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A thin uniform rod of length L and mass M is bent into a complete circle. A point mass m is placed at the center of this circular ring. What is the net gravitational force exerted on m?

1. 4*pi²*G*M*m/L²
2. G*M*m/(4*pi²*L²)
3. 2*pi*G*M*m/L²
4. zero

Correct answer: zero

Solution

Each element of the ring attracts m toward itself, but for every element there is a diametrically opposite element exerting an equal and opposite force on m. By symmetry all contributions cancel, so the net gravitational force at the center is zero.

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