Question

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

• A. $1.5\text{hours}$.
• B. $1.6\text{hours},$ if it is rotating from west to east.
• C. $24/17$ hours, if it is rotating from east to west.
• D. $24/17$ hours, if it is rotating from west to east.

Answer 1

Answer: (B), (C)

The correct options are
B $1.6\text{hours},$ if it is rotating from west to east.
C $24/17$ hours, if it is rotating from east to west.

Angular velocity of earth about its axis is $\frac{2\pi }{24}$
Angular velocity of satellite is $\frac{2\pi }{1.5}$

For a satellite rotating from west to east (same as earth), the relative angular velocity
${\omega }_{\text{rel}}=\frac{2\pi }{1.5}-\frac{2\pi }{24}=2\pi \left(\frac{15}{24}\right)$
$T^{\prime }=\frac{2\pi }{1.5}+\frac{24}{15}=1.6\text{hrs}$

For a satellite rotating from east to west (same as earth), the relative angular velocity
${\omega }_{\text{rel}}=\frac{2\pi }{1.5}+\frac{2\pi }{24}=2\pi \left(\frac{17}{24}\right)$
$\Rightarrow T^{\prime }=\frac{2\pi }{{\omega }_{\text{rel}}}=\frac{24}{17}\text{hrs}$