Question

Two 110 kg lead spheres are suspended from 200 m -long massless cables. The tops of the cables have been carefully anchored exactly 1.0 m apart.

By how much is the distance between the centers of the spheres less than 1.0 meter?

Answer 1

Gravitational force

Gravitational force F = Gm^2/r^2 = mg tan theta Where tan theta = Gm/gr^2 Distance change on each side L sin theta Total distance s = 2L sin theta For very small angles sin theta = tan theta = theta S = 2L theta S = 2L Gm/gr^2 = 2(200)(6.67x10^-11 * 110/9.8 * (1)^2) S = 2.99x10^-7 m

Where

$\tan \theta= \frac{Gm}{gr^2}$

Distance change on each side $L\sin \theta$

Total distance $s=2L \sin \theta$

For very small angles $\sin \theta = \tan \theta= \theta$

$S=2L \theta$

$S=2L \frac{Gm}{gr^2}=2(200)(\frac{6.67x10^{-11}*110}{9.8*(1)^2})$