The distance between the centre of the Earth and the Moon is $384000 \, km$. If the mass of the Earth is $6 \times 10^{24} \, kg$ and $G = 6.67 \times 10^{-11} \, N m^2/kg^2$,the orbital speed of the Moon is nearly ......... $km/s$.

The time period of a geostationary satellite is

Earth revolves around the sun in a circular orbit of radius $R$. The angular momentum of the revolving earth is directly proportional to

The time period of a geostationary satellite is (in $\text{ h}$)

$A$ geostationary satellite above the equator is orbiting around the earth at a fixed distance $r_1$ from the center of the earth. $A$ second satellite is orbiting in the equatorial plane in the opposite direction to the earth's rotation,at a distance $r_2$ from the center of the earth,such that $r_1 = 1.21 r_2$. The time period of the second satellite as measured from the geostationary satellite is $\frac{24}{p}$ hours. The value of $p$ is $....$

$A$ satellite is to be placed in an equatorial geostationary orbit around the Earth for communication.
$(a)$ Calculate the height of such a satellite.
$(b)$ Find out the minimum number of satellites that are needed to cover the entire Earth,so that at least one satellite is visible from any point on the equator.
Given: $M = 6 \times 10^{24} \ kg$,$R = 6400 \ km$,$T = 24 \ h$,$G = 6.67 \times 10^{-11} \ N \cdot m^2/kg^2$.