First slide

Gravitational force and laws

Question

A uniform ring of mass m and radius r is placed directly above a uniform sphere of mass M and of equal radius. The centre of the ring is directly above the centre of the sphere at a distance $r \sqrt{3}$ as shown in the figure. The gravitational force exerted by the sphere on the ring will be

Moderate

A

$\frac{GMm}{8 r^{2}}$
B

$\frac{GMm}{4 r^{2}}$
C

$\sqrt{3} \frac{GMm}{8 r^{2}}$
D

$\frac{GMm}{8 r^{3} \sqrt{3}}$

Solution

$\begin{array}{l} dF = G \frac{Mdm}{4 r^{2}} \\ F = ΣdFcos θ \\ = Σ \frac{GMdm}{4 r^{2}} \cos θ \\ = \frac{GM}{4 r^{2}} \times \frac{\sqrt{3} r}{2 r} Σdm \\ = \frac{\sqrt{3} GMm}{8 r^{2}} \end{array}$
Alternative solution:
The gravitational field due to the ring at a distance $\sqrt{3} r$ is given by
$E = \frac{Gm (\sqrt{3} r)}{{[r^{2} + (\sqrt{3} r)^{2}]}^{3 / 2}} or E = \frac{\sqrt{3} Gm}{8 r^{2}}$
The required force is EM, i.e., $(\sqrt{3} Gm) M / 8 r^{2}$

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