Ncert Exercises & Solutions

Q. 33 A mass m is placed at P at 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?

Hint: Use the concept of gravitational field along the axis of a ring.

Step 1: Find the initial force on the object.

Consider the diagram in which a system consisting of a ring and a point mass m is shown.

The gravitational force acting on an object of mass m placed at point P at a distance h along the normal through the centre of a circular ring of mass M and radius r is given by,

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

(along PO) .........(I)

Step 2: Find the final force on the object.

When the mass is displaced up to a distance of 2h, then,

F' = \frac{GMm \times 2 h}{[r^{2} + {(2 h)}^{2}]^{3 / 2}} = \frac{2 GMmh}{{(h^{2} + 4 h^{2})}^{3 / 2}} = \frac{2 GMm}{5 \sqrt{5} h^{2}} . . . . . . . . . . . . . . . . . . (ii)

Step 3: Find the ratio of the forces.

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

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