploring, this is good--you can drop exploration teams anywhere along the Martian equator and recover them a few days later.

4. Nov 14, 2009

### JANm

Hello Eachus
Ok the term beanstalk is nice and you mention the L_1 point where Earth gravitation is equal to the moon gravitation. Am I correct with this Lagrange_1 point?
Still I try to find d_ms (moon-o-static distance) take object x:

1 GMm/d_xs^2=mv^2/d_xs

dividing left and right by m, because a brick and a feather fall equally in vacuum.
dividing left and right by d_xs because v/dxs is omega

you get 1b GM/d_xs^3=omega^2,so d_xs=(GM/omega^2)^1/3.
so d_es=26000=(GM_earth/day^2)

We know omega_moon, so we need to know GM_moon:
GM_moon*400000=GM_earth*3800 so GM_moon=38*GM_earth/4000, approx hundredst of earthmass...

d_ms=26000*38/(900*4000)=1,3*1,9/9=300 meter, indeed rotation of the moon is too slow to have a moonstationary orbit.

This was not what I expected to be...
greetings Janm

5. Nov 14, 2009

### JANm

Hello
Sorry I forgot the 1/3 power. So actual d_ms=(1,3*1,9/9)^(1/3)=0,27^(1/3)=0,65, so 650meter nevertheless too small to be important.

I am interested in the distance from the moon where gravity of the moon is equal to that of the earth. The turning point as they say...
greetings

6. Nov 14, 2009

### JANm

Hello
My reply is still no good. I guess my takeoff was too fast.
d_ms=d_es*((GM_m/Omega_m^2)/(GM_e/Omega_e^2))^(1/3)
So much for a field mathematic...
d_ms=26000*(38/(4000*900))^(1/3)
=26000*(10,5/1000.000)^(1/3)=520 KM
Still small but a more realistic answer. Now we need the radius of the moon at least...
Greetings Janm

7. Nov 14, 2009

### pervect

Staff Emeritus
8. Nov 15, 2009

### JANm

Thank you for this article. Indeed L1 is the point in between. I see L1=R(1-(alpha/3)^(1/3))
with alpha = M_m/(M_m+M_m)=1/100, R=1 1/3 lightsec it becomes
L1=1 1/3(1-(1/300)^(1/3))=4/3(1-11/75)=1,15, so 55.000 km of the moon, and 245.000 km of the earth.
The moon o static is still not good I used omega_moon=30 while it should be 1/30...
greetings Janm

9. Nov 25, 2009

### JANm

Hello
Found some data:
Moon: M_m=7,353E22kg, P_m=27,4day, R_m=1738km
Earth: M_e=5,976E24kg, P_e=23,96hour, R_e=6378km
their mean distance d=384,4E3km, G=6,6726E-11Nm^2/kg^2.
Hope this will help for calculating length of cord and
Langrange-point1
greetings Janm

10. Nov 28, 2009

### JANm

Hello All
It is becoming a little of a monologe. I have calculated the moon-o-static satellite radius:
GMm/r^2=mv^2/r (1)
Deviding (1) by mr one gets:
GM/r^3=v^2/r^2=w^2 (2)
muliply (2) with r^3/w^2 and powering to 1/3 one gets:
r=(GM/w^2)^(1/3)