The time period is given by, T=\\dfrac{2\\pi}{\\omega} Angular Velocity is given by, \\omega=\\dfrac{2\\pi}{T} and then find relative angular velocityGiven 8=\\dfrac {2\\pi}{\\omega_1+\\omega_2}=\\frac {2\\pi}{\\dfrac {2\\pi}{T_1}+\\dfrac {2\\pi}{T_2}} \\Rightarrow T_1=24 h for the earthSubstituting this value in the above equation we get T_2=12 h (T_2 being the time period of the satellite, it will remain the same as the distance from the centre of the earth remains constant)Now when the sense of rotation of the earth and the satellite is same we get \\Rightarrow T=\\dfrac {2\\pi}{\\omega_2-\\omega_1}=\\dfrac {2\\pi}{\\dfrac {2\\pi}{T_1}-\\dfrac {2\\pi}{T_2}}=24 h