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

Two satellites are moving in a common plane along circular paths. with one having angular velocity $w_1=1.09 \times 10^{-3} rad/s$ and another $w_2=1.08 \times 10^{-3}rad /s$. Find time interval which seperates the periodic approaches of the satellites to each other over minimum distance, if they are revolving in opposite sense.

a) 0.6 hours b) 0.8 hours c) 0.2 hours d) 1 hour

1 Answer

Let they be at closest distance after a time $t_1$
$w_1t_1+w_2t_2=\large\frac{2}{T}$
$t_1= \large\frac{2 \pi}{w_1+w_2}$
$\qquad= \large\frac{2 \pi}{(1.09+1.08) \times 10^{-3}}s$
$\qquad= \large\frac{2 \pi}{2.17 \times 10^{-3} \times 3600}$$hours$
$\qquad=0.8$ hours
Hence b is the correct answer

 

answered Aug 26, 2013 by meena.p
edited Feb 18, 2014 by meena.p
 

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