Distance of satellite from the surface of the earth is given to be 3 times its radius

Now, including the radius of the earth, the satellite is at a distance of 4 times the radius of the earth from the centre of the earth

T² is proportional to r³

Under normal condition for existing satellite close to the surface of the earth, T₁ is proportional to r₁^3/2

For the other satellite which is 4 times away from the radius of the earth, T₂ is proportional to r₂^3/2

Let the radius of the earth be r

T₁ = k r₁^3/2

60 = k r^3/2  ----------------------Eqn (1)

Hence, substituting the condition that the satellite is at a distance of 4 times the radius of the earth, we get

T₂ = k (4r)^3/2  ----------------------Eqn (2)

Eqn (1) / Eqn (2) is 60/T₂ = 1/ (4)^3/2 which is 1/ 4√4 =1/ 4 * 2 = 1/8

Hence, new time period is T₂ = 60 * 8 = 480 minutes