Q: A satellite close to earth is in orbit above the equator with a period of rotation of 1.5 hrs. If it is above a point P on the equator at some time, it will be above P again after time
(A) 1.5 hrs.
(B) 1.6 hrs. if it is rotating from west to east.
(C) 24/17 hrs. if it is rotating from west to east.
(D) 24/17 hrs. if it is rotating from east to west.
Solution : Let ωo = the angular velocity of earth above its axis = 2π/24 rad/hr
Let ω = angular velocity of satellite.
For a satellite rotating from west to east. (same as earth), the relative angular velocity ω1 = ω – ωo
Time period of rotation relative to earth =2π/ω1 = 1.6 h
Now for a satellite rotating from east to west (opposite to earth) the relative angular velocity ω2 = ω + ωo
Time period of rotation relative to earth =2π/ω2 = 24/17 hrs.