A geostationary satellite is orbiting the earth at a height of `5R` above the surface of the earth, `2R` being the radius of the earth. The time period of another satellite in hours at a height of `2R` form the surface of the earth is

Options

A. 5
B. 10
C. `6sqrt2`
D. `(6)/(sqrt2)`

Answer

C

Explanation

According to Kepler.s third law `T^2 prop r^(3//2)`
` therefore T_2/T_1 = (r_2/r_1)^(3//2) = ((R + 2R)/(R+ 5R))^(3//2) = (1)/(2^(3//2))`
`(T_2)/(24) = (1)/(2^(3//2)) " or " T_2 = (24)/(2^(3//2) ) = (24)/(2sqrt2) = 6 sqrt2` hours