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Question

(A) Let the angle made by each string with the vertical be $	heta$. The distance between the two spheres is $r = L + 2L \sin 	heta$. Since the angle

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$	an^{-1}\left[\frac{GM}{gL^2}ight]$

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(A) Let the angle made by each string with the vertical be $	heta$. The distance between the two spheres is $r = L + 2L \sin 	heta$. Since the angle $	heta$ is very small,$\sin 	heta \approx 	heta$,so $r \approx L$. The gravitational force of attraction between the spheres is $F = \frac{GM^2}{r^2} \approx \frac{GM^2}{L^2}$. For equilibrium of one sphere,the forces acting are tension $T$,gravitational force $F$,and weight $Mg$. Resolving forces: $T \sin 	heta = F = \frac{GM^2}{L^2}$ $T \cos 	heta = Mg$ Dividing the two equations: $	an 	heta = \frac{F}{Mg} = \frac{GM^2 / L^2}{Mg} = \frac{GM}{gL^2}$ Therefore,$	heta = 	an^{-1}\left[\frac{GM}{gL^2}ight]$.

Two metallic spheres each of mass $M$ are suspended by two strings each of length $L$. The distance between the upper ends of the strings is $L$. The angle which the strings will make with the vertical due to the mutual attraction of the spheres is

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