If the gravitational force between two objects were proportional to \(\frac{1}{R}\) (and not as $1/R^2$) where \(R\) is the separation between them, then a particle in circular orbit under such a force would have its orbital speed v proportional to:

1. $1/R^2$                        

2. $R^{\circ }$

3. $R^1$                             

4. \(1/R\)

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

Rohini satellite is at a height of 500 km and Insat-B is at a height of 3600 km from the surface of the earth. The relation between their orbital velocity (\(v_R,~v_i\)) is:

1. \(v_R>v_i\)

2. \(v_R<v_i\)

3. \(v_R=v_i\)

4. No relation 

Two satellites A and B go around the earth in circular orbits at heights of ${\mathrm{R}}_{\mathrm{A}} \mathrm{and} {\mathrm{R}}_{\mathrm{B}}$ respectively from the surface of the earth. Assuming earth to be a uniform sphere of radius $R_e$, the ratio of the magnitudes of their orbital velocities is:

1.  $\sqrt{\frac{R_B}{R_A}}$

2.  $\frac{R_B+R_e}{R_A+R_e}$

3.  $\sqrt{\frac{R_B+R_e}{R_A+R_e}}$

4.  ${\left(\frac{R_A}{R_B}\right)}^2$

The radii of the circular orbits of two satellites A and B of the earth are \(4R\) and \(R,\) respectively. If the speed of satellite A is \(3v,\) then the speed of satellite B will be: