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

The time period of a geostationary satellite is 24 h, at a height 6R_(

1.	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
Answer» `(12)/(2.5)h` `6sqrt(2)h` `12sqrt(2) h` `(24)/(2.5) h` Solution :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.