Question

Two$100kg$lead spheres are suspended from $100-m$-long massless cables. The tops of the cables have been carefully anchored exactly $1m$apart. By how much is the distance between the centers of the spheres less than $1m?$

Short answer

the distance between the centers of the spheres is$1.4\times {10}^{-5}m$

Step-by-step solution

Given information

mass of sphere $1$is $m_1=100kg$

mass of sphere $2$is $m_2=100kg$

Length of the cable$L=100m$

Explanation

The gravitational force $$\begin{gathered} F=\frac{G{m}^2}{r^2}=mg\tan \left(\theta \right) \\ \tan \left(\theta \right)=\frac{G{m}^2}{g{r}^2} \end{gathered}$$

Distance change on each side is $L\sin \left(\theta \right)$

so, the total distance is ,$S=2L\sin \left(\theta \right)$

For small angle we have,

$\sin \left(\theta \right)=\tan \left(\theta \right)=\theta$

The above equation can be written as

role="math" localid="1649926341518" $$\begin{gathered} S=2L\sin \left(\theta \right)=2L\theta \\ S=2L\left(\frac{G{m}^2}{g{r}^2}\right) \\ =\frac{2\times 100\times 6.67\times {10}^{-11}\times {\left(100\right)}^2}{=9.8\times {\left(1\right)}^2}=1.4\times {10}^{-5}m \end{gathered}$$

So, the distance between the centers of the spheres is$1.4\times {10}^{-5}m$