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

Respuesta :

Olajidey

Answer:

35870474.30504 m

Explanation:

Given that,

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 × 10⁶ m

The gravitational force will balance the centripetal force

[tex]\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}}}[/tex]

From Kepler's law we have relation

[tex]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[/tex]

Distance from the center of the Earth would be

[tex]42250474.30504-6.38\times 10^6\\[/tex]

= 35870474.30504 m

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