The potential energy of a satellite is $-8 \times 10^9 \ J$. What is its binding energy (escape energy)?

$A$ body is projected vertically from the Earth's surface with a velocity equal to half the escape velocity. The maximum height reached by the body is ($R =$ radius of the Earth).

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 the meteorite should have at $O$ is: (Take Gravitational constant $G = 6.67\times10^{-11} \ Nm^2 \ kg^{-2}$)

The escape velocity of a rocket launched from the surface of the earth:

If $V, R$ and $g$ denote respectively the escape velocity from the surface of the earth,the radius of the earth,and the acceleration due to gravity,then the correct equation is:

The radius of the planet is double that of the earth,but their average densities are same. $V_{p}$ and $V_E$ are the escape velocities of planet and earth respectively. If $\frac{V_p}{V_E}=x$,the value of '$x$' is