$A$ geostationary satellite is orbiting the earth at a height of $5R$ above the surface of the earth,where $R$ is the radius of the earth. The time period of another satellite in hours at a height of $2R$ from the surface of the earth is:

The earth (mass $M = 6 \times 10^{24} \ kg$) revolves around the sun with an angular velocity $\omega = 2 \times 10^{-7} \ rad/s$ in a circular orbit of radius $R = 1.5 \times 10^8 \ km$. The force exerted by the sun on the earth in newtons is:

$A$ satellite is placed in a circular orbit around the Earth at an altitude of $1000 \,km$. The time period of the satellite in minutes is approximately (mass of the Earth $= 6 \times 10^{24} \,kg$, radius of the Earth $= 6.4 \times 10^6 \,m$, $G = 6.67 \times 10^{-11} \,Nm^2 \,kg^{-2}$).

$A$ satellite revolving around the earth at a certain height experiences acceleration due to gravity equal to $\frac{16}{49} g_0$,where $g_0$ is the acceleration due to gravity on the earth's surface. If $R$ is the radius of earth,then the square of time period of the satellite's revolution is equal to $K\left[\frac{\pi^2 R^3}{G M}\right]$. The value of $K$ is

$A$ satellite of the earth is revolving in a circular orbit with a uniform speed $v$. If the gravitational force suddenly disappears,the satellite will

$A$ 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 $= 6.67 \times 10^{-11} \ Nm^2 \ kg^{-2}$)