Time Period of Satellite.

It is the time taken by satellite to go once around the earth.

therefore                T = (Circumference of the orbit/orbital velocity)

          T = 2 πr/ν = 2πr (r/GM)^1/2                       [As ν = (GM/r)^1/2]

      T = 2π(r³/GM)^1/2 2π (r3/gR²)^1/2                 [As GM = gR²]

[As r = R + h]]

Note:

a)             T = 2 π (r³/GM)

    T² = (4 π²/GM) i.e., T² ∝ r³

This is in accordance with Kepler’s third law of planetary motion  becomes a (semi major axis) if the orbit is elliptic.

(b) Time period of nearby satellite,

From 

                                       [As  and GM = gR²]

For earth R = 6400 km and g = 9.8 m/s²

                T = 84.6 minute ≈ 1.4 hr

(iv) Time period of nearby satellite in terms of density of planet can be given as

(v) If the gravitational force of attraction of the sun on the planet varies as F ∝ (1/rⁿ) then the time period varies as

(vi) If there is a satellite in the equatorial plane rotating in the direction of earth’s rotation from west to east, then for an observer, on the earth, angular velocity of satellite will be . The time interval between the two consecutive appearances overhead will be

 

If ,  i.e. satellite will appear stationary relative to earth. Such satellites are called geostationary satellites.

Height of Satellite.

As we know, time period of satellite

By squaring and rearranging both sides          (gR²T²/4π²) = (R+H)³

                h = (T²gR²/4π²)^1/3 - R

By knowing the value of time period we can calculate the height of satellite the surface of the earth.