For the moon to cease as the earth's satellite, its orbital velocity has to be increased by a factor of -

1.	2	2.	\(\sqrt{2}\)
3.	\(1/\sqrt{2}\)	4.	4

The height of a point vertically above the earth’s surface, at which the acceleration due to gravity becomes 1% of its value at the surface is: (Radius of the earth = R)

1. 8R           

2. 9R

3. 10R         

4. 20R

The escape velocity of an object from the earth depends upon the mass of the earth (M), its mean density, its radius (R) and the gravitational constant (G). Thus the formula for escape velocity is:

(1) $v=R\sqrt{\frac{8\mathrm{\pi }}{3}Gp}$                                         

(2) $v=M\sqrt{\frac{8\mathrm{\pi }}{3}GR}$

(3) $v=\sqrt{2GMR}$                                             

(4) $v=\sqrt{\frac{2GM}{R^2}}$

Escape velocity on a planet is $v_e$. If radius of the planet remains same and mass becomes 4 times, the escape velocity becomes

(1)$4v_e$                                               

(2)$2v_e$

(3)$v_e$                                                 

(4)$\frac{1}{2}{v}_e$

The mass of the earth is 81 times that of the moon and the radius of the earth is 3.5 times that of the moon. The ratio of the escape velocity on the surface of earth to that on the surface of moon will be

(1)0.2                                                             

(2)2.57

(3)4.81                                                             

(4)0.39

If radius of earth is R then the height h’ at which value of ‘g’ becomes one-fourth is 

(1)$\frac{R}{4}$               

(2)$\frac{3R}{4}$

(3)R                 

(4)$\frac{R}{8}$

The escape velocity from the surface of the earth is ${\mathrm{v}}_{\mathrm{e}}$. The escape velocity from the surface of a planet whose mass and radius are 3 times those of the earth will be:

(1) ${\mathrm{v}}_{\mathrm{e}}$                                                             

(2) 3${\mathrm{v}}_{\mathrm{e}}$

(3) 9${\mathrm{v}}_{\mathrm{e}}$                                                           

(4) 27${\mathrm{v}}_{\mathrm{e}}$

(a) $About 9.8\times {10}^{6}J$                        (b) $About 6.4\times {10}^{8}J$

(c) $About 3.1\times {10}^{10}J$                      (d) $About 27.4\times {10}^{12}J$

Two planets have the same average density but their radii are $R_1$  and $R_2$. If acceleration due to gravity on these planets be $g_1$ and $g_2$ respectively, then

(1)    $\frac{g_1}{g_2}$= $\frac{R_1}{R_2}$                   

(2) $\frac{g_1}{g_2}$=$\frac{R_2}{R_1}$

(3)    $\frac{g_1}{g_2}$= $\frac{{R_1}^2}{{R_2}^2}$                  

(4) $\frac{g_1}{g_2}$= $\frac{{R_1}^3}{{R_2}^3}$

Assume that the acceleration due to gravity on the surface of the moon is 0.2 times the acceleration due to gravity on the surface of the earth. If $R_E$ is the maximum range of a projectile on the earth’s surface, what is the maximum range on the surface of the moon for the same velocity of projection ?

(a) 0.2$R_{\epsilon }$                     

(b) 2$R_{\epsilon }$

(c) 0.5$R_{\epsilon }$                     

(d) 5$R_{\epsilon }$