Two 99 kg lead spheres are suspended from 130.3 m long massless cables. The top of the cables have been carefully anchored exactly 1.02 m apart. What is the distance between the centers of the spheres?

Respuesta :

Answer:

the distance between the centers of the spheres is 1.72 × 10⁻⁷m

Explanation:

The mass of the two spheres ,m = 99kg

The length of the cable is L= 130.3 m

The gravitational force

[tex]F = \frac{Gm^2}{r^2} = mg\tan\theta[/tex]

[tex]\tan\theta=(\frac{Gm}{gr^2} )[/tex]

Distance change on each side is [tex]L \sin\theta[/tex]

so, the total distance is ,S = [tex]2L\sin\theta[/tex]

For small angle we have,

[tex]\sin\theta=\tan\theta=\theta[/tex]

The above equation can be written as

[tex]S = 2L\sin\theta=2l\theta[/tex]

[tex]S = 2L(\frac{Gm}{gr^2}) \\\\= 2(130.3)(\frac{(6.67\times10^-^1^1)(99)}{(9.8\times1.02)} )\\\\= 1.72\times10^-^7m[/tex]

So, the distance between the centers of the spheres is 1.72 × 10⁻⁷m

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