1well i also was thinknin the same thing but here i think there is no restriction ..
Suppose you find a motionless rope reaching vertically up in the sky (without anybody holding the other end) and hanging down nearly to the ground. What would be the length of the rope?
-
UP 0 DOWN 0 3 18
18 Answers
11Kaymant 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!)
33also I think there may be an issue of stable equilibrium?--Nishant Bhaiya...
I think the Rope will be in stable equilibrium..because... centrifugal force we will take to ct at COM while gravity will act at Center of gravity... which is lower than COM here....so the point of force acting upward is above the point of application of downward force..
21I 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
66If 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.
66No 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.
62Then may be we will need to omuch steel to be produced?
also I think there may be an issue of stable equilibrium?
1but wont the withstanding power also depend on the cross section
?
adjusting the cross section we can handle it
1There 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!
1well 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 ...
1Kaymant 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.
66@satan
what Nishant sir said is right. Gravitational forces depend on the distance and and so does the centripetal (or centrifugal, depending upon your choice of reference frame). Hence, you cannot find the forces by simply considering the center of mass. And when I say "just touching the ground", it means exactly that--- practically it touches the ground. This is not a difficult question. However, it has been modeled upon a practical problem: Can we have a lift to the sky?
62@satan..
in that case should it not have beenintegral of
∫Gdm/R2 limit from x-l/2 to x+l/2??
62@satan
there is one more thing
the gravitational foce could vary with height
because i think height will be very large..
dont know what kaymant sir has in mind
62good work satan.. let Kaymant sir confirm..
Today someone was asking me if there is some user on targetiit who can get into top 100. :)