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Q.

A satellite is seen every $8$ hours over equator at a place on the Earth when its sense of rotation is opposite to the Earth. The time interval (in hour) after which it can be seen at the same place when the sense of rotation of Earth and satellite is same will ______

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a

28

b

24

c

25

d

26

answer is A.

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Detailed Solution

Initially, relative velocity of satellite with respect to earth

$ω' = ω_{δ} + ω_{e} \frac{2 π}{T^{'}} = \frac{2 π}{T_{δ}} + \frac{2 π}{T_{e}} o r T^{'} = 8 h, T_{e} = 24 h o n s o l v i n g T_{δ} = 12 h, I n s e c o n d c a s e : T = \frac{T_{e} T_{δ}}{T_{e} - T_{δ}} T = \frac{(24) (12)}{24 - 12} = \frac{(24) (12)}{12} \Rightarrow T = 24 h$

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