Answer
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Hint: A satellite which appears to be in a fixed position at a definite height ( i.e. , stationary) to an observer on earth is called a geostationary satellite. This satellite is also called the geosynchronous satellite as its angular speed is synchronized with the angular speed of the earth about its axis, i.e. This satellite revolves around the earth with the same angular speed in the same direction as is done by earth around its axis.
Complete step by step solution:
All the three essential conditions for satellite motion should be followed.
It rotates in the equatorial plane.
Its height from earth surface is $36000km(\sim 6R)$
Its angular velocity and time period should be the same as that of earth.
Its rotating direction should be the same as that of earth (west to east).
Its orbit is called parking orbit and its orbital velocity is $3.1km/\sec $
Maximum latitude at which message can be received by geostationary satellite is
$\theta ={{\cos }^{-1}}(\dfrac{{{R}_{e}}}{{{R}_{e}}+h})$
The area of earth’s surface covered by geostationary satellite is $S=\Omega {{R}^{2}}=\dfrac{2\pi h{{R}^{2}}}{{{R}_{e}}+h}$
Distance of a geostationary satellite varies from different part of the earth. But the distance of geostationary satellite from the earth is kept to be $35,956km$ to maintain the time period of 24 hours.
Note: Geostationary satellites are used for communication. Its distance from the centre of the earth is nearly $7R$ where $R=6400km$. Things inside the satellite appear to be weightless because weight of the satellite is used for centripetal force required for its rotation.
Complete step by step solution:
All the three essential conditions for satellite motion should be followed.
It rotates in the equatorial plane.
Its height from earth surface is $36000km(\sim 6R)$
Its angular velocity and time period should be the same as that of earth.
Its rotating direction should be the same as that of earth (west to east).
Its orbit is called parking orbit and its orbital velocity is $3.1km/\sec $
Maximum latitude at which message can be received by geostationary satellite is
$\theta ={{\cos }^{-1}}(\dfrac{{{R}_{e}}}{{{R}_{e}}+h})$
The area of earth’s surface covered by geostationary satellite is $S=\Omega {{R}^{2}}=\dfrac{2\pi h{{R}^{2}}}{{{R}_{e}}+h}$
Distance of a geostationary satellite varies from different part of the earth. But the distance of geostationary satellite from the earth is kept to be $35,956km$ to maintain the time period of 24 hours.
Note: Geostationary satellites are used for communication. Its distance from the centre of the earth is nearly $7R$ where $R=6400km$. Things inside the satellite appear to be weightless because weight of the satellite is used for centripetal force required for its rotation.
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