How many times is escape velocity, of orbital velocity for a satellite revolving near earth?

(a)$\sqrt{2}$ times                    (b) 2 times

(c) 3 times                       (d) 4 times

The acceleration due to gravity near the surface of a planet of radius R and density d is

proportional to

1. $\frac{d}{R^2}$                            

2. $d{R}^2$

3. dR                             

4. $\frac{d}{r}$

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

be

1. $V_e \propto pR$                         

2. $V_e \propto R\sqrt{\rho }$

3. $V_e\propto \frac{\sqrt{p}}{R}$                        

4. $V_e \propto \frac{1}{\sqrt{p}R}$

If the distance between two masses is doubled, the gravitational attraction between them:

1. Is doubled                               

2. Becomes four times

3. Is reduced to half                     

4. Is reduced to a quarter

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. $\frac{R}{4}$

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.

If R is the radius of the earth and g the acceleration due to gravity on the earth's surface,

the mean density of the earth is:

1. $4\pi G/3gR$                                 

2. $3\pi R/4gG$

3. $3g/4\pi RG$                               

4.$\pi RG/12G$        

A planet has twice the radius but the mean density is $\frac{1}{4}th$ as compared to earth. What is the ratio of escape velocity from earth to that from the planet ?

1. 3:1                                                                 

2. 1:2

3. 1:1                                                                 

4. 2:1

3 particles each of mass m are kept at vertices of an equilateral triangle of side L. The

gravitational field at centre due to these particles is

1. zero                                                                       

2. $\frac{3GM}{L^2}$

3. $\frac{9GM}{L^2}$                                                                     

4. $\frac{12}{\sqrt{3}}\frac{GM}{L^2}$

Four particles each of mass M, are located at the vertices of a square with side L. The

gravitational potential due to this at the centre of the square is

1. $-\sqrt{32}\frac{GM}{L}$                                   

2. $-\sqrt{64}\frac{GM}{L^2}$

3. zero                                             

4. $\sqrt{32}\frac{GM}{L}$