Two satellites A and B revolve round the same planet in coplanar circular orbits lying in the same plane. Their periods of revolutions are 1 h and 8 h, respectively. The radius of the orbit of A is 104 km. The speed of B relative to A, when they are close in km/h is [AIIMS 2018]

Given,   TA=1 h,TB=8 hand                  rA=104 kmFrom Kepler's law,   TA2TB2=rA3rB3⇒1282=1043rB3⇒  rB3=64×1043∴  rB=4×104 kmSpeed of satellIte  A,vA=2πrATA=2π×1041                                      =2π×104 km/hSpeed of satellIte   B,vB=2πrBTB=2π×4×1048                                       =π×104 km/hThe speed of B relative to A when they are close,vBA=vA-vR=2π×104-π×104      =π×104 km/h