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:

1.	\(36v_d\)	2.	\(12v_d\)
3.	\(6v_d\)	4.	\(20v_d\)

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 \(9.8~\text{m/s}^2\). 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)

The earth (mass = $6\times {10}^{24}kg$ ) revolves round the sun with angular velocity

$2\times {10}^{‐7}rad/s$ in a circular orbit of radius $1.5\times {10}^{8}km$ . The force exerted by the sun on

the earth in Newtons, is

1. $18\times {10}^{25}$                               

2. Zero

3. $27\times {10}^{39}$                               

4. $36\times {10}^{21}$

The gravitational force between two point masses $m_1$  and $m_2$  at separation r is given by $F=k\frac{m_1{m}_2}{r^2}$  The constant k                                                                

1. Depends on system of units only

2. Depends on medium between masses only

3. Depends on both 1 and 2

4. Is independent of both 1 and 2

The density of a newly discovered planet is twice that of earth. The acceleration due to

gravity at the surface of the planet is equal to that at the surface of the earth. If the

radius of the earth is R, the radius of the planet would be

1. 2R                   

2. 4R

3. $\frac{1}{4}R$               

4. $\frac{1}{2}R$