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

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

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 

The radii of the circular orbits of two satellites \(A\) and \(B\) of the earth are \(4R\) and \(R,\) respectively. If the speed of the satellite \(A\) is \(3v,\) then the speed of the 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 remote sensing satellite of the earth revolves in a circular orbit at a height of \(0.25 \times10^6~\text{m}\) above the surface of the earth. If Earth’s radius is \(6.38\times10^6~\text{m}\) and \(g=9.8~\text{ms}^{-2},\) then the orbital speed of the satellite is:
1. \(7.76~\text{kms}^{-1}\)
2. \(8.56~\text{kms}^{-1}\)
3. \(9.13~\text{kms}^{-1}\)
4. \(6.67~\text{kms}^{-1}\)

An artificial satellite revolves around a planet for which gravitational force \((F)\) varies with the distance \(r\) from its centre as \(F \propto r^{2}.\) If \(v_0\) is its orbital speed, then:

A comet orbits the sun in a highly elliptical orbit. The comet has a constant:
1. linear speed
2. angular speed
3. angular momentum
4. kinetic energy