Give the binding energy of body lying on the earth at distance $r$ from the centre of earth $\left(+\frac{ GM _{ E } m}{r}=\frac{ GM _{e} m}{ R _{ E }+h}\right)$

Two stars of masses $3\times10^{31}\, kg$ each, and at distance $2\times10^{11}\, m$ rotate in a plane about their common centre of mass $O. A$ meteorite passes through $O$ moving perpendicular to the star’s rotation plane. In order to escape from the gravitational field of this double star, the minimum speed that meteorite should have at $O$ is : (Take Gravitational constant $G = 6.67\times10^{-11}\, Nm^2\, kg^{-2}$)

A mass $m$ , travelling at speed $V_0$ in a straight line from far away is deflected when it passes near a black hole of mass $M$ which is at a perpendicular distance $R$ from the original line of flight. $a$ , the distance of closest approach between the mass and the black hole is given by the relation 

An object of mass m is raised from the surface of the earth to a height equal to the radius of the earth, that is, taken from a distance $R$ to $2R$ from the centre of the earth. What is the gain in its potential energy ?

Two bodies of masses ${m_1}$ and ${m_2}$ are initially at rest at infinite distance apart. They are then allowed to move towards each other under mutual gravitational attraction. Their relative velocity of approach at a separation distance $r$ between them is