Question

The particles A and B of mass m each are separated by a distance r. Another particle C of mass M is placed at the midpoint of A and B. Find the work done in taking C to a point equidistant r from A and B without acceleration (G= Gravitational constant and only gravitational interaction between A, B and C is considered)

• A. $2GmM\times (\frac{2}{R}-\frac{1}{r})$
• B. $2GmM\times (\frac{2}{R}+\frac{1}{r})$
• C. $\frac{3GMm}{r}$
• D. $\frac{4GMm}{r}$

Answer 1

The correct option is A $2GmM\times (\frac{2}{R}-\frac{1}{r})$

Work done by external agent (without changing kinetic energy of the system of particles)

$=$Final potential energy of the system$-$Initial petential energy of the system

$=U_f-U_i$

$U_i=\frac{-GmM}{\frac{R}{2}}+\frac{-GmM}{\frac{R}{2}}+\frac{-Gmm}{R}$

$U_f=\frac{-GmM}{r}+\frac{-GmM}{r}+\frac{-Gmm}{r}$

$W_{externalagent}=(dfrac-GmMr+\frac{-GmM}{r}+\frac{-Gmm}{r})-(\frac{-GmM}{\frac{R}{2}}+\frac{-GmM}{\frac{R}{2}}+\frac{-Gmm}{R})$

$=(dfrac-2GmMr+\frac{-Gmm}{r})-(\frac{-2GmM}{\frac{R}{2}}+\frac{-Gmm}{R})$

$=\frac{-2GmM}{r}+\frac{4GmM}{R}$

$=2GmM\times (\frac{2}{R}-\frac{1}{r})$