Question

A satellite close to earth is in orbit above the equator with a period of rotation of $1.5hrs$. If it is above a point $P$ on the equator at some time, it will be above $P$ again after time.

• A. $1.5hrs.$
• B. $1.6hrs.$ if it 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

Answer 1

Answer: (B), (D)

The correct options are
B $1.6hrs.$ if it rotating from west to east
D $24/17$ hrs. if it is rotating from east to west

Let ${\omega }_0=$ the angular velocity of earth above its axis $=\frac{2\pi }{24}rad/hr$
Let $\omega =$ angular velocity, of satellite $=\frac{2\pi }{1.5}$
For a satellite rotating from west to east. (same as earth), the relative angular velocity
${\omega }_1=\omega -{\omega }_0$
$\therefore$ Time period of rotation relative to earth $=\frac{2\pi }{{\omega }_1}=1.6h$
Now for a satellite rotating from east to west (opposite to earth) the relative angular velocity ${\omega }_2=\omega +{\omega }_0$
Time period of rotation relative to earth $=\frac{2\pi }{{\omega }_2}=\frac{24}{17}hrs.$
Hence $\left(B\right)$ and $\left(D\right)$ are correct.