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Orbital Velocity of a Satellite | Geostationary Satellite | Physics Grade XI


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Orbital Velocity of a Satellite | Geostationary Satellite | Physics Grade XI

Orbital Velocity of a Satellite | Geostationary Satellite

Orbital Velocity of a Satellite:
The velocity which is required to keep the satellite revolves around its orbit is called orbital velocity of a satellite. It depends upon the radius of the orbit in which it revolves.
Suppose a satellite of mass 'm' revolving around the earth at a height 'h' from the surface. Let v0 be the orbital velocity of the satellite. Then the centripetal force acting on the planet is,
     Fc = mv02/r + h
And the gravitational force between the earth and the satellite is
     Fg = GMm/ (R+h)2
Where M is the mass of the earth. As the required centripetal force is provided by the gravitational force,
     Fa = Fc
     GMm/ (R+h)2 = mv02/ (R+h)
     v0 = √GM/R+h  ……………. (i)
if g is the acceleration due to gravity on the earth’s surface, then
     g = GM/R2
     GM = gR2
     And v0 = √g R2/ R+h
     V0 = R √g/R+h ………(ii)
This is the expression for orbital velocity of satellite.
Period of satellite:
The period of a satellite is the time required to complete one revolution round the earth around its orbit. It is denoted by T.

T = circumstance of circular orbit/ orbital velocity

Or, T = 2πr/ v0 = 2πr √r/ GM

= 2π(R+h)√(R+h)/GM

= 2π√(R+h)3/GM  ..……. (iii)

Height of satellite
The distance between the earth surface and the satellite.
From equation (iii),
     T = 2π√(R+h)3/GM
Squaring both sides, we get
     (R+h)9 = T2 GM/ 4π2
Taking cube root on both sides,
     (R+h) = (T2 GM/4π2)1/3
     H = (T2 GM/4π2)1/3 – R
Geostationary satellite
The satellite which seems to be stationary from the earth surface is called geostationary satellite. The orbit in which it revolves around the earth is called parking orbit.
Height of geostationary satellite
As we know the period of satellite is,
     T = 2π√(R+h)3/GM
     T = 2π/R √(R+h)3/g   [g = GM/R2]
On solving the expression,
     H [ T2R2/ 4π2 * g]1/3 – R  …………… (iv)
Substituting the values, R = 6.4 * 106, T = 24 * 3600sec, g = 9.8 ms-2 in equation (iv),
     We get, h = 36,000km
Speed of Satellite: The speed of satellite of the geostationary satellite is
     V0 = 2πr/T
     T = 24* 3600s, and r = 36000 + 6400 = 42400km
     V0 = 2π * 424000/ 24 * 3600 = 3.1 km/s

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