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# How to mount Satellite in orbit

What is satellite?  These are body which move around the moon or Some of them are natural while others are artificial.  To put satellites in orbit at certain height above, it must enter the orbit with correct speed where the centripetal force needed to keep it in orbit is equal to gravitational force Satellite

Let consider a satellite of mass, m, which circles the Earth of mass M, in an orbit A, close to its surface. If the Earth is spherical and its radius r, the centripetal force is equal to the $gravitational\quad force.\quad that\quad is\quad \frac { m{ v }^{ 2 } }{ r } \quad =\frac { GMm }{ { r }^{ 2 } } \quad =mg$

Where g = acceleration due to gravity, v = the velocity of mass m in orbit that is ${ v }^{ 2 }=gr\quad ,\quad therefore\quad v=\quad \sqrt { gr }$

Assume, that r = 6.4 x 106m and g = 9.8ms-2

Then  v = 8kms-1

# Satellite place in Parking orbit

If the satellite of mass m is circling the Earth in the plane of equator in an orbit B, concentric with the Earth surface and if it  moves with velocity, v in the same direction of rotation of Earth at distance, R, from the center of Earth, then Satellite $\frac { m{ v }^{ 2 } }{ R } \quad =\quad \frac { GMm }{ { R }^{ 2 } } ,\quad but\quad GM\quad =g{ r }^{ 2 }\quad where\quad r=radius\quad of\quad earth$ $therefore\quad \frac { m{ v }^{ 2 } }{ R } \quad =\frac { mg{ r }^{ 2 } }{ { R }^{ 2 } }$ $then\quad v=\sqrt { \frac { g{ r }^{ 2 } }{ R } }$

If the period of the satellite in orbit is T, then $v\quad =\frac { 2\Pi R }{ { T }_{ 1 } }$ $therefore\quad \frac { 4{ \Pi }^{ 2 }{ R }^{ 2 } }{ { T }_{ 1 }^{ 2 } } \quad ={ v }^{ 2 }\quad =\frac { g{ r }^{ 2 } }{ R }$ ${ T }_{ 1 }^{ 2 }=\frac { 4{ \pi }^{ 2 }{ R }^{ 3 } }{ g{ r }^{ r } }$ ${ T }_{ 1 }=\sqrt { \frac { 4{ \pi }^{ 2 }{ R }^{ 3 } }{ g{ r }^{ 2 } } } \quad or\quad 2\pi \sqrt { \frac { { R }^{ 3 } }{ g{ r }^{ 2 } } }$

If the period of satellite in orbit is equal to the period of Earth as it turns about its axis which is 24 hours, the satellite will stay at the same place above the Earth as the Earth rotates. This orbit is called parking orbit. Satellites place in this parking orbit can be used to transmit television programmes continuously from one part of the world to another

The radius R, can be found as follows $R\quad =\quad \sqrt [ 3 ]{ \frac { { T }^{ 2 }g{ r }^{ 2 } }{ 4{ \pi }^{ 2 } } } where\quad g\quad =9.8m{ s }^{ -2 }\quad and\quad r\quad =6.4\times { 10 }^{ 6 }m$ $therefore\quad R\quad =\sqrt [ 3 ]{ \frac { { (24\quad \times 3600) }^{ 2 }\times 9.8({ 6.4\times 10 }^{ 6 }{ ) }^{ 2 } }{ 4{ \pi }^{ 2 } } } =42340km$

The height of parking orbit above the surface of Earth = R – r

so (42340- 6400)km = 35940

Also the velocity of the satellite in the parking orbit is $\frac { 2\pi R }{ { T }_{ 1 } } \quad =\quad \frac { 2\pi \quad \times 42340km }{ 24\quad \times 3600s } \quad =3.1km{ s }^{ -1 }$ Satellite

In reality the orbit of satellite and that of planet is not spherical but an eclipse. The Earth shape is not spherical; it is flat at the pole. Therefore the gravitational force acting on different part of satellite’s orbit is not the same. Air which extend to great height affect the velocity of satellite. Therefore because of above observation there are variation in satellite orbit. The direction of satellite is changed by a small angle each time it moves around the Earth. So at Small at intervals, small correction of direction of satellite have to be made to keep it in correct orbit.  