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Question

Two satellites S1 and S2 are revolving around a planet in the opposite sense in coplanar circular concentric orbits. At time t=0, the satellites are farthest apart. The periods of revolution of S1 ad S2 and 3h and 24h respectively. The radius of the orbit of S1 is 3×104km. Then the orbital speed of S2 as observed from.

A
The planet is 4π×104km h1 when S2 is closest from S1
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B
The planet is 2π×104km h1 when S2 is farthest from S1
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C
S1 is π×104km h1 when S2 is closest from S1
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D
S1 is 3π×104km h1 when S2 is closest to S1
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Solution

The correct option is D S1 is 3π×104km h1 when S2 is closest to S1
From Kepler's Laws,
R31R32=T21T22
R1R2=14
R2=12×104 km

Relative to the planet,
v1=2πR1T1=2π×3×1043=2π×104 km/h
v2=2πR2T2=2π×12×10424=π×104 km/h

At closest approach distance between S1 & S2, relative orbital speed of S2 as seen from S1 will be v1+v2 as they are moving in the opposite directions.
Hence, the above orbital speed is :
v=v1+v2=3π×104 km/hr

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