The time period of a geostationary satellite is 24 h, at a height `6R_(E)` (`R_(E)` is radius of earth) from surface of earth. The time period of another satellite whose height is `2.5 R_(E)` from surface will be

Options

A. `(12)/(2.5)h`
B. `6sqrt(2)h`
C. `12sqrt(2) h`
D. `(24)/(2.5) h`

Answer

B

Explanation

We know that square of time period is proportional to cube of the radius.
`T^(2) prop r^(3)`
`T^(2) prop (R_(E) + h)^(3)`
`(T_(1)^(2))/(T_(2)^(2)) = ((R_(E) + 6R_(E))^(3))/((R_(E) + 2.5 R_(E))^(3))`
`(T_(1)^(2))/(T_(2)^(2)) = (7^(3))/((7)/(2))^(3)`
`(T_(1)^(2))/(T_(2)^(2)) = 8`
`T_(2) = (T_(1))/(2sqrt(2))`
`T_(2) = (24)/(2sqrt(2))`
`T_(2) = 6sqrt(2) h`
Hence option (b) is correct.