The altitude of a geosynchronous orbit is [tex]3.59\cdot 10^7 m[/tex]
Explanation:
A geostationary (or geosynchronous) orbit is the orbit of a satellite that stays over the same spot on the equator of the rotating Earth.
This means that the period of a geostationary satellite is equal to the period of rotation of the Earth, which is 24 hours:
[tex]T=24 h \cdot 3600 s/h = 86400 s[/tex]
We can find the altitude of the orbit in the following way. First, we notice that the orbital speed of the satellite is given by
[tex]v=\frac{2\pi r}{T}[/tex]
where r is the radius of the orbit.
Then we also notice that the gravitational force between the satellite and the Earth is equal to the centripetal force, so we can write:
[tex]\frac{GMm}{r^2}=m\frac{v^2}{r}[/tex]
where
G is the gravitational constant
M is the mass of the Earth
m is the mass of the satellite
Re-arranging the equation,
[tex]\frac{GM}{r}=v^2[/tex]
And substituting the expression for the velocity,
[tex]\frac{GM}{r}=(\frac{2\pi r}{T})^2=\frac{4\pi^2 r^2}{T^2}[/tex]
Solving for r,
[tex]r=\sqrt[3]{\frac{GMT^2}{4\pi^2}}[/tex]
And substituting:
[tex]G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2}\\M=5.98\cdot 10^{24} kg\\T=86400 s[/tex]
we find:
[tex]r=\sqrt[3]{\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24})(86400)^2}{4\pi^2}}=4.225\cdot 10^7 m[/tex]
And since the radius of the Earth is
[tex]R=6.37\cdot 10^6 m[/tex]
The altitude of the satellite is
[tex]h=r-R=4.225\cdot 10^7 - 6.37\cdot 10^6 = 3.59\cdot 10^7 m[/tex]
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