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. $\frac{R}{4}$

The value of ‘g’ at a particular point is $9.8$ $\mathrm{m}/{\mathrm{s}}^2$. Suppose the earth suddenly shrinks uniformly to half its present size without losing any mass then value of ‘g’ at the same point will now become: (assuming that the distance of the point from the centre of the earth does not shrink)

If both the mass and the radius of the earth is decreased by 1%, then the value of the acceleration due to gravity will:

1. decrease by 1%.                 

2. increase by 1%.

3. increase by 2%.                    

4. remain unchanged.