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I've been reading about the Shell Theorem and how the gravitational force inside of a sphere is equal to zero. I was wondering if the same was true for a circle, but couldn't find any definitive answer as to how to calculate it and whether the net gravity would also be zero.

How do you calculate the gravity inside of a circle of mass?

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    $\begingroup$ It’s zero only at the center of the circle. $\endgroup$
    – G. Smith
    Commented Jul 7, 2019 at 0:32
  • $\begingroup$ You can treat the ring as an infinite collection of infinitesimal masses and add up their contributions by doing an integral over the circle. If I remember correctly, the integral is simple only on the axis that passes through the middle and is perpendicular to the circle. $\endgroup$
    – G. Smith
    Commented Jul 7, 2019 at 0:37
  • $\begingroup$ Integration always lets you express the potential and field of any mass distribution, and numerically calculate it on a computer, but the integral often cannot be done analytically. $\endgroup$
    – G. Smith
    Commented Jul 7, 2019 at 0:55
  • $\begingroup$ isn't a circle two dimensional, and would have no mass? $\endgroup$
    – Adrian Howard
    Commented Jul 7, 2019 at 1:45
  • $\begingroup$ @AdrianHoward Since the OP was talking about spherical shells, I think the question is about a circular ring, not a circular disk. In either case, physicists idealize one- and two-dimensional mass distributions by thinking in terms of a linear mass density (i.e., mass per unit length) or an areal mass density (mass per unit area). If $\lambda$ is the linear mass density of a ring of radius $a$, then a segment of angular size $d\theta$ has mass $ dm=\lambda ds=\lambda a d\theta$. $\endgroup$
    – G. Smith
    Commented Jul 7, 2019 at 2:09

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Let $M$ mass of ring, $R_1$ radius of ring and $R_2$ field point with gravitational acceleration

$$g ={\frac {Gm}{r^{2}}} = \oint_l Gdm/r^2 = \\ \displaystyle{\frac{GM}{R_1^2}\int_0^{2\pi} \frac{d\phi}{\sin^2\phi + \left(\cos\phi - \frac{R_2}{R_1}\right)^2}} = \\ \displaystyle{\frac{GM}{R_1^2}\int_0^{2\pi} \frac{d\phi}{1 - 2\frac{R_2}{R_1}\cos\phi + \left(\frac{R_2}{R_1}\right)^2}} = \\ \displaystyle{\frac{GM}{R_1^2} \frac{2\pi}{\sqrt{\left(1+\frac{R_2}{R_1}^2\right)^2 - 4\left(\frac{R_2}{R_1}\right)^2}}}$$

A relative plot etc. at Wolfram Alpha where $x = R_2/R_1$.

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ok then as a thought experiment, inside a two dimensional circle considered to have constant mass the same net gravitational pull, for an object inside it, in the same plane, would be the same as inside a sphere, In a hollow circle having mass only around the outer edge (as a ring), net gravity anywhere inside would be zero, because even if the object were off center it would be pulled equally in all directions, because of the inverse square law of gravitational distance. in a solid circle, having constant mass all across it (as a plate), net gravity would be zero only in the center.

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  • $\begingroup$ Could you provide some reference deriving the amount? Any citations? $\endgroup$
    – ThePiachu
    Commented Jul 7, 2019 at 7:52

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