The time period of a geo-stationary satellite is 24 h, at a height 6RE (RE is the radius of earth) from surface of earth. The time period of another satellite whose height is 2.5 RE from surface will be            [NEET (Odisha) 2019]

From Kepler's third law, the time period of revolution of satellite around earth isT2∝r3 or T∝r3/2               …(i)where, r is the radius of satellite's orbit.Here,    r1=6RE+RE,T1=24 h             r2=2.5RE+RE,T2=?So, from Eq. (i), we get      T1T2=r1r23/2      24T2=6RE+RE2.5RE+RE3/2=73.53/2⇒  T2=24(2)3/2=2422=122=62 h