If the radius of a planet is \(\mathrm{R}\) and its density is $\rho$ , the escape velocity from its surface will be:

1. $$ ${\mathrm{v}}_{\mathrm{e}}\propto \mathrm{\rho R}$

2. $$ ${\mathrm{v}}_{\mathrm{e}}\propto \sqrt{\mathrm{\rho }}\mathrm{R}$

3. $$ ${\mathrm{v}}_{\mathrm{e}}\propto \frac{\sqrt{\mathrm{\rho }}}{\mathrm{R}}$

4. $$ ${\mathrm{v}}_{\mathrm{e}}\propto \frac{1}{\sqrt{\mathrm{\rho }}\mathrm{R}}$

For a planet having mass equal to the mass of the earth but radius equal to one-fourth of the radius of the earth, its escape velocity will be:

A black hole is an object whose gravitational field is so strong that even light cannot escape from it. To what approximate radius would Earth (mass \(= 5.98\times 10^{24}~\text{kg}\)) have to be compressed to be a black hole?
1. \(10^{-9}~\text{m}\)
2. \(10^{-6}~\text{m}\)
3. \(10^{-2}~\text{m}\)
4. \(100​~\text{m}\)

A projectile is fired upwards from the surface of the earth with a velocity ${\mathrm{kv}}_{\mathrm{e}}$ where ${\mathrm{v}}_{\mathrm{e}}$ is the escape velocity and k < 1. If r is the maximum distance from the center of the earth to which it rises and R is the radius of the earth, then r equals:
1. \(\frac{R}{k^2}\)

2. \(\frac{R}{1-k^2}\)

3. \(\frac{2R}{1-k^2}\)

4. \(\frac{2R}{1+k^2}\)

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

1.  R/3                            

2.  R/2

3.  R/4                            

4.  R/5

A particle is located midway between two point masses each of mass \(\mathrm{M}\) kept at a separation \(2\mathrm{d}.\) The escape speed of the particle is: (neglect the effect of any other gravitational effect)

1.  $\sqrt{\frac{2\mathrm{GM}}{\mathrm{d}}}$

2.  $2\sqrt{\frac{\mathrm{GM}}{\mathrm{d}}}$

3.  $\sqrt{\frac{3\mathrm{GM}}{\mathrm{d}}}$

4.  $\sqrt{\frac{\mathrm{GM}}{2\mathrm{d}}}$