question_answer 58)

                  A mass m is placed at P a distance h along the normal through the centre O of a thin circular ring of mass M and radius r (Fig.).                 If the mass is removed further away such that OP becomes 2h, by what factor the force of gravitation will decrease, if \[h=r\]?

Answer:

                  Gravitational force on an object of mass m at point P on the axis of die circular ring of radius r and mass M is given by                 \[F=\frac{GMmh}{{{({{r}^{2}}+{{h}^{2}})}^{3/2}}}\]                         Where h = 2h, then                 \[F'\,=\,\frac{GMm\,\times \,2h}{{{({{r}^{2}}+4{{h}^{2}})}^{3/2}}}\]                 If \[h=r,\] then                 \[F=\frac{GMmr}{2\sqrt{2}\,{{r}^{3}}}\,=\frac{GMm}{2\sqrt{2}\,{{r}^{2}}}\]                 and \[F'=\frac{2GMmr}{{{(5{{r}^{2}})}^{3/2}}}\,=\frac{2\,GMm}{5\sqrt{5}\,{{r}^{2}}\,}\]                 \[\therefore \]\[\frac{F'}{F}=\,\,\frac{4\sqrt{2}}{5\sqrt{5}}\] or \[F'\,=\frac{4\sqrt{2}}{5\sqrt{5}}\,F\]