\(F\) = \(\frac{GMm}{d^2}\)

Parametric form for $x^2 + y^2 = 1^2$ is $(\cos \theta, \sin \theta).$

$\tan \phi$ = $\displaystyle \frac{\sin \theta}{2 - \cos \theta}$ {$\phi$ is an angle at (2, 0)}

$\implies \cos \phi$ = $\displaystyle \frac{2 - \cos \theta}{\sqrt{5 - 4 \cos \theta}}$

D = $\sqrt{(\cos \theta - 2)^2 + (\sin \theta - 0)^2}$ = $\sqrt{5 - 4 \cos \theta}$

Therefore, D $\cos \phi$ = d = 2 - $\cos \theta$

$\overline{d} = \displaystyle \frac{1}{\pi} \int_0^{\pi} (2 - \cos \theta) d \theta$ = $\displaystyle \frac{1}{\pi} [2 \theta - \sin \theta]_0^{\pi} = 2$

There is nothing wrong with center of gravity to have an average distance of 2 from (0, 0) to (2, 0).

$F$ = $G m$ $\displaystyle \frac{1}{2 \pi}$ $\displaystyle \int_{-\pi}^{\pi} \frac{d M}{d \theta} \frac{d \theta}{d^2}$

$\implies F = G m \frac{d M}{d \theta} \frac{1}{2 \pi}$ $\displaystyle \int_{-\pi}^{\pi} \frac{d \theta}{d^2}$ = $G m$ $\frac{d M}{d \theta} \frac{1}{\pi}$ $\displaystyle \int_{0}^{\pi} \frac{d \theta}{(2 - \cos \theta)^2}$ = $G m$ $\frac{d M}{d \theta} \frac{1}{\pi}$ [1.20919957615588]

$\implies F = G m$ $\frac{d M}{d \theta} {2 \pi}$ [0.061258766157963] = $G m M [0.061258766157963]$ = $\displaystyle \frac{G M m}{4.04032106126624^2}$

I am facing a little bit of confusion until here. It seems that point mass of M to m which takes d = 2 becomes d = 4.04 for scattered mass of M to m. Cancellation of vertical forces is caused by similarity between two hemispheres. Although I am not sure whether a stronger force or a weaker force suppose to act, the point is the fate of average distance is not equivalent to the fate of inverse square law.

Please help to get the correct equivalent of effective distance which may not be 4.04 as described above.