Q.A point mass $m$ is placed at a point P lying on the axis of a thin uniform circular ring of mass $M$ and radius $r$ — that is, P is on the straight line through the centre $O$ of the ring and perpendicular to its plane. Initially the distance of the mass from the centre is $OP = h$. The mass is then moved further away along the same axis until its distance from the centre becomes $OP = 2h$. Taking $h = r$, by what factor does the gravitational force of attraction on $m$ due to the ring decrease?

Newton's Law of Universal Gravitation

Every particle of matter attracts every other particle with a force directed along the line joining them. For two point masses $m_1$ and $m_2$ separated by a distance $r$, the magnitude of the attraction is

where $G = 6.67 \times 10^{-11}\ \text{N m}^{2}\,\text{kg}^{-2}$ is the universal gravitational constant. The force is attractive, acts along the line of centres, and obeys Newton's third law (equal and opposite on the two bodies). Its defining feature is the inverse-square dependence on separation.

Superposition. When several masses act on a body, the net gravitational force is the vector sum of the individual forces. For an extended body — a rod, ring or shell — the body is treated as a collection of point-mass elements and the contributions are integrated. For a uniform ring of mass $M$ and radius $R$, the force on a point mass $m$ placed on its axis at distance $h$ from the centre is directed along the axis and has magnitude

$F = \frac{G M m\, h}{\left(R^{2} + h^{2}\right)^{3/2}},$ …