Question

A satellite that goes around the earth once every 24 hours iscalled a geosynchronous satellite. If a geosynchronoussatellite is in an equatorial orbit, its position appearsstationary with respect to a ground station, and it is known as ageostationary satellite
Find the radius Rof the orbit of a geosynchronous satellite that circles the earth.(Note that Ris measured from the center of the earth, not the surface.) You mayuse the following constants:
The universal gravitational constant G is 6.67 \times 10^{-11}\;{\rm N \; m^2 / kg^2}.
The mass of the earth is 5.98 \times 10^{24}\;{\rm kg}.
The radius of the earth is 6.38 \times 10^{6}\;{\rm m}.

Answer 1

Answer:

35870474.30504 m

Explanation:

r = Distance from the surface

T = Time period = 24 h

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

m = Mass of the Earth =  5.98 × 10²⁴ kg

Radius of Earth = $6.38\times 10^6\ m$

The gravitational force will balance the centripetal force

$\dfrac{GMm}{R^2}=m\dfrac{v^2}{R}\\\Rightarrow v=\sqrt{\dfrac{GM}{R}}$

$T=\dfrac{2\pi r}{v}\\\Rightarrow T=\dfrac{2\pi r}{\sqrt{\dfrac{GM}{r}}}$

From Kepler's law we have relation

$T^2=\dfrac{4\pi^2r^3}{GM}\\\Rightarrow r^3=\dfrac{T^2GM}{4\pi^2}\\\Rightarrow r=\left(\dfrac{(24\times 3600)^2\times 6.67\times 10^{-11}\times 5.98\times 10^{24}}{4\pi^2}\right)^{\dfrac{1}{3}}\\\Rightarrow r=42250474.30504\ m$

Distance from the center of the Earth would be

$42250474.30504-6.38\times 10^6=\mathbf{35870474.30504\ m}$