Question

A uniform ring of mass m is lying at a distance $1.73a$ from the centre of a sphere of mass $M$ just over the sphere where $a$ is the small radius of the ring as well as that of the sphere. Then, the gravitational force exerted by one on the other is

• A. $\frac{GMm}{8a^2}$
• B. $\frac{GMm}{(1.73a{)}^2}$
• C. $\sqrt{3}\frac{GMm}{a^2}$
• D. $1.73\frac{GMm}{8a^2}$

Answer 1

Answer: (D)

$1.73\frac{GMm}{8a^2}$

In this case the expression of gravitational force is $F=GMm\frac{r}{(r^2+R^2{)}^{\frac{3}{2}}}$
here, $r=1.73a,R=a$
$\therefore F=GMm\frac{1.73a}{(2.99a^2+a^2{)}^{\frac{3}{2}}}=GMm\frac{1.73a}{a^3(4{)}^{\frac{3}{2}}}=1.73\frac{GMm}{8a^2}($take $2.99\sim 3)$