$A$ mass of $6 \times 10^{24} \,kg$ is to be compressed in the form of a solid sphere such that the escape velocity from its surface is $3 \times 10^4 \,ms^{-1}$. The radius of the sphere is (Universal gravitational constant $G = 6.66 \times 10^{-11} \,N \,m^2 \,kg^{-2}$) (in $\,km$)

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

An object is thrown directly away from the surface of the earth with an initial speed $v$. The object reaches up to a height of $\frac{4}{5} R_E$ from the earth's surface,where $R_E$ is the radius of the earth. If the escape velocity of the object is $v_E$,then the value of $\frac{v}{v_E}$ is:

$A$ particle of mass $m$ is thrown upwards from the surface of the earth with a velocity $u$. The mass and the radius of the earth are $M$ and $R$, respectively. $G$ is the gravitational constant and $g$ is the acceleration due to gravity on the surface of the earth. The minimum value of $u$ so that the particle does not return back to earth 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 kinetic energy needed to project a body of mass $m$ from the earth's surface (radius $R$) to infinity is: