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

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

Answers (1)

Let us consider a small element of a ring of mass dM at point A and the distance between dM and m is x.

x^{2}=r^{2}+h^{2}

The gravitational force between dM and m, dF=\frac{G(dM)m}{x^{2}}

dF has 2 components: one along PO and another perpendicular to PO

\int \; \sin \theta =0 (due to the symmetry of the ring)

F=\int dF\; \cos \theta =\int \frac{G(dM)m}{x^{2}}\frac{h}{x}

=\frac{Gmh}{x^{3}}\int dM=\frac{GMmh}{x^{3}}=\frac{GMmh}{(r^{2}+h^{2})^{\frac{3}{2}}}

When the mass is displaced to a distance of 2 h F'=\frac{GMm(2h)}{(r^{2}+(2h)^{2})^{\frac{3}{2}}}=\frac{2GMmh}{(r^{2}+4h^{2})^{\frac{3}{2}}}

Where h =r                                      F=\frac{GMmr}{(r^{2}+r^{2})^{\frac{3}{2}}}=\frac{GMm}{2\sqrt{2}r^{2}}

F'=\frac{GMmr}{(5r^{2})^{\frac{3}{2}}}=\frac{2GMm}{5\sqrt{5}r^{2}}

\frac{F^{'}}{F}=\frac{4\sqrt{2}}{5\sqrt{5}}

F'=\frac{4\sqrt{2}}{5\sqrt{5}}F

by a factor of \frac{4\sqrt{2}}{5\sqrt{5}}, the force of gravitation will decrease.

Posted by

infoexpert23

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