A mass of  $6\times {10}^{24} kg$ is to be compressed in a sphere in such a way that the escape

velocity from the sphere is $3\times {10}^{8 }m/s$. Radius of the sphere should be 

$\left(G=6.67\times {10}^{-11} N-m^2/k{g}^2\right)$

1. 9 km                     

2. 9 m

3. 9 cm                     

4. 9 mm

The mass and diameter of a planet have twice the value of the corresponding parameters

of earth. Acceleration due to gravity on the surface of the planet is

1. $9.8m/se{c}^2$                          

2. $4.9m/se{c}^2$ 

3. $980m/se{c}^2$                         

4. $19.6m/se{c}^2$           

Force of gravity is least at

1. The equator

2. The poles

3. A point in between the equator and any pole

4. None of these

The velocity with which a projectile must be fired so that it escapes earth’s gravitation

does not depend on:

1. Mass of the earth

2. Mass of the projectile

3. Radius of the projectile’s orbit

4. Gravitational constant

The gravitational force between two stones of mass 1 kg each separated by a distance of

1 metre in vacuum is

1. Zero                                                 

2. 6.675$\times {10}^{‐5}newton$

3. $6.675\times {10}^{‐11}newton$                 

4. $6.675\times {10}^{‐8}newton$

The escape velocity for the Earth is taken \(v_d\). Then, the escape velocity for a planet whose radius is four times and the density is nine times that of the earth, is:

Two particles of equal mass go round a circle of radius R under the action of their mutual

gravitational attraction. The speed of each particle is

1. $\nu =\frac{1}{2R}\sqrt{\frac{1}{Gm}}$                             

2. $\nu =\sqrt{\frac{Gm}{2R}}$

3. $\nu =\frac{1}{2}\sqrt{\frac{Gm}{R}}$                               

4. $\nu =\sqrt{\frac{4Gm}{R}}$   

The acceleration of a body due to the attraction of the earth (radius R) at a distance 2R

from the surface of the earth is (g = acceleration due to gravity at the surface of the

earth)

1. $\frac{g}{9}$                       

2. $\frac{g}{3}$

3. $\frac{g}{4}$                       

4.  g

 If V, R, and g denote respectively the escape velocity from the surface of the earth, the

radius of the earth, and acceleration due to gravity, then the correct equation is:

1. $V=\sqrt{gR}$                                 

2. V=$\sqrt{\frac{4}{3}g{R}^3}$

3. V = R$\sqrt{g}$                                   

4. V=$\sqrt{2gR}$

The depth at which the effective value of acceleration due to gravity is $\frac{g}{4}$, is:

1. R                              

2. $\frac{3R}{4}$

3. $\frac{R}{2}$                             

4. $\frac{R}{4}$