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 the radius of a planet is $R$ and density is $\rho$,then the escape velocity $v_{e}$ of any body from its surface will be proportional to:

$A$ 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)

The masses and radii of the Earth and Moon are $M_1, R_1$ and $M_2, R_2$ respectively. Their centers are at a distance $d$ apart. The minimum velocity with which a particle of mass $m$ should be projected from a point midway between their centers so that it escapes to infinity is:

$A$ rocket is fired vertically with a speed of $4 \ km/s$ from the Earth's surface. How far from the Earth does the rocket go before returning to the Earth (in $km$)? (Take radius of Earth $R = 6.4 \times 10^6 \ m$ and $g = 10 \ m/s^2$)

What is the escape velocity for the surface of a black hole?