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. `6sqrt(2)`
B. `(6)/(sqrt(2))`
C. `5`
D. `10`

Answer

A

Explanation

According to Kepler's third law, `T^(2) prop r^(3)`
`(T_(2))/(T_(1))=((r_(2))/(r_(1)))^(1//2)`
Give `T_(1)=24h`
`r_(1)+R+h_(1)=R+5R=6R`
`r_(2)=R+h_(2)=R+2R=3R`
On solving we get
`T_(2)=6sqrt(2)h`