Satellite a orbits the earth at a height 3.00 times the earth's radius. satellite b orbits the earth at a height 2.00 times the earth's radius. what is the ratio of the period of satellite b to the period of satellite a?

a) 1 to √(3/2)^3

b) 1 to 3√(3/2)^2

c) 1 to √(2/3)^3

d) 1 to 3√(2/3)^2

*the "3√" represents a cubic root*
i think the answer is b, but i am not sure

The correct answer to this question is this one: "B) 1 to 3√(3/2)^2 "

The fact that Satellite A orbits the earth at a height 3.00 times the earth's radius and  Satellite B orbits the earth at a height 2.00 times the earth's radius. The ratio of the of the period of Satellite B to the period of Satellite A is 1 : 3√(3/2)^2 

A)1: √(3/2)^3

Explanation:

The formula for the period of the satellite is given by:

T = 2π √(r^3/GM)  

For satellite A,

T1 = 2π √((3r)^3/GM)  ………. (i)

For satellite B,

T2 = 2π √((2r)^3/GM) ……….. (ii)

Dividing equation (ii) by (i), we get

T2/T1  = √(2)^3/√(3)^3

Further Simplification results in

T2:T1 =1: √(3/2)^3

stretching the net tighter