Ring Gravity

Classical Mechanics Level 3

There is a circular ring in the x y xy plane of radius 1 1 centered on the origin. It has a uniform linear mass density of 1 1 . The universal gravitational constant is G G .

There is a test point at ( x , y ) = ( 1 2 , 0 ) (x,y) = \Big(\frac{1}{2} , 0 \Big) .

If the absolute value of the gravitational field strength at the test point is α \alpha , give your answer as α G \large{\frac{\alpha}{G}} .

Note: There is no ambient gravity


The answer is 2.167.

Steven Chase by Steven Chase , 37, USA

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1 solution

Otto Bretscher
Dec 21, 2018

Doing it with "brute force" (and using symmetry) we find α G = ∫ C x − 1 2 ( ( x − 1 2 ) 2 + y 2 ) 3 2 d s = ∫ 0 2 π cos ⁡ θ − 1 2 ( ( cos ⁡ θ − 1 2 ) 2 + sin ⁡ 2 θ ) 3 2 d θ ≈ 2.167 \frac{\alpha}{G}=\int_{C}\frac{x-\frac{1}{2}}{\left((x-\frac{1}{2})^2+y^2\right)^{\frac{3}{2}}}ds=\int_0^{2\pi}\frac{\cos\theta-\frac{1}{2}}{\left((\cos\theta-\frac{1}{2})^2+\sin^2\theta\right)^{\frac{3}{2}}}d\theta\approx \boxed{2.167}

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