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Two satellite `S_1 and S_2` revolve round a planet in coplanar circular orbits in the same sense. Their periods of revolutions are 1 h nd 8 h respectively. The radius of the orbit of `S_1` is `10^4 km`. When `S_2` is closet to `S_1`., find (a). The speed of `S_2` relative to `S_1` and (b). the angular speed of `S_2` as observed by an astronaut in `S_1`.

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The correct Answer is:
(i)4 (ii
(i) According to Kepler’s third law
`Tpropr^(3//2)`
`T_1^2/T_2^2=R_1^3/R_2^3 rArr R_2^3=R_1^3 xx T_2^2/T_1^2 therefore R_2^3=10^12xx8^2/1^2=64xx10^12 rArr R_2=4xx10^4` km
The speed of satellite `S_1 V_1=(2piR_1)/T_1 =(2pixx10^4)/1` km/hr
The speed of satellite `S_2V_2=(2piR_2)/T_1=(2pixx4xx10^4)/8 = pixx10^4`
The speed of satellites `S_2` relative to `S_1`.

The speed of satellite `S_2V_2=(2piR_2)/T_1 = (2pixx4xx10^4)/8=pixx10^4`
The speed of satellites `S_2` relative to `S_1`.
`|V_(2.1)|=|V_2-V_1|=pixx10^4-2pixx10^4 = pixx10^4` km/hr
(ii) The angular speed of satellite `S_2` relative to `S_1`
`S_1=V_r/R_r =(|V_2-V_1|)/(R_2-R_1) =(pixx10^4)/((4xx10^(4)-10^4))=pi/3` rad/hr
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