The time period of a satellite of earth is 5 hours. If the separation between the earth and the satellite is increased speartion between the earth and the satellite is increased to 4 times the previous value, the new time period will becomes

Options

A. 10 hours
B. 80 hours
C. 40 hours
D. 20 hours

Answer

C

Explanation

(c ) According to Kelper's law of planetary motion `T^2 prop R^3`
`:. T_2 = T_1((R_2)/(R_1))^(3//2) = 5xx[(4R)/(R )]^(3//2) = 5xx2^3 =40 hour`