Find the gravitational force of attraction between the ring and sphere as shown in the diagram, where the plane of the ring is perpendicular to the line joining the centres. If √8R is the distance between the centres of a ring (of mass 'm') and a sphere (mass 'M') where both have equal radius 'R'.

Solution

Physics sir · 4 steps

Shrink the sphere to a point at its centre, then use the ring's field on its own axis.

1. Sphere as point. Outside a uniform sphere, its gravity acts as if all of M sits at its centre Y. So the problem becomes a point mass M on the axis of the ring, at distance x = √8R from the ring's centre X.
2. Ring on its axis. Every bit of the ring is the same distance √(R² + x²) from the point. The sideways pulls cancel in pairs, and only the components along the axis add up. That gives the standard axial result:

   F = (GMm x)/(R² + x²)^(3/2)

3. Put in numbers. With x = √8R, we get R² + x² = R² + 8R² = 9R², and (9R²)^(3/2) = 27R³. So F = (GMm × √8R)/27R³ = √8/27·GMm/R².
4. So. The force of attraction is √8/27·GmM/R², which is option (c).

The idea to keep. A ring is not a point mass, even from far away. On its axis use Gmx/(R²+x²)^(3/2), and treat any uniform sphere as a point at its centre.

Answer from the NTA final answer key. Solution by PrepWiser.