Going up!

Let $R$ be earth radius, $M$ be its mass, and $\omega$ be its angular velocity.Since, the time period of revolution of pipe is same as earth's , its angular velocity would be same as that of earth ( $\omega$ )

Let $T(r)$ be tensile stress at distance $r$ from center of earth.Let $\rho$ be density of pipe.

Consider forces on a small element of mass $\rho A dr$ at a distance $r$ having thickness $dr$ ,

$\frac{GM\rho Adr}{r^2} - AdT = \rho A dr {\omega}^{2} r$

$\Rightarrow \frac{dT}{dr} = \rho r(\frac{GM}{r^3} - \omega^2)$

When $T$ is max, the pipe should be thickest, and at that point $\frac{dT}{dr} = 0$

$\Rightarrow r = \sqrt[3] {\frac{GM}{\omega^2}} = \boxed{\sqrt[3] {\frac{gR^2}{\omega^2}}}$

Substitute $\omega = \frac{2 \pi}{86400} rad/sec$ to get $r = \boxed{4.23 \times 10^7}$

Consider a small element of the tether at a distance $r$ from the center of the Earth, having length $dr$ . Let's analyze the forces acting on this element. Let $M$ be the mass of the Earth, and let $\omega$ be its angular velocity. Note that the tension throughout the tether varies as a function of $r$ . Let $T(r)$ denote the tension of the tether at height $r$ . The forces acting on the small portion of the tether are:

i) Gravity: The gravitational force is directed downwards (towards the center of the Earth). From Newton's Law of Gravitation, we know that this force has magnitude $\dfrac{Gm(r)M}{r^2}$ .

ii) Centrifugal force: Note that we are working in a rotating frame, so we will encounter a centrifugal force. This force is directed upwards (away from the center of the Earth). The magnitude of this force is given by $m(r)\omega^2r$ .

iii) Tension: The force exerted by tension comes from the difference between the tensions at the upper and lower edges of this element. The magnitude of the tension, thus, is given by $T(r+dr)-T(r)$ , and is directed upwards.

Notice the subtlety of the previous comment. If $T(r+dr)>T(r)$ , from commonsense we know that the force should be directed upwards. Indeed, note that $T(r+dr)>T(r) \implies T(r+dr)-T(r)>0$ , so we have a positive force acting upwards. If $T(r+dr)<T(r)$ , from commonsense we know that the force will be directed downwards. Again, $T(r+dr) < T(r) \implies T(r+dr)-T(r)<0$ , so we have a negative force directed upwards, which is equivalent to a positive force directed downwards. Either way, the expression $T(r+dr)-T(r)$ changes its magnitude and sign accordingly.

Let's try to play with the term $T(r+dr)-T(r)$ a bit. Note that $T(r+dr)-T(r)= \frac{T(r+dr)-T(r)}{(r+dr)-r} \times dr = T'(r)dr$

See the free body diagram below:
If the image doesn't load, go to:-http://s21.postimg.org/xh2fp7v1j/Untitled.png

Balancing the forces, we obtain: $m(r) \omega^2 r + T'(r) d(r)= \frac{GMm(r)}{r^2}$ $\implies T'(r) d(r)= \frac{GMm(r)}{r^2} - m(r)\omega^2 r$ Now, let $\rho (r)$ be the density of this element, and let $A$ be its circular cross-sectional area (note that $A$ remains constant throughout the rope). The mass of the element is given by: $m(r)= \rho(r)Adr$ Plugging these values, we obtain: $T'(r) d(r)= \frac{GM\rho(r)Adr}{r^2} - \rho(r)Adr\omega^2r$ $\implies T'(r) \frac{GM \rho(r) A}{r^2} - \rho(r)A\omega^2r$ Notice that $\rho(r)$ is maximized when $T'(r)=0$ , so $\frac{GM \rho(r) A}{r^2} - \rho(r)A\omega^2r = 0$ $\implies r= \sqrt[3]{\frac{GM}{\omega^2}}$ Let $T$ be the time period of the rotation of the Earth, so $\omega = \dfrac{2 \pi}{T}$ . We then have $r= \sqrt[3]{\frac{GMT^2}{4 \pi^2}}$ Plugging the values from the question gives $r \approx \boxed{4.23 \times 10^{7} \text{ meter} }$ .

The cable would need to be thickest at the point where the tension in the cable is greatest. Consider the infinitesimal segment right at the bottom of the cable. This section of the cable has tension directed upward to counteract the force of gravity and to maintain a circular orbit. The next infinitesimal part of the cable, moving upward, will, by Newton's third law, have a tension force equal in magnitude directed downward plus a tension force that needs to counteract both the other tension force and gravity in order to maintain a circular orbit. Following this pattern, the net tension in the rope will increase until the tension force no longer has to counteract gravity and instead needs to work with gravity to maintain a circular orbit. This point occurs when the infinitesimal segment of the rope would be in a geostationary orbit since gravity is providing the only net force (the two tension forces, although great in magnitude, cancel each other here). Now we just need to calculate at which height this orbit occurs. Letting $R$ be the radius of geostationary orbits and $\omega$ be the angular frequency of the earth and the cable, the net force, provided solely by gravity must be: $F = \omega^{2}Rm$ We know this force is being provided solely by gravity, and using Newton's law of Gravitation: $\frac{ GMm }{R^{2}} = \omega^{2}Rm$ canceling $m$ and rearranging $R^{3} = \frac{ GM}{\omega^{2}}$ To find $\omega$ we use the relation $T = \frac{ 2\pi}{\omega}$ and we know that the period $T = 86400$ seconds solving for $\omega$ gives $\omega = 7.27 \cdot 10^{-5}$ plugging in $G = 6.67 \cdot 10^{-11}$ and $M = 6 \cdot 10^{24}$ and $\omega$ $R^{3} = 7.57 \cdot 10^{22}$ $R = 4.23 \cdot 10^{7}$ Which, by the logic stated above, is the point of greatest tension, so it must be the thickest part.

Let us assume the space tether to be a non uniform cable of variable mass distribution such that its center of mass is at a height 'h' from the center of the earth. We may assume the total mass of the tether to be concentrated at that point. Let it be 'm'. The mass of the earth M may also be assumed to be concentrated at its center. Thus we must simply consider the equilibrium of the mass m by balancing the gravitational pull of the Earth with the centrifugal pseudo force.

Thus, we have, GmM/h^2 = mv^2/h.

But here v = 2 x pi x h /T, where T is orbital time, T=24x 60 x 60 in seconds.

Thus, after plugging the values and solving the above equations, we get the answer as 4.23 x 10^7 m.

Consider a small mass at a distance r from the centre of the earth Let T[r] be the tension in the string directed toward the centre of the earth. So the net tension on an elemental mass is [T+dT]-T=dT

tention =T(r)

dT/dr =T' therefore *dT=T'dr *
let the elemental partical have mass m{r} at a distance r

m{r}=DAdr where D is the density and A

is the crosssectional area of the elemental part Gm{r}M/r^2 = m{r}rw^2 + T'dr

we get by solving *T'=DA[GM/r^2 - rw^2] *

* The tension in the sring is maximum where the area is maximum so that the tensile stress remains contant * T' = 0 we get r^3= GM/w^2

on putting values we obtain the final answer as * r=4.23 E+7 *

For ordinary tension problems in newtonian mechanics, the mass of the string is neglected, therefore the tension is constant in both end of the string; however since we consider the mass of the string, it would have a variable tension.

$dT = dm_{s} a$ (1)

where $dT$ is the differerntial tension on a segment $dm_{s}$ of the string; experiencing a net acceleration $a$ .

The net acceleration $a$ is the sum of acceleratons due to gravity and due to earth's rotation,

$a = ω^{2} r - \frac{GM_{e}}{r^{2}}$ (2)

where $ω$ is the angular speed of earth's rotation $\frac{2π}{Period}$ ; {r) is the distance from the earth's center; $G$ is the universal gravitational constant and $M_{e}$ is the earth's mass.

Assuming that the mass of the string depends on it's length;

$dm = λdr$ (3)

where $λ$ is the mass per unit length of the string. Putting it all together;

$dT = λdr [ω^{2} r - \frac{GM_{e}}{r^{2}}]$ (4)

$\frac{dT}{dr} = λ [ω^{2} r - \frac{GM_{e}}{r^{2}}]$ (5)

to find the maximum tension (where the string is thickest);

$\frac{dT}{dr} = 0$ (6)

$λ [ω^{2} r - \frac{GM_{e}}{r^{2}}] = 0$ (7)

$[\frac{GM_{e}}{ω^{2}}]^{\frac{1}{3}} = r$ (8)