The time period of a geo-stationary satellite is 24 ~h , at a height 6 R_E-R_E is the radius of earth) from surface of earth. The time period of another satellite whose height is 2.5 R _ E from surface will be

Step-by-step solution

From Kepler's third law, the time period of revolution of satellite around earth is T ^2 r ^3 or T r ^ 3 / 2 where, r is the radius of satellite's orbit. Here, r ₁=6 R _ E + R _ E , T ₁=24 ~h r ₂=2.5 R _ E + R _ E , T ₂= ? where R _ E = radius of earth So, from Eq. (i), we get aligned T ₁ ~T ₂ & = ( r ₁ r ₂ )^ 3 / 2 24 ~T ₂ & = ( 6 R _ E + R _ E 2.5 R _ E + R _ E )^ 3 / 2 = ( 7 3.5 )^ 3 / 2 T ₂ & = 24 (2)^ 3 / 2 = 24 2 2 = 12 2 =6 2 ~h aligned