If the gravitational force between two objects were proportional to \(\frac{1}{R}\) (and not as\(\frac{1}{R^2}\)${\mathrm{}}^{}$) 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. \(\frac{1}{R^2}\)
2. \(R^{0}\)
3. \(R^{1}\)
4. \(\frac{1}{R}\)

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 \(R_A ~\text{and}~R_B\) 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: