A hook in the sky

well i also was thinknin the same thing but here i think there is no restriction ..

Kaymant Sir,
I was wrong, I admit.
But then, how do you physically explain the limit on the length of the rope. I mean, what will be the problem if the rope is just a bit longer, and why?

( The answer might be quite apparent, but I'm just just not being able to grasp it!)

I was wondering y such wonderful discussion wasnt painted all over, then I noticed :

#17 Posted 03:03am 18-03-09
Re: A hook in the sky

#18 Posted 04:03am 18-03-09
Re: A hook in the sky

If this rope can be set up, it will orbit the Earth and will be entirely stable. However, an entirely new trouble unfolds, if we try to actually set the rope up (which will require some ground breaking technology).
Let us try to calculate the tension T(x) as a function of (our already defined) x. At some x, we look up a mass element of length \delta x. It is moving in a circle of radius x; the required centripetal acceleration (I am back to an inertial frame) will be supplied by the gravitational acceleration and the resultant tension:
T(x)+\dfrac{GM\lambda \,\delta x}{x^2}-T(x+\delta x)=(\lambda \delta x)\omega^2 x
A slight rearrangement gives us
T(x+\delta x)-T(x)=\dfrac{GM\lambda \,\delta x}{x^2}-(\lambda \delta x)\omega^2 x
Dividing throughout by \delta x and taking limit as \delta x→0, we get
\dfrac{\mathrm{d}T}{\mathrm{d}x}=\dfrac{GM\lambda }{x^2}-\lambda\omega^2 x\qquad[1]
This differential equation together with the initial condition T(R)=0 give us
T(x)=GM\lambda \left(\dfrac{1}{R}-\dfrac{1}{x}\right)+\lambda\omega^2(x^2-R^2)
Dividing throughout by the cross-section area of the rope and using \rho as the density of the material, we obtain the stress that the material will have to withstandS=GM\rho\left(\dfrac{1}{R}-\dfrac{1}{x}\right)+\rho\omega^2(x^2-R^2)
T(x)=GM\rho \left(\dfrac{1}{R}-\dfrac{1}{x}\right)+\rho\omega^2(x^2-R^2)
The maximum of the stress S will be when dT/dx =0 , which using (1) imlpies, when x=x_m=\sqrt[3]{\dfrac{GM}{\omega^2}} which is approximately 42000 km (the distance from the center where the telecommunication satellites are generally orbiting). It turns out that the maximum stress exceeds by huge margin the tensile strength of steel or even that of carbon fiber, so such an endeavor is bound to fail with current technology.

No Abhirup aka metal, to find the length of the rope, we don't need to know the maximum tension. As already observed by Satan, this rope must be at the equator. Suppose its length is L.
Take the reference frame that is fixed to Earth rigidly and rotating at the same angular speed. In this reference frame the gravitation force on the rope is balanced by the centrifugal force.
Take the x-axis running along the rope away from the Earth (mass M, radius R), the center of the Earth being the origin. If the linear mass density of the rope be \lambda, the gravitational force on the rope is
F_g = \int_R^{L+R} \dfrac{GM\lambda }{x^2}\, \mathrm{d} x= GM\lambda\left( \dfrac{1}{R}-\dfrac{1}{R+L}\right)
while the centrifugal force is
F_c = \int_R^{L+R} \lambda \omega^2 x \, \mathrm{d} x= \dfrac{\lambda \omega^2}{2}\left((R+L)^2-R^2\right)
It is required that F_g=F_c, which gives us (after a bit of simplification) a quadratic in L:
L^2 +3LR + 2R^2-\dfrac{GM}{\omega^2R}=0
with the permitted solution
L=\dfrac{R}{2}\left(-3+\sqrt{1+\dfrac{8GM}{\omega^2R^3}}\right)
Using the terrestrial data, we get the length as a staggering L ≈ 145 000 km.

but wont the withstanding power also depend on the cross section
?

adjusting the cross section we can handle it

There must be some restriction.
I have solved this problem before.
It turns out that tension will increase with increase in height from earth's surface, and after a particular height, it will be so high that not even steel can withstand it.
If there is no restriction then the rope can be as long as we like!

well the rope will be very long (can be calculated - a finite value ) there will be two forces acting on it i)the force of gravity ii)the outward force (centripetal) when the rope if reaching vertically up in the sky these two forces will be equal so

GM/(r+x)(r) = ω²((x/2)+r)
{
the expression at the let we get after integrating
from r to r+x = Gλd(R+x)/(R+x)²
}
x= length of rope
r= radius of earth
ω= angular velocity
G= universal gravitational constant
solving the equation we get the value of x
further this equation is valid if the body is at the equator it is because at any other latitude the above two forces will not balance each other as they woudnt be oppsite

sorry sir for my earlier casualness ...

Kaymant Sir,
I don't think I should post the answer---- we've already discussed it.
But, I woulkd like to mention that to find out the length of the rope,
one needs to know the maximum "tension" that the stongest material on earth can withstand.

yes i didntsee that..

forgot to mention it will be x/2 being the location of its com