If the height of a satellite from the earth is negligible in comparison to the radius of the earth $R$,the orbital velocity of the satellite is

Two satellites $S_{1}$ and $S_{2}$ are revolving around a planet in the opposite sense in coplanar circular concentric orbits. At time $t=0$,the satellites are farthest apart. The periods of revolution of $S_{1}$ and $S_{2}$ are $3 \,h$ and $24 \,h$,respectively. The radius of the orbit of $S_{1}$ is $3 \times 10^{4} \,km$. Then,the orbital speed of $S_{2}$ as observed from

$A$ satellite of mass $\frac{M}{2}$ is revolving around the Earth in a circular orbit at a height of $\frac{R}{3}$ from the Earth's surface. The angular momentum of the satellite is $M \sqrt{\frac{GMR}{x}}$. The value of $x$ is . . . . . . ,where $M$ and $R$ are the mass and radius of the Earth,respectively. ($G$ is the gravitational constant)

$A$ satellite of mass $m$ is revolving close to the surface of a planet of density $d$ with time period $T$. The value of the universal gravitational constant $G$ in terms of $d$ and $T$ is given by:

$A$ planet is revolving around the sun in an elliptical orbit. Its closest distance from the sun is $r_{min}$,and the farthest distance from the sun is $r_{max}$. If the orbital angular velocity of the planet when it is nearest to the sun is $\omega$,then the orbital angular velocity at the point when it is at the farthest distance from the sun is:

$A$ satellite is moving in a low nearly circular orbit around the earth. Its radius is roughly equal to that of the earth's radius $R_e$. By firing rockets attached to it,its speed is instantaneously increased in the direction of its motion so that it becomes $\sqrt{\frac{3}{2}}$ times larger. Due to this,the farthest distance from the centre of the earth that the satellite reaches is $R$. The value of $R$ is $....R_e$.