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?

Explanation

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 = 42180 x `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.