Which of the following is the evidence to show that there must be a force acting on earth

and directed towards the sun?

1. Deviation of the falling bodies towards east

2. Revolution of the earth around the sun

3. Phenomenon of day and night

4. The apparent motion of the sun around the earth

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=$\sqrt{2gR}$