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

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

i wanna say it’s between b and d, leaning more towards b.

answer; how far out of place which the object finished relative to the starting location;