If the Earth is considered a point mass of $6 \times 10^{24} \ kg$ revolving around the Sun at a distance of $1.5 \times 10^8 \ km$ in a period of $T = 3.14 \times 10^7 \ s$,then the angular momentum of the Earth around the Sun will be:

Two satellites $A$ and $B$ of masses $200 \, kg$ and $400 \, kg$ are revolving around the Earth at heights of $600 \, km$ and $1600 \, km$ respectively. If $T_{A}$ and $T_{B}$ are the time periods of $A$ and $B$ respectively,then find the value of $T_{B} - T_{A}$.
[Given: Radius of Earth $R = 6400 \, km$,Mass of Earth $M = 6 \times 10^{24} \, kg$,$G = 6.67 \times 10^{-11} \, Nm^{2}/kg^{2}$]

$A$ 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 $....R_e$.

$A$ planet is moving in a circular orbit. It completes $2$ revolutions in $360$ days. What is its angular frequency?

Two particles of equal mass $m$ go around a circle of radius $R$ under the action of their mutual gravitational attraction. The speed of each particle with respect to their centre of mass is

$A$ remote-sensing satellite of Earth revolves in a circular orbit at a height of $0.25 \times 10^6 \, m$ above the surface of Earth. If Earth's radius is $6.38 \times 10^6 \, m$ and $g = 9.8 \, m s^{-2}$,then the orbital speed of the satellite is ...... $km s^{-1}$.