The escape velocity for a body projected vertically upwards from the surface of the Earth is $11 \ km/s$. If the body is projected at an angle of $45^o$ with the vertical,the escape velocity will be ........... $km/s$.

If ${v_e}$ and ${v_o}$ represent the escape velocity and orbital velocity of a satellite corresponding to a circular orbit of radius $R$,then

The escape velocity of a body from the surface of the Earth:

The escape velocity of an object from the Earth depends upon the mass of the Earth $(M)$,its mean density $(\rho)$,its radius $(R)$,and the gravitational constant $(G)$. Thus,the formula for escape velocity is:

If a planet of mass $6.4 \times 10^{23} \ kg$ can be compressed into a sphere such that the escape velocity from its surface is $8 \times 10^4 \ m/s$,then what should be the radius of the sphere (in $km$)? (Gravitational constant,$G = 6.67 \times 10^{-11} \ N \cdot m^2/kg^2$)

$A$ 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).