A uniform string of large length L of uniform cross section, mass per unit length λ and free at both ends is revolving round the earth at equator in vertical position in such a way that its position r

Question

A uniform string of large length L of uniform cross section, mass per unit length λ and free at both ends is revolving round the earth at equator in vertical position in such a way that its position relative to the earth (mass of earth = M, radius = R) is always the same and its lower end is slightly above the earth's surface.

TestHub · Accepted Answer

For an element dy of the string at distance y from earth's centre, $d U = - \frac{GM λ d y}{y}$ , so $U = - GM λ ln \frac{R + L}{R}$ - (A) is correct.

Net gravitational force: $F = \int_{R}^{R + L} \frac{GM λ d y}{y ^{2}} = \frac{GM λ L}{R ( R + L )}$ - (B) is correct.

The string revolves with the earth, so the centripetal force is $F_{CP} = \int_{R}^{R + L} λ ω^{2} x d x = \frac{λ ω ^{2}}{2} [(R + L)^{2} - R^{2}] = \frac{λ ω ^{2}}{2} L (2 R + L)$ .

Equating with F: $ω = \frac{2 GM}{R ( R + L ) ( 2 R + L )}$ - (C) is correct and (D) is wrong.

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