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Question

Two satellites S1 and S2 are revolving around a planet in coplanar and concentric circular orbits of radii R1 and R2 in the same direction respectively. Their respective periods of revolution are 1hour and 8hour. The radius of the orbit of satellite S1 equal to 104 km. Their relative speed when they are closest, in kmph is:
  1. π2×104
  2. π×104
  3. 2π×104
  4. 4π×104

A
π×104
B
4π×104
C
π2×104
D
2π×104
Solution
Verified by Toppr

Time period of revolution is T2=4π2GMr3
T21T22=r31r32
Given T1=1 hr, T2=2 hr and r1=104 km.
r2=r1×T2/32T2/31=104×82/312/3=4×104 km
Now,
v1=r1ω1=r12πT1=104×2π1=2π×104 km/hr ( since we want the answer in kmph, we didn't convert km to m and hr to s while substituting )
v2=r2ω2=r22πT2=4×104×2π8=π×104 km/hr
Since we want the relative speed when the satellites are closer, direction of linear speed is in the same direction.
Hence, relative speed=(v1v2)=(2ππ)×104=π×104 km/hr

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