Newton's Law Of Universal Gravitation

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If the distance between two interacting masses is reduced to one-third of its original value while the masses stay fixed, by what factor does the gravitational force change?

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NCERT Class 11, Chapter 8 (Gravitation) emphasises the inverse-square character of gravity: with the masses held constant, $F \propto \dfrac{1}{r^2}$. This nonlinear dependence is central, because it means equal changes in distance produce unequal changes in force, and small reductions in separation can strongly amplify the attraction. Writing the ratio of the new force to the old force, $\dfrac{F'}{F} = \dfrac{r^2}{r'^2}$, and substituting $r' = \dfrac{r}{3}$ gives $\dfrac{F'}{F} = \dfrac{r^2}{(r/3)^2} = \dfrac{r^2}{r^2/9} = 9$. Hence the force becomes nine times larger when the distance is cut to one-third. The option 'Decreases 9 times' reverses the relationship, mistakenly treating force as proportional to $r^2$ rather than its inverse. 'Increases 3 times' ignores the squaring of the distance factor and treats the dependence as linear in $r$. 'Remains the same' wrongly assumes only the masses determine the force. As a plausibility check, bringing masses closer must strengthen the attraction, so the force has to increase, and squaring the factor of three confirms the precise ninefold rise. This same inverse-square reasoning recurs throughout gravitation and electrostatics.

Newton's Law Of Universal Gravitation

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