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Home  >>  JEEMAIN and NEET  >>  Physics  >>  Class11  >>  Gravitation

A satellite revolving in a circular equatorial orbit of radius $r= 2 \times 10 ^4 km$ from west to East appears over a certain point at the equator every $11.6$ hours. Calculate the actual angular velocity of the satellite

\[(a)\;5.62 \times 10^{-4} rad/s \quad (b)\;2.23 \times 10^{-4} rad/s \quad (c)\;1.23 \times 10^{-4} rad/s \quad (d)\;6.21 \times 10^{-4} rad/s \]

1 Answer

angular velocity of earth about its own axis
$w_e =\large\frac{2 \pi}{24 \times 3600}$$rad/s$
$w_s=\large\frac{2\pi}{3600}\bigg[\large\frac{1}{24}+\large\frac{1}{11.6}\bigg]$$rad/s$
$\quad= 2.23 \times 10^{-4}\; rad/s$
Hence b is the correct answer.

 

answered Aug 22, 2013 by meena.p
edited Feb 17, 2014 by meena.p
 

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