The escape velocity of a body from a planet of mass $M$ and radius $R$ is $14 \,km \,s^{-1}$. The escape velocity of the body from another planet having same mass and diameter $8R$ (in $km \,s^{-1}$) is

$A$ body of mass $m$ is dropped from a height $h = \frac{R}{2}$ above the surface of the Earth,where $R$ is the radius of the Earth. Find its speed when it hits the Earth's surface. (Given: $v_e$ is the escape velocity from the Earth's surface).

Earth has mass $M_1$ and radius $R_1$,and the moon has mass $M_2$ and radius $R_2$. The distance between their centers is $r$. $A$ body of mass $M$ is placed on the line joining them at a distance $r/3$ from the center of the Earth. To project the mass $M$ to escape to infinity,the minimum speed required is:

The escape velocity for a body projected vertically upwards from the surface of earth is $11.2 \ km/s$. If the body is projected at an angle of $45^o$ with the vertical,the escape velocity will be ......... $km/s$.

The escape velocity from the surface of Earth of mass $M$ and radius $R$ is $V_{e}$. The escape velocity from the surface of a planet whose mass and radius are $3$ times that of the Earth will be:

$A$ body of mass $m$ is projected with velocity $\lambda v_{e}$ in a vertically upward direction from the surface of the earth into space. It is given that $v_{e}$ is the escape velocity and $\lambda < 1$. If air resistance is considered to be negligible,then the maximum height from the center of the earth,to which the body can go,will be: ($R$: radius of earth)