Question

Find the gravitational force of attraction between the ring and sphere as shown in the diagram, where the plane of the ring is perpendicular to the line joining the centres. If $ \surd 8R$ is the distance between the centres of a ring (of mass $ ‘m’$) and a sphere (of mass $ ‘M’$) where both have equal radius $ ‘R’$.

• A. $\left(\frac{\sqrt{8}}{9}\right)\frac{GmM}{R}$
• B. $\left(\frac{\sqrt{8}}{27}\right)\frac{GmM}{R^2}$
• C. $\left(\frac{\sqrt[2]{2}}{3}\right)\frac{GmM}{R^2}$
• D. $\left(\frac{1}{\sqrt[3]{8}}\right)\frac{GmM}{R^2}$

Answer 1

The correct option is B

$\left(\frac{\sqrt{8}}{27}\right)\frac{GmM}{R^2}$

Step 1. Given data

$\surd 8R$ is the distance between the centers of a ring (of mass $‘m’$) and a sphere (of mass $‘M’$) where both have an equal radius $‘R’$.

Step 2. Calculating gravitational force of attraction

We know the gravitational field of the ring is given as $=-\frac{Gmx}{{\left(R^2+x^2\right)}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}$

Also, the force between sphere and ring is

$$\begin{gathered} =\frac{GmM\left(\sqrt{8}R\right)}{{\left(R^2+8R^2\right)}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}} \\ =\frac{GmM}{R^2}\times \frac{\sqrt{8}}{27} \end{gathered}$$

Where, $G$ is the universal gravitational constant

$m$is the mass of the ring

$M$ is the mass of the sphere

$R$is the distance between ring and sphere

Hence, the correct option is (B).