Physics - Gravitation Question with Solution | TestHub

A. The gravitational potential energy of a system of three masses is the sum of potential energies of all unique pairs.$$U=-\frac{G{m}^2}{a}-\frac{G{m}^2}{a}-\frac{G{m}^2}{a}=-\frac{3G{m}^2}{a}$$ So A is correct.

B. Initial potential energy $U_{initial}=-3G{m}^2/a$. When one mass is removed to infinity, the remaining two masses have potential energy $U_{final,system}=-G{m}^2/a$. The work done by an external agent is $W_{ext}=U_{final,system}-U_{initial}=(-G{m}^2/a)-(-3G{m}^2/a)=2G{m}^2/a$. So B is correct.

C. The net gravitational force on any one mass due to the other two masses is the vector sum of two forces of magnitude $G{m}^2/a^2$ with an angle of $60^{\circ }$ between them. The resultant force is $2(G{m}^2/a^2)\cos (30^{\circ })=\sqrt{3}G{m}^2/a^2$. By symmetry, this resultant force acts along the median (and altitude) towards the midpoint of the opposite side, which passes through the center (centroid) of the equilateral triangle. Thus, it will move towards the center of the triangle. So C is correct.

D. If two masses are fixed, the potential energy of the third mass at a vertex is $U_m=-G{m}^2/a-G{m}^2/a=-2G{m}^2/a$. For the mass to escape, its total mechanical energy must be non-negative. $K+U_m\ge 0\Rightarrow \frac{1}{2}m{v}^2-\frac{2G{m}^2}{a}\ge 0\Rightarrow v^2\ge \frac{4Gm}{a}$. Thus, the minimum escape velocity is $\sqrt{4Gm/a}$. So D is correct.