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If a satellite in the equatorial plane rotating in the direction of earth’s rotation from west to east. An observer which is on the earth the time interval between the two consecutive appearances overhead will be:

Option: 1

T = \frac{T_{E}T_{S}}{T_{E}-T_{S}}


Option: 2

T = \frac{T_{E}T_{S}}{T_{S}-T_{E}}


Option: 3

T = \frac{2T_{E}T_{S}}{T_{E}-T_{S}}


Option: 4

T = \frac{4T_{E}T_{S}}{T_{E}-T_{S}}


Answers (1)

best_answer

As we learn

Time period of satellite for appearance overhead -

T=\frac{2\pi }{W_{S}-W_{E}}=\frac{T_{S}T_{E}}{T_{E}-T_{S}}

if W_{S}=W_{E}, T=D

T\rightarrow Time \: period

- wherein

if satellite in the equilateral planes moves from west to east Angular velocity of satellite will 

\left ( W_{S}-W_{E} \right )

W_{S}\rightarrow South velocity

W_{E}\rightarrow East velocity

 

 

Here TE = time period of earth

            Ts = time period of satellite

For an observer on the earth, angular velocity of satellite will be (ws – wE). So the interval between two consecutive appearances overhead will be

T = \frac{2\pi }{w_{s}- w_{E}} = \frac{2\pi }{\frac{2\pi }{T_{S} }-\frac{2\pi }{T_{E}}}

T = \frac{T_{S}T_{E}}{T_{E}-T_{S}}

 

Posted by

Anam Khan

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