Question

A satellite close to surface of earth is orbiting in equatorial plane with time period of 1.5 hrs. If it is above a point P on the equator at some time it will be above P again after x hrs. Here x if satellite rotates from west to east is (Answer upto two digits after the decimal point.)

Answer 1

${\omega }_1=\frac{2\pi }{1.5},{\omega }_2=\frac{2\pi }{24}$
Since the earth and the satellite are rotating in the same direction, the relative angular velocitiy is ${\omega }_1-{\omega }_2$.
${\omega }_1-{\omega }_2=\frac{2\pi }{t}\Rightarrow \frac{2\pi }{1.5}-\frac{2\pi }{24}=\frac{2\pi }{t}\Rightarrow t=\frac{24}{15}=1.60$