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 specific 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. \(\left(\frac{R_{A}}{R_{B}}\right)^{2}\)

Two particles of equal masses go around a circle of radius \(R\) under the action of their mutual gravitational attraction. The speed of each particle is:
1. \(v = \frac{1}{2 R} \sqrt{\frac{1}{Gm}}\)
2. \(v = \sqrt{\frac{Gm}{2 R}}\)
3. \(v = \frac{1}{2} \sqrt{\frac{G m}{R}}\)
4. \(v = \sqrt{\frac{4 Gm}{R}}\)

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:

A satellite \(S\) is moving in an elliptical orbit around the earth. If the mass of the satellite is very small as compared to the mass of the earth, then:

A satellite is revolving around the earth with speed \(v_0\). If it is stopped suddenly, then with what velocity will the satellite hit the ground? (\(v_e\)= escape velocity from the earth's surface)
1. \(\sqrt{v_{e}^{2} - v_{0}^{2}}\)
2. \(\sqrt{v_{e}^{2}-2 v_{0}^{2}}\)
3. \(\sqrt{v_{e}^{2}-3 v_{0}^{2}}\)
4. \(\sqrt{v_{e}^{2}-\frac{v_{0}^{2}}{2}}\)

A remote sensing satellite of the earth revolves in a circular orbit at a height of \(0.25\times 10^{6}\) m above the surface of the earth. If the earth’s radius is \(6.38\times 10^{6}\) m and \(g = 9.8\) ms^-1, then the orbital speed of the satellite is:
1. \(7.76\) kms^-1
2. \(8.56\) kms^-1
3. \(9.13\) kms^-1
4. \(6.67\) kms^-1