$A$ geostationary satellite is orbiting the earth at a height of $6R$ above the surface of the earth,where $R$ is the radius of the earth. The time period of another satellite at a height of $2.5R$ from the surface of the earth is

$A$ 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 in a circular orbit of radius $R$ about the center of the Earth. $A$ meteorite of the same mass $M$,falling towards the Earth,collides with the satellite completely inelastically. The speeds of the satellite and the meteorite are the same,just before the collision. The subsequent motion of the combined body will be:

$A$ satellite of mass $1000 \ kg$ is launched to revolve around the earth in an orbit at a height of $270 \ km$ from the earth's surface. Kinetic energy of the satellite in this orbit is . . . . . . $\times 10^{10} \ J$. (Mass of earth $= 6 \times 10^{24} \ kg$,Radius of earth $= 6.4 \times 10^6 \ m$,Gravitational constant $= 6.67 \times 10^{-11} \ Nm^2 \ kg^{-2}$)

Two satellites $A$ and $B$ are revolving around the earth in orbits of heights $1.25 R_E$ and $19.25 R_E$ from the surface of the earth respectively,where $R_E$ is the radius of the earth. The ratio of the orbital speeds of the satellites $A$ and $B$ is (in $: 1$)

The radii of circular orbits of two satellites $A$ and $B$ of the earth are $4R$ and $R$ respectively,where $R$ is the radius of the earth. If the speed of satellite $B$ is $6V$,then the speed of satellite $A$ will be: