FF = GMmd2\frac{GMm}{d^2}

Parametric form for x2+y2=12x^2 + y^2 = 1^2 is (cosθ,sinθ).(\cos \theta, \sin \theta).

tanϕ\tan \phi = sinθ2cosθ\displaystyle \frac{\sin \theta}{2 - \cos \theta} {ϕ\phi is an angle at (2, 0)}

    cosϕ\implies \cos \phi = 2cosθ54cosθ\displaystyle \frac{2 - \cos \theta}{\sqrt{5 - 4 \cos \theta}}

D = (cosθ2)2+(sinθ0)2\sqrt{(\cos \theta - 2)^2 + (\sin \theta - 0)^2} = 54cosθ\sqrt{5 - 4 \cos \theta}

Therefore, D cosϕ\cos \phi = d = 2 - cosθ\cos \theta

d=1π0π(2cosθ)dθ\overline{d} = \displaystyle \frac{1}{\pi} \int_0^{\pi} (2 - \cos \theta) d \theta = 1π[2θsinθ]0π=2\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).

FF = GmG m 12π\displaystyle \frac{1}{2 \pi} ππdMdθdθd2\displaystyle \int_{-\pi}^{\pi} \frac{d M}{d \theta} \frac{d \theta}{d^2}

    F=GmdMdθ12π\implies F = G m \frac{d M}{d \theta} \frac{1}{2 \pi} ππdθd2\displaystyle \int_{-\pi}^{\pi} \frac{d \theta}{d^2} = GmG m dMdθ1π\frac{d M}{d \theta} \frac{1}{\pi} 0πdθ(2cosθ)2\displaystyle \int_{0}^{\pi} \frac{d \theta}{(2 - \cos \theta)^2} = GmG m dMdθ1π\frac{d M}{d \theta} \frac{1}{\pi} [1.20919957615588]

    F=Gm\implies F = G m dMdθ2π\frac{d M}{d \theta} {2 \pi} [0.061258766157963] = GmM[0.061258766157963]G m M [0.061258766157963] = GMm4.040321061266242\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.

Note by Lu Chee Ket
3 years, 11 months ago

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Hi could u specify your question?

Aditya Kumar - 3 years, 11 months ago

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My question is actually in a process of determining the force between a hollow sphere and a point mass. I were to determine that they are not equivalent to force between point mass and point mass.

Radius of 1 for the circle or semicircle with mass distributed along while a distance of 2 from its center to a point mass is fixed.

I am wondering what is the correct equivalent distance between the circle or semi circle to the point mass, in term of distance for the inverse square law of force of attraction between them.

Lu Chee Ket - 3 years, 11 months ago

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