From Kepler's third law, the time period of revolution of satellite around earth is T2∝r3 or T∝r3/2 where, r is the radius of satellite's orbit. Here, r1=6RE+RE,T1=24 h r2=2.5RE+RE,T2= ? where RE= radius of earth So, from Eq. (i), we get T2T1 T224⇒T2=(r2r1)3/2=(2.5RE+RE6RE+RE)3/2=(3.57)3/2=(2)3/224=2224=212=62 h