Question

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 (figure).

If the mass is moved further away such that OP becomes 2h, by what factor the force of gravitation will decrease, if h = r?

Answer 1

Gravitational force F at P

F_h = \(\frac{GMm\cos\theta}{AP^2}\) = \(\frac{GMhh}{(r^2+h^2)^{\frac{3}{2}}}\)

\(\because\) cosθ = \(\frac{h}{(r^2+h^2)^{\frac{1}{2}}}\)

\(\frac{F_r}{F_{2r}}\) = \(\frac{\frac{GMm.r}{(r^2+r^2)^{\frac{3}{2}}}}{\frac{GMm2r}{[r^2+(2r)^2]^{\frac{3}{2}}}}\)

or \(\frac{(r^2+4r^2)^{\frac{3}{2}}}{2(2r^2)^{\frac{3}{2}}}\)

\(\frac{F_r}{F_{2r}}\) = \(\frac{(5r^2)^{\frac{3}{2}}}{2\sqrt2r^3}\)

= \(\frac{5\sqrt5r^3}{4\sqrt2r^3}\)

\(\frac{F_r}{F_{2r}}\) = \(\frac{5}{4}\sqrt{\frac{5}{2}}\)

or \(\frac{F_{2r}}{F_{r}}\) = \(\frac{4}{5}\sqrt{\frac{2}{5}}\)

or F_2r = \(\frac{4}{5}\sqrt{\frac{2}{5}}\)F_r