$A$ 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: $M = 6 \times 10^{24} \ kg$,$R = 6400 \ km$,$T = 24 \ h$,$G = 6.67 \times 10^{-11} \ N \cdot m^2/kg^2$.

(N/A) Given:
Mass of the Earth,$M = 6 \times 10^{24} \ kg$
Radius of the Earth,$R = 6400 \ km = 6.4 \times 10^6 \ m$
Time period,$T = 24 \ h = 24 \times 3600 \ s = 86400 \ s$
Gravitational constant,$G = 6.67 \times 10^{-11} \ N \cdot m^2/kg^2$
$(a)$ For a geostationary satellite,the orbital period $T$ is related to the orbital radius $r = R + h$ by Kepler's Third Law:
$T^2 = \frac{4 \pi^2 r^3}{GM}$
$r^3 = \frac{T^2 GM}{4 \pi^2}$
$r = \left( \frac{T^2 GM}{4 \pi^2} \right)^{1/3}$
Substituting the values:
$r = \left( \frac{(86400)^2 \times 6.67 \times 10^{-11} \times 6 \times 10^{24}}{4 \times (3.14)^2} \right)^{1/3} \approx 4.22 \times 10^7 \ m$
Height $h = r - R = 4.22 \times 10^7 \ m - 0.64 \times 10^7 \ m = 3.58 \times 10^7 \ m = 35800 \ km$.
$(b)$ To cover the entire equator,each satellite must cover an angular width of $120^\circ$ (or $2\pi/3$ radians) to ensure global coverage with three satellites. Thus,the minimum number of satellites required is $3$.