Swinging Disco-Ball and applying Newton's Law of Gravitation

Question

Imagine you have two swinging disco balls of 180 kgdangling 1 m. horizontal distance apart from each other suspended on 100 m. massless cables. How much is the distance between the center of the sphere less than 1.0 m?

Here's what I think:

Let theta be the degree measurement between the string when ball is at rest and when the right ball swings to the left and the left to the right.

$F_{gravity}=6.67x10^{-11}(180^2)/1=2.16*10^{-6}$

Since a=F/m,

$a=2.16*10^{-6}/180\\ F_x=Tsin\theta-Fg=ma_{x}\\ F_Y=Tcos\theta-mg=0\\ $

This gives us:

$Tcos\theta=180*9.8=1764 N; T=1764/cos\theta\\ 1764 tan\theta-2.16*10^{-6}=180(1.2*10^{-8})\\ tan\theta=2.45*10^{-9}\\ \theta=1.4*10^{-7}$

We know what theta is now so because we also know the length of the string, the distance displaced by one side is $180 sin(1.4*10^{-7})=4.4*10^{-7}$. Because the question asks for the distance less than 1 m. we double that amount to get $8.8*10^{-7}$

To my surprise this answer was wrong. Am I not understanding the question correctly?

Answer 1

You've mixed distance (100m) and mass (180 kg). I don't see any other reason to multiply the mass and a trigonometric function

Answer 2

I see several mistakes. First, your force equation doesn’t represent an equilibrium: there’s a residual acceleration for some reason. Moreover, you derive this acceleration from $F_{gravity}$, so the $x$-direction equation is effectively $T\sin\theta = 2 F_{gravity}$, making $\sin\theta$ about twice as large as it should be. Then, having arrived at a very small value for $\tan\theta$ ($2.45\times10^{-9}$), you somehow end up with a value of $\theta$ that’s two orders of magnitude larger. This is not plausible, and indeed, several different sources all give $\theta = 2.45\times10^{-9}$, consistent with the small-angle approximation. After that, as Andrei points out in his answer, you multiply $\sin\theta$ by 180, the mass of the balls, instead of by 100, the length of the string. Finally, there’s another order-of-magnitude problem: $\sin(1.4\times10^{-7})$ should be on the order of $10^{-7}$, so 180 or 100 times that should be on the order of $10^{-5}$, but you’ve got $4.4\times10^{-7}$. This is at least consistent with the two orders of magnitude anomaly in the earlier arctangent calculation, so it appears that you’re making some systematic error when computing these trigonometric functions and their inverses.