How much time a satellite in an orbit of height 35780 km above earth's surface would take, if the mass of the earth would have been four times its original mass?

How much time a satellite in an orbit of height 35780 km above earth's

1.	How much time a satellite in an orbit of height 35780 km above earth's surface would take, if the mass of the earth would have been four times its original mass?
Answer» Solution :Given : Height of the SATELLITE above the earth's SURFACE = 35780 km Velocity of the satellite = 3.08 km/sec. To find: TIME (T) Solution:`G=6.67xx10^(-11) Nm^2 // kg^2 , M=4xx6xx10^24` kg (for earth) R=6400 km (for earth) `=6.4xx10^6` m, h=height of the satellite above the earth's surface = 35780 km R+h =6400 + 35780=42180 km = 42180x `10^3` m `THEREFORE v=sqrt((G4M)/(R+h))=sqrt(((6.67xx10^(-11))xx(4xx6xx10^24))/(42180xx10^3 m))= sqrt((4xx40.02xx10^13)/(42180xx10^3 m))` `therefore ` v=6160.489 m/s = 6.16 km/s `T=(2pir)/v =(2pi(R+h))/v=(2xx3.14(6400+35780))/6.16`=43001.68 sec =11.94 hrs. = 12 hrs. (approx .) The satellite will take 12 hrs. to revolve in the orbit.