Q. 335.0( 2 Votes )

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. 8.3).



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 :

Given:


Mass of the ring = M


Layout of the ring = r


Mass of the object = m


Initial distance of the object h = r


Final distance of the object 2h = 2r


To find the gravitational force on the object of mass m due to the ring let us consider an elemental mass on the ring of mass dM. The force due to this elemental mass is then,




where x is the distance of the object from the elemental mass m and



Now at the point P the force dF can be resolved into a vertical component and a horizontal component. If we consider two elemental masses diametrically opposite to each other, then the vertical component due to the two forces cancel each other as they are equal and opposite and the horizontal forces add up. Therefore, we only need to consider the horizontal component of the forces, therefore the net force on the object,



Now from the triangle OPA in the figure we have,



therefore,




When h = r



When h = 2h = 2r




Therefore, the force decreases by a factor of .


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