$A$ body is projected vertically upwards from the surface of the earth with a velocity equal to half the escape velocity. If $R$ is the radius of the earth,the maximum height attained by the body from the surface of the earth is

The ratio of the radii of planets $A$ and $B$ is ${k_1}$ and the ratio of acceleration due to gravity on them is ${k_2}$. The ratio of escape velocities from them will be

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 for the Earth is $11 \ km/s$. The escape velocity for a planet having twice the radius and the same density as the Earth is .......... $km/s$.

$A$ body is projected vertically from earth's surface with $\left(\frac{1}{3}\right)^{\text{rd}}$ of escape velocity. The maximum height reached by the body is ($R=$ radius of earth).

Explain escape energy and provide its definition. Also, explain escape speed.