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. `6 sqrt(2)`
D. `6//sqrt(2)`

Answer

C

Explanation

From Kepler's third law, `T^(2)prop r^(3)`
Hence, `T_(1)^(2)propr_(1)^(3) and T_(2)^(2)propr_(2)^(3)`
So, `(T_(2)^(2))/(T_(1)^(2))=(r_(2)^(3))/(r_(1)^(3))=((3R)^(3))/((6R)^(3))or (T_(2)^(2))/(T_(1)^(2))=(1)/(8)`
`T_(2)^(2)=(1)/(8)T_(1)^(2)rArrT_(2)=(24)/(2sqrt(2))=6sqrt(2)`h