Register to reply 
A rope in tension between Earth and Moon 
Share this thread: 
#55
Feb2114, 06:37 PM

Physics
Sci Advisor
PF Gold
P: 6,040

Are there any other effects we've missed? 


#56
Feb2114, 06:50 PM

Mentor
P: 11,580

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


#57
Feb2114, 06:51 PM

Physics
Sci Advisor
PF Gold
P: 6,040

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


#58
Feb2114, 07:23 PM

Physics
Sci Advisor
PF Gold
P: 6,040




#59
Feb2114, 07:43 PM

Mentor
P: 11,580




#60
Feb2114, 07:52 PM

Sci Advisor
Thanks
P: 3,438

Consider the end of the rope when we've started in the straight up and down configuration. The boundary condition is, as you say: 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 upanddown configuration and that energy (PE plus KE) is conserved, I think that is enough to completely determine the equilibrium configuration  which is not straight upanddown. 88BitTRex, this is a really fun problem even without relativity. 


#61
Feb2114, 08:04 PM

Physics
Sci Advisor
PF Gold
P: 6,040

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. 


#62
Feb2114, 08:09 PM

Physics
Sci Advisor
PF Gold
P: 6,040




#63
Feb2114, 11:48 PM

P: 3,908




#64
Feb2214, 01:40 AM

Sci Advisor
Thanks
P: 3,438




#65
Feb2214, 04:30 AM

P: 529

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/((Rr)^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. 


#66
Feb2214, 12:34 PM

Mentor
P: 15,065

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 EarthMoon 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})/R^{3}, not GM_{E}/R^{3}. 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 $$\frac {dT}{dr} = \frac{\mu}{1+k} \Bigl( \Bigl(\frac 1 {r^2}  r\Bigr)  k \Bigl(\frac 1 {(1r)^2}  1\Bigr) \Bigr)$$ The tension in the rope is thus $$T(r) = \frac{\mu}{1+k}(rr_0)\Bigl(\Bigl(\frac 1 {r_0r}\frac{r+r_0}2\Bigr) k*\Bigl(\frac 1 {(1r)(1r_0)}  1\Bigr)\Bigr)$$ 


#67
Feb2214, 03:13 PM

Physics
Sci Advisor
PF Gold
P: 6,040




#68
Feb2214, 05:36 PM

Mentor
P: 15,065

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).



Register to reply 
Related Discussions  
Calculate tension in a rope (pulleyropemass system)  Introductory Physics Homework  1  
If the tension in the rope steadily increases, at what value of the tension does the  Introductory Physics Homework  4  
Person suspended on a rope standing on a beam (torque and rope tension)  Introductory Physics Homework  0  
Rope and moon  General Physics  28  
Moon and tides (tide on the moon instead of earth)  General Physics  7 